Maharashtra Board Class 11 Chemistry Chapter 13 Nuclear and Radio Solutions

Nuclear Chemistry And Radioactivity Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 13 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 13 Exercise Solutions

Exercise 13(A)

1. Choose Correct Option.

Question A.Identify nuclear fusion reaction
α. \(_{1}^{1}\text{H} + _{1}^{1}\text{H} \longrightarrow _{1}^{2}\text{H} + _{+1}^{0}\text{e}\)
b. \(_{1}^{2}\text{H} + _{1}^{1}\text{H} \longrightarrow _{2}^{3}\text{He}\)
c. \(_{1}^{3}\text{H} + _{1}^{1}\text{H} \longrightarrow _{1}^{3}\text{H} + _{1}^{1}\text{p}\)
Answer: Among the given options, reactions (i) and (ii) represent nuclear fusion reaction wherein lighter nuclei combine to form a heavy nucleus.
In simple words: Nuclear fusion is a process where two lighter atomic nuclei combine to form a single heavier nucleus, releasing a large amount of energy. Options (a) and (b) show such combinations.

🎯 Exam Tip: Nuclear fusion involves lighter nuclei combining, while nuclear fission involves a heavy nucleus splitting. Remember the key characteristics of each to differentiate them in exams.

Question B.The missing particle from the nuclear reaction is
\(_{13}^{27}\text{Al} + _{2}^{4}\text{He} \longrightarrow \text{?} + _{0}^{1}\text{n}\)
α. \(_{15}^{30}\text{P}\)
b. \(_{16}^{32}\text{S}\)
c. \(_{10}^{14}\text{Ne}\)
d. \(_{14}^{30}\text{Si}\)
Answer: (A) \(_{15}^{30}\text{P}\)
In simple words: To find the missing particle, we balance the atomic numbers (bottom numbers) and mass numbers (top numbers) on both sides of the nuclear reaction. Aluminum-27 plus Helium-4 gives a total of 31 mass and 15 atomic number. Subtracting one neutron (1 mass, 0 atomic number) leaves 30 mass and 15 atomic number, which corresponds to Phosphorus-30.

🎯 Exam Tip: Always balance both the mass number (sum of superscripts) and the atomic number (sum of subscripts) in nuclear reactions to identify missing particles. Conservation laws are crucial here.

Question C.\(_{27}^{60}\text{Co}\) decays with half-life of 5.27 years to produce \(_{28}^{60}\text{Ni}\). What is the decay constant for such radioactive disintegration?
a. 0.132 y\(^{-1}\)
b. 0.138
c. 29.6 y
d. 13.8%
Answer: (a) 0.132 y\(^{-1}\)
In simple words: The decay constant (\(\lambda\)) is a measure of how quickly a radioactive isotope decays. It is related to the half-life (t\(_{1/2}\)) by the formula \(\lambda = 0.693 / \text{t}_{1/2}\). Plugging in the given half-life of 5.27 years, we get approximately 0.132 y\(^{-1}\).

🎯 Exam Tip: Remember the fundamental relationship between decay constant and half-life: \(\lambda = 0.693 / \text{t}_{1/2}\). This formula is frequently used in radioactivity problems.

Question D.The radioactive isotope used in the treatment of Leukemia is
a. \(^{60}\text{Co}\)
b. \(^{226}\text{Ra}\)
c. \(^{32}\text{P}\)
d. \(^{131}\text{I}\)
Answer: (c) \(^{32}\text{P}\)
In simple words: Phosphorus-32 is a radioactive isotope that is specifically used in the medical treatment of Leukemia, a type of blood cancer, because of its unique decay properties.

🎯 Exam Tip: Be familiar with common radioisotopes and their specific applications, especially in medicine and industry, as these are often tested.

Question E.The process by which nuclei having low masses are united to form nuclei with large masses is
a. chemical reaction
b. nuclear fission
c. nuclear fusion
d. chain reaction
Answer: (c) nuclear fusion
In simple words: Nuclear fusion is the process where two light atomic nuclei combine to form a heavier nucleus, releasing a large amount of energy. This is the opposite of nuclear fission, where a heavy nucleus splits.

🎯 Exam Tip: Clearly distinguish between nuclear fusion (combining light nuclei) and nuclear fission (splitting heavy nuclei). Both release energy but through opposite mechanisms.

2. Explain

Question A.On the basis of even-odd of protons and neutrons, what type of nuclides are most stable?
Answer:
• Nuclides with even number of protons (Z) and even number of neutrons (N) are most stable.
• These nuclides tend to form proton-proton and neutron-neutron pairs.
• This impart stability to the nucleus.
In simple words: Atomic nuclei with an even number of both protons and neutrons are generally the most stable because these even numbers allow nucleons to pair up, leading to a more energetically favorable and stable arrangement.

🎯 Exam Tip: The even-odd rule for nuclear stability is a key concept. Remember that even-even nuclei are the most stable, followed by even-odd, odd-even, and finally odd-odd nuclei being the least stable.

Question B.Explain in brief, nuclear fission.
Answer:
i. Nuclear fission: It is a process which involves splitting of the heavy nucleus of an atom into two nearly equal fragments accompanied by release of the large amount of energy.
e.g. Nuclear fission of \(^{235}\text{U}\)
ii. When a uranium nucleus absorbs neutron, it breaks into two lighter fragments and releases energy (heat), more neutrons, and other radiation. This can be given as,
\(_{92}^{235}\text{U} + _{0}^{1}\text{n} \longrightarrow _{56}^{142}\text{Ba} + _{36}^{91}\text{Kr} + 3_{0}^{1}\text{n} + \text{Energy}\)
iii. Characteristics of nuclear fission reactions:
• The mass of the fission products is less than the parent nucleus. A large amount of energy corresponding to the mass loss is released in each fission.
• When one uranium-235 nucleus undergoes fission, three neutrons are emitted, which subsequently disintegrate three more uranium nuclei and thereby produce nine neutrons. Such a chain continues by itself.
• In a very short time enormous amount of energy is liberated, which can be utilized for destructive or peaceful purposes.
• Energy released per fission is approximately 200 MeV.
Note:
• Each fission may lead to different products.
• There is no unique way for fission of \(^{235}\text{U}\) that produces Ba and Kr. There are 400 ways for fission of \(^{235}\text{U}\) leading to 800 fission products.
• Many of these fission products are radioactive which undergo spontaneous disintegrations giving rise to new elements in the periodic table.
In simple words: Nuclear fission is the process where a heavy atomic nucleus, like Uranium-235, splits into two smaller nuclei when hit by a neutron, releasing a tremendous amount of energy, more neutrons, and other radiation. This process can lead to a self-sustaining chain reaction.

🎯 Exam Tip: When explaining nuclear fission, include the definition, a balanced example reaction (like Uranium-235), and key characteristics such as energy release, neutron emission, and the possibility of a chain reaction.

Question C.The nuclides with odd number of both protons and neutrons are the least stable. Why?
Answer:
• The nuclides with odd number of both protons and neutrons are the least stable because, odd number of protons and neutrons results in the presence of two unpaired nucleons.
• These unpaired nucleons result in instability. Hence, such nuclides are the least stable.
In simple words: Nuclei with an odd number of both protons and neutrons are the least stable because the presence of these unpaired nucleons leads to higher energy states and greater instability, making them more prone to radioactive decay.

🎯 Exam Tip: Understanding nucleon pairing is crucial for stability. Odd numbers of both protons and neutrons mean more unpaired nucleons, which increases nuclear instability. Contrast this with the high stability of even-even nuclei.

Question D.Referring the stability belt of stable nuclides, which nuclides are \(\beta^{-}\) and \(\beta^{+}\) emitters? Why?
Answer:
• Beta decay occurs when an unstable nucleus emits a beta particle and energy. A beta particle is either an electron or a positron. An electron is a negatively charged particle, and a positron is a positively charged electron (or anti-electron).
• When the beta particle is an electron, the decay is called beta-minus (\(\beta^{-}\)) decay. In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus.
• When the beta particle is a positron, the decay is called beta-plus (\(\beta^{+}\)) decay. In beta-plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.
• Thus, beta-minus decay occurs when a nucleus has too many neutrons relative to protons (i.e., N/Z > 1) and beta-plus decay occurs when a nucleus has too few neutrons relative to protons (i.e., N/Z < 1).
• By referring the stability belt of stable nuclides, nuclides with N/Z > 1 are to the left of the stability zone. Such nuclides are beta-minus emitters as they become stable when a neutron converts to a proton.
• Nuclides with N/Z < 1 are to the right of the stability zone. Such nuclides are beta-plus emitters as they become stable when a proton converts to a neutron.
In simple words: Nuclides to the left of the stability belt (high N/Z ratio) undergo \(\beta^{-}\) decay to convert a neutron to a proton, while nuclides to the right of the belt (low N/Z ratio) undergo \(\beta^{+}\) decay to convert a proton to a neutron, both aiming for a more stable N/Z ratio.

🎯 Exam Tip: Correlate the N/Z ratio with the type of beta decay. High N/Z (neutron-rich) leads to \(\beta^{-}\) emission, while low N/Z (proton-rich) leads to \(\beta^{+}\) emission, both shifting the nucleus towards the band of stability.

Question E.Explain with an example each nuclear transmutation and artificial radioactivity. What is the difference between them?
Answer:
i. Nuclear transmutation: It involves transformation of a stable nucleus into another nucleus takes place which can be either stable or unstable.
ii. Artificial (induced) radioactivity: It is nuclear transmutation where the product nucleus is radioactive. The product nucleus decays spontaneously with emission of radiation and particles.
e.g.
Step I: \(_{5}^{10}\text{B} + _{2}^{4}\text{He} \longrightarrow _{7}^{13}\text{N} + _{0}^{1}\text{n}\)
Stable Radioactive
Step II: \(_{7}^{13}\text{N} \longrightarrow _{6}^{13}\text{C} + _{+1}^{0}\text{e}\)
Radioactive (spontaneous emission of positron)
Step-I can be considered as nuclear transmutation as it produces a new nuclide \(_{7}^{13}\text{N}\).
However, the new nuclide is unstable (radioactive). Hence, step-I involves artificial (induced) radioactivity. Thus, in artificial transmutation, a stable element is collided with high speed particles to form another radioactive element.
In simple words: Nuclear transmutation is the change of one element into another via nuclear reactions, which can produce either stable or unstable products. Artificial radioactivity is a specific type of nuclear transmutation where the product formed is itself radioactive, meaning a stable nucleus is transformed into an unstable, decaying nucleus.

🎯 Exam Tip: The key difference is that nuclear transmutation is the general term for changing one element to another, while artificial radioactivity specifically refers to transmutation that results in a *radioactive* product. Use a clear example to illustrate both concepts.

Question F.What is binding energy per nucleon? Explain with the help of diagram how binding energy per nucleon affects nuclear stability?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख द्रव्यमान संख्या (A) के फलन के रूप में प्रति न्यूक्लियॉन बंधन ऊर्जा (MeV में) को दर्शाता है। यह दिखाता है कि प्रति न्यूक्लियॉन बंधन ऊर्जा कैसे बढ़ती है, 56Fe के आसपास अधिकतम तक पहुँचती है, और फिर बहुत भारी नाभिकों के लिए घट जाती है, जो परमाणु स्थिरता को दर्शाता है। फ्यूजन और विखंडन प्रक्रियाओं के ऊर्जा उत्पादन को भी रेखांकन में दर्शाया गया है।
i. Binding energy per nucleon (B), for nucleus containing (A) nucleons with binding energy (B.E.) is given as,
\(B = \text{B.E./A}\)
ii. Mean binding energy per nucleon (B) for the most stable isotopes as a function of mass number is shown above. This plot leads to the following inferences:
a. Light nuclides: (A < 30)
The peaks with A values in multiples of 4. For example, \(_{2}^{4}\text{He}\), \(_{6}^{12}\text{C}\), \(_{8}^{16}\text{O}\) are more stable.
b. Medium mass nuclides: (30 < A < 90)
B increases typically from 8 MeV for A = 16 to nearly 8.3 MeV for A between 28 and 32 and it remains nearly constant 8.5 MeV beyond this and shows a broad maximum. The nuclides falling on the maximum are most stable which turns possess high values. \(^{56}\text{Fe}\) with B value of 8.79 MeV is the most stable.
c. Heavy nuclides (A > 90)
B decreases from maximum 8.79 MeV to 7.7 MeV for A \(\approx\) 210, \(^{209}\text{Bi}\) is the stable nuclide. Beyond this, all nuclides are radioactive (\(\alpha\)-emitters).
In simple words: Binding energy per nucleon is the average energy required to remove a single nucleon from a nucleus, indicating its stability. A higher binding energy per nucleon generally means a more stable nucleus, with medium-mass nuclei (like iron) being the most stable, as seen in the binding energy curve.

🎯 Exam Tip: When describing binding energy per nucleon, define it clearly and explain how its value on the stability curve (peaking around iron-56) directly correlates with nuclear stability, explaining why both fusion and fission can release energy.

Question G.Explain with example \(\alpha\)-decay.
Answer:
i. The emission of \(\alpha\)-particle from the nuclei of an radioelement is called \(\alpha\)-decay.
ii. The charge on an \(\alpha\)-particle is +2 with a mass of 4 u.
It is identical with helium nucleus and hence an \(\alpha\)-particle is designated as \(_{2}^{4}\text{He}\).
iii. In the \(\alpha\)-decay process, the parent nucleus \(_{Z}^{A}\text{X}\) emits an \(\alpha\)-particle and produces daughter nucleus \(_{Z-2}^{A-4}\text{Y}\). The parent nucleus thus loses two protons (charge +2) and two neutrons. The total mass lost is 4 u. The daughter nucleus will therefore, have mass 4 units less and charge 2 units less than its parent.
iv. General equation for \(\alpha\)-decay process can be given as:
\(_{Z}^{A}\text{X} \longrightarrow _{Z-2}^{A-4}\text{Y} + _{2}^{4}\text{He}\)
Parent Daughter Emitted particle
nuclei
e.g. Radium 226 decays to form radium 222:
\(_{88}^{226}\text{Ra} \longrightarrow _{86}^{222}\text{Rn} + _{2}^{4}\text{He}\)
Parent Daughter Emitted particle
nuclei
In \(\alpha\)-decay process of radium, radon (daughter nuclei) is formed with loses of two protons (charge +2) and two neutrons. The total mass lost is 4 u.
Thus, radon has a mass of 4 units less and charge 2 units less than its parent radium.
In simple words: Alpha decay is a type of radioactive decay where an unstable heavy nucleus emits an alpha particle (which is a helium nucleus, \(_{2}^{4}\text{He}\)), transforming into a new nucleus with its mass number reduced by 4 and its atomic number reduced by 2.

🎯 Exam Tip: For \(\alpha\)-decay, remember that the emitted particle is always a helium nucleus (\(_{2}^{4}\text{He}\)). This means the parent nucleus loses 4 units from its mass number and 2 units from its atomic number to form the daughter nucleus.

Question H.Energy produced in nuclear fusion is much larger than that produced in nuclear fission. Why is it difficult to use fusion to produce energy?
Answer:
• Nuclear fusion involves the fusion of lighter nuclei to form a heavy nucleus which is accompanied by an enormous amount of energy (heat).
• Fusion reaction requires extremely high temperature typically of the order of \(10^{8}\) K.
In simple words: Nuclear fusion releases significantly more energy than fission because the binding energy per nucleon increases sharply for light nuclei, leading to a larger mass defect when they combine. However, it's difficult to harness because fusion requires incredibly high temperatures (millions of Kelvin) to overcome the electrostatic repulsion between positively charged nuclei and force them to merge.

🎯 Exam Tip: Highlight the immense energy release of fusion due to a large mass defect for light nuclei. The primary difficulty in harnessing fusion is achieving and maintaining the extreme temperatures and pressures needed to overcome electrostatic repulsion, making it a challenging engineering feat.

Question I.How does N/Z ratio affect the nuclear stability? Explain with a suitable diagram.
Answer:
• When the graph of number of neutrons (N) against protons (Z) is drawn, and all the stable isotopes are plotted on it, there is quite a clear correlation between N and Z. This graph is shown in the adjacent figure.
• A large number of elements have several stable isotopes and hence, the curve appears as a belt or zone called stability zone. All stable nuclides fall with this zone and the nuclei that are to the left or to the right of the stability zone are unstable and exhibit radioactivity. Below the belt, a straight line which represents the ratio N/Z to be nearly unity (i.e., N = Z) is shown.
• For nuclei lighter than \(_{20}^{40}\text{Ca}\), the straight line (N = Z) passes through the belt. The lighter nuclides are therefore stable (N/Z being 1).
• The N/Z ratio for the stable nuclides heavier than calcium gives a curved appearance to the belt with gradual increase of N/Z (> 1). The heavier nuclides therefore, need more number of neutrons than protons to attain stability. The heavier nuclides with increasing number of protons render large coulombic repulsions. With increased number of neutrons, the protons within the nuclei get more separated, which renders them stable.
• Thus, nuclear stability is linked to the number of nucleons (neutrons and protons). In general, the lighter stable nuclei have equal numbers of protons and neutrons while heavier stable nuclei have increasingly more neutrons than protons.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख न्यूट्रॉन की संख्या (N) बनाम प्रोटॉन की संख्या (Z) का एक ग्राफ है, जिसे "स्थिरता का बैंड" कहा जाता है। नीले बिंदु स्थिर नाभिकों को दर्शाते हैं, और यह दिखाते हैं कि हल्के तत्वों के लिए N/Z अनुपात 1 के करीब होता है, जबकि भारी तत्वों के लिए यह स्थिरता बनाए रखने के लिए 1 से अधिक हो जाता है। आरेख स्वाभाविक रूप से होने वाले और कृत्रिम रूप से उत्पादित रेडियोआइसोटोपों को भी प्रदर्शित करता है।
[Note: Atoms with unstable nuclei are radioactive (exhibit radioactivity). To become more stable, the nuclei undergo radioactive decay.]
In simple words: The N/Z ratio is critical for nuclear stability; stable light nuclei typically have an N/Z ratio near 1, while stable heavy nuclei require an N/Z ratio greater than 1 to counteract proton-proton repulsion. Deviations from this "band of stability" indicate an unstable nucleus that will undergo radioactive decay.

🎯 Exam Tip: Focus on explaining the "band of stability" concept and how the ideal N/Z ratio changes with increasing atomic number. Emphasize that nuclei outside this band are unstable and decay to achieve a more favorable N/Z ratio.

Question J.You are given a very old sample of wood. How will you determine its age?
Answer:The age of the wood sample can be determined by radiocarbon dating as \(^{14}\text{C}\) becomes a part of a plant due to the photosynthesis reaction (i.e., absorption of [\(^{14}\text{CO}_{2}\) + \(^{12}\text{CO}_{2}\)]).
i. The activity (N) of given wood sample and that of fresh sample of live plant (\(\text{N}_{0}\)) is measured, where, \(\text{N}_{0}\) denotes the activity of the given sample at the time of death.
ii. The age of the given wood sample. can be determined by applying following Formulae:
\(t = \frac{2.303}{\lambda} \text{log}_{10} \frac{\text{N}_{0}}{\text{N}}\)
where \(\lambda = \frac{0.693}{5730 \text{ y}} = 1.21 \times 10^{-4} \text{ y}^{-1}\).
Note: The oldest rock found so far in Northern Canada is 3.96 billion years old.
In simple words: The age of an old wood sample can be determined using radiocarbon dating, which involves measuring the remaining activity of carbon-14 (N) in the sample compared to its initial activity (\(\text{N}_{0}\)) when it was alive. Using the carbon-14 half-life and the decay formula, the time since the organism died can be calculated.

🎯 Exam Tip: When explaining radiocarbon dating, mention the role of carbon-14, its incorporation into living organisms, and how its decay after death allows for age determination using the decay constant and the ratio of current to initial activity.

3. Answer The Following Question

Question A.Give example of mirror nuclei.
Answer:Example of mirror nuclei: \(_{1}^{3}\text{H}\) and \(_{2}^{3}\text{He}\)
In simple words: Mirror nuclei are pairs of isobars (nuclei with the same mass number) where the number of protons in one is equal to the number of neutrons in the other, and vice versa. \(_{1}^{3}\text{H}\) (1 proton, 2 neutrons) and \(_{2}^{3}\text{He}\) (2 protons, 1 neutron) are an example.

🎯 Exam Tip: Remember the definition of mirror nuclei: same mass number (A), but the number of protons in one equals the number of neutrons in the other. A classic example is tritium and helium-3.

Question B.Balance the nuclear reaction:
\(_{54}^{118}\text{Xe} \longrightarrow \text{?} + _{54}^{118}\text{I}\)
Answer:
\(_{54}^{118}\text{Xe} \longrightarrow _{-1}^{0}\text{e} + _{53}^{118}\text{I}\)
In simple words: To balance the reaction, we need to ensure that the mass numbers and atomic numbers are conserved. Xenon-118 transforms into Iodine-118, meaning the mass number remains the same. The atomic number changes from 54 to 53, so a particle with an atomic number of -1 and a mass number of 0, which is a beta particle (electron), must be emitted.

🎯 Exam Tip: For balancing nuclear reactions, sum the mass numbers (superscripts) and atomic numbers (subscripts) on both sides of the equation. Any missing values reveal the identity of the unknown particle.

Question C.Name the most stable nuclide known. Write two factors responsible for its stability.
Answer:The most stable nuclide known is lead (\(^{208}\text{Pb}\)).
Two factors responsible for its stability are as follows:
• It is a nuclide with even number of both protons (Z) and neutrons (N).
• It has two magic numbers i.e., 82 (for protons) and 126 (for neutrons).
In simple words: The most stable nuclide is Lead-208 because it has an even number of both protons and neutrons, which contributes to stability through nucleon pairing, and crucially, both its proton number (82) and neutron number (126) are "magic numbers," indicating a completely filled nuclear shell structure, similar to how noble gases have stable electron shells.

🎯 Exam Tip: When asked about nuclear stability, remember the "magic numbers" (2, 8, 20, 28, 50, 82, 126 for protons and neutrons) as they indicate exceptionally stable nuclei due to filled nuclear shells. The even-even nucleon count is also a primary factor.

Question D.Write relation between decay constant of a radioelement and its half life.
Answer:Relation between decay constant of a radioelement and its half-life is given as,
\(\lambda = \frac{0.693}{\text{t}_{1/2}}\)
Where, \(\lambda\) = Decay constant, \(\text{t}_{1/2}\) = Half-life of a radioelement
In simple words: The relationship between the decay constant (\(\lambda\)) and the half-life (\(\text{t}_{1/2}\)) of a radioelement is inversely proportional, expressed as \(\lambda = 0.693 / \text{t}_{1/2}\), meaning a shorter half-life corresponds to a larger decay constant.

🎯 Exam Tip: This formula is fundamental in radioactive decay calculations. Make sure to recall that 0.693 is approximately the natural logarithm of 2 (\(\ln 2\)).

Question E.What is the difference between an \(\alpha\)-particle and helium atom?
Answer:
• Helium atom is composed of 2 protons and 2 neutrons (or 1 neutron) along with 2 electrons in the outer shell.
• On the other hand, \(\alpha\)-particle constitutes 2 protons and 2 neutrons bound together to form a particle which is similar to helium (except presence of electrons).
• Helium is one of the inert gas which is stable (duplet complete) whereas \(\alpha\)-particle is unstable and highly reactive.
In simple words: An alpha particle is essentially a helium nucleus, consisting of two protons and two neutrons, and carries a +2 charge without any electrons. A neutral helium atom, however, has the same nucleus but also contains two electrons orbiting it, making it electrically neutral and chemically stable as an inert gas.

🎯 Exam Tip: The crucial difference lies in the presence of electrons. An \(\alpha\)-particle is an ionized helium *nucleus* (He\(^{2+}\)), while a helium atom is a neutral atom with electrons. This affects charge, chemical reactivity, and stability.

Question F.Write one point that differentiates nuclear reactions from chemical reactions.
Answer:Chemical reactions:
• Rearrangement of atoms by breaking and forming of chemical bonds.
• Different isotopes of an element have same behaviour.
Nuclear reactions:
• Elements or isotopes of one element are converted into another element in a nuclear reaction.
• Isotopes of an element behave differently.
In simple words: Chemical reactions involve the rearrangement of atoms through the breaking and forming of chemical bonds, where elements retain their identity. Nuclear reactions, conversely, involve changes within the nucleus itself, leading to the transformation of one element into another, and isotopes can behave differently due to their nuclear properties.

🎯 Exam Tip: The fundamental distinction is that chemical reactions involve *electrons* and atom rearrangement, while nuclear reactions involve *nuclei* and element transformation. Changes in mass and energy are also significantly different.

Question G.Write pairs of isotones and one pair of mirror nuclei from the following:
\(_{5}^{10}\text{B}\), \(_{6}^{12}\text{C}\), \(_{13}^{27}\text{Al}\), \(_{6}^{11}\text{C}\), \(_{14}^{28}\text{S}\)
Answer:Isotones: i. \(_{5}^{10}\text{B}\) and \(_{6}^{11}\text{C}\)
ii. \(_{13}^{27}\text{Al}\) and \(_{14}^{28}\text{S}\)
Mirror nuclei: Since there are no isobars the given set of nuclides does not contain a pair of mirror nuclei.
In simple words: Isotones are atoms of different elements that have the same number of neutrons. From the given list, \(_{5}^{10}\text{B}\) and \(_{6}^{11}\text{C}\) (both have 5 neutrons) are isotones, and \(_{13}^{27}\text{Al}\) and \(_{14}^{28}\text{S}\) (both have 14 neutrons) are also isotones. Mirror nuclei are pairs of isobars where the proton and neutron counts are swapped, but no such pair exists in the provided list.

🎯 Exam Tip: To identify isotones, calculate the number of neutrons (N = A - Z) for each nuclide and look for matches. For mirror nuclei, first find isobars (same A), then check if their proton and neutron numbers are interchanged.

Question H.Derive the relationship between half life and decay constant of a radioelement.
Answer:Equation for the decay constant is given as,
\(\lambda = \frac{2.303}{t} \text{log}_{10} \frac{\text{N}_{0}}{\text{N}}\) ...(i)
Where, \(\lambda\) = Decay constant
N = Number of nuclei (atoms) present at time t
At t = 0, N = \(\text{N}_{0}\).
Hence, at t = \(\text{t}_{1/2}\), N = \(\text{N}_{0}/2\)
Substitution of these values of N and t in equation (i) gives,
\(\lambda = \frac{2.303}{\text{t}_{1/2}} \text{log}_{10} \frac{\text{N}_{0}}{\text{N}_{0}/2}\)
\( = \frac{2.303}{\text{t}_{1/2}} \text{log}_{10}2\)
\( = \frac{2.303}{\text{t}_{1/2}} \times 0.3010 = \frac{0.693}{\text{t}_{1/2}}\)
Hence, \(\lambda = \frac{0.693}{\text{t}_{1/2}}\)
or \(\text{t}_{1/2} = \frac{0.693}{\lambda}\)
In simple words: Starting from the integrated rate law for radioactive decay, we define half-life (\(\text{t}_{1/2}\)) as the time it takes for half of the initial radioactive nuclei (\(\text{N}_{0}\)) to decay, leaving N = \(\text{N}_{0}/2\). Substituting these values into the rate law and using the logarithm of 2 (\(\ln 2 \approx 0.693\)), we derive the relationship \(\text{t}_{1/2} = 0.693 / \lambda\), showing that half-life is inversely proportional to the decay constant.

🎯 Exam Tip: The derivation of \(\text{t}_{1/2} = 0.693 / \lambda\) is a common exam question. Ensure you understand the steps involved, especially the substitution of \(\text{N} = \text{N}_{0}/2\) at \(\text{t} = \text{t}_{1/2}\) and the use of \(\ln 2\).

Question I.Represent graphically \(\text{log}_{10}\) (activity /dps) versus t/s. What is its slope?
Answer:Equation for a decay constant (\(\lambda\)) is given as,
\(\lambda = \frac{2.303}{t} \text{log}_{10} \frac{\text{N}_{0}}{\text{N}}\) ...(i)
From equation (i),
\(\text{log}_{10}\text{N} = -\frac{\lambda}{2.303}t + \text{log}_{10}\text{N}_{0}\)
On comparing with y = mx + c, y = \(\text{log}_{10}\text{N}\), \(m = -\frac{\lambda}{2.303}\), x = t, c = \(\text{log}_{10}\text{N}_{0}\)
Now, \(N \propto (-\frac{\text{dN}}{\text{dt}})\)
Hence, instead if \(\text{log}_{10}\text{N}\) versus t, \(\text{log}_{10}(-\frac{\text{dN}}{\text{dt}})\) which is \(\text{log}_{10}\) (activity) is plotted.
The graph of \(\text{log}_{10}\) (activity/dps) versus t/s gives a straight line which can be represented as follows:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक सीधी रेखा ग्राफ को दर्शाता है जहाँ Y-अक्ष पर \(\text{log}_{10}\) (गतिविधि/dps) है और X-अक्ष पर समय (t/s) है। रेखा का ढलान ऋणात्मक होता है, जिसका मान \(-\frac{\lambda}{2.303}\) के बराबर होता है, जो रेडियोधर्मी क्षय की घातीय प्रकृति को दर्शाता है।
Thus, slope will be \(-\frac{\lambda}{2.303}\)
In simple words: When the logarithm (base 10) of radioactivity (activity/dps) is plotted against time (t/s), a straight line is obtained. The slope of this line is equal to \(-\frac{\lambda}{2.303}\), where \(\lambda\) is the decay constant, indicating an exponential decrease in activity over time.

🎯 Exam Tip: Remember that a plot of \(\text{log}_{10}\) (activity) vs. time for radioactive decay yields a straight line with a negative slope. This slope is directly proportional to the decay constant (\(\lambda\)), allowing for its determination from experimental data.

Question J.Write two units of radioactivity. How are they interrelated?
Answer:The unit of radioactivity is curie (Ci).
1 Ci = \(3.7 \times 10^{10}\) dps
ii. Other unit of radioactivity is Becquerel (Bq).
1 Bq = 1 dps
Thus, 1 Ci = \(3.7 \times 10^{10}\) dps = \(3.7 \times 10^{10}\) Bq
In simple words: Two primary units of radioactivity are the Curie (Ci) and the Becquerel (Bq). One Curie is defined as \(3.7 \times 10^{10}\) disintegrations per second (dps), while one Becquerel is simply 1 disintegration per second. Therefore, 1 Curie is equivalent to \(3.7 \times 10^{10}\) Becquerel.

🎯 Exam Tip: Know the definitions and interrelation of Curie and Becquerel. Becquerel is the SI unit and is directly equal to 1 dps, making it easier to conceptualize, while Curie represents a much larger activity.

Question K.Half life of \(^{24}\text{Na}\) is 900 minutes. What is its decay constant?
Answer:Given: \(\text{t}_{1/2}\) = 900 minutes
To find: \(\lambda\)
Formula: \(\lambda = \frac{0.693}{\text{t}_{1/2}}\)
Calculation: \(\lambda = \frac{0.693}{\text{t}_{1/2}} = \frac{0.693}{900 \text{ min}}\)
\(\lambda = 7.7 \times 10^{-4} \text{ min}^{-1}\)
Ans: Decay constant for \(^{24}\text{Na}\) is \(7.7 \times 10^{-4} \text{ min}^{-1}\)
In simple words: To find the decay constant (\(\lambda\)) from the half-life (\(\text{t}_{1/2}\)), we use the formula \(\lambda = 0.693 / \text{t}_{1/2}\). Given a half-life of 900 minutes, the decay constant for Sodium-24 is calculated to be \(7.7 \times 10^{-4}\) per minute.

🎯 Exam Tip: Always ensure unit consistency. If half-life is in minutes, the decay constant will be in min\(^{-1}\). Be careful with calculator precision for the constant 0.693.

Question L.Decay constant of \(^{197}\text{Hg}\) is 0.017 h\(^{-1}\). What is its half life?
Answer:Given: \(\lambda\) = 0.017 h\(^{-1}\)
To find: \(\text{t}_{1/2}\)
Formula: \(\text{t}_{1/2} = \frac{0.693}{\lambda}\)
Calculation: \(\text{t}_{1/2} = \frac{0.693}{\lambda} = \frac{0.693}{0.017 \text{ h}^{-1}}\)
\( = 40.77 \text{ h}\) (by using log table)
Ans: Thus, the half-life of \(^{197}\text{Hg}\) is 40.77 h.
Calculation using log table:
\(\frac{0.693}{0.017}\)
\( = \text{Antilog}_{10}[\text{log}_{10}0.693-\text{log}_{10}0.017]\)
\( = \text{Antilog}_{10}[\overline{1}.8407 - \overline{2}.2304]\)
\( = \text{Antilog}_{10}[1.6103]\)
\( = 40.77\)
In simple words: To find the half-life (\(\text{t}_{1/2}\)) from the decay constant (\(\lambda\)), we rearrange the formula to \(\text{t}_{1/2} = 0.693 / \lambda\). Given a decay constant of 0.017 h\(^{-1}\), the half-life for Mercury-197 is calculated to be 40.77 hours.

🎯 Exam Tip: This is a direct application of the half-life formula. Pay attention to unit conversions if the given decay constant or half-life is in different units (e.g., seconds vs. hours). Log table calculations are rarely required in modern exams, but direct calculation is expected.

Question M.The total binding energy of \(^{58}\text{Ni}\) is 508 MeV. What is its binding energy per nucleon?
Answer:Given: B.E. of \(^{58}\text{Ni}\) = 508 MeV,
A = 58
To find: Binding energy per nucleon \(\overline{B}\)
Formula: \(\overline{B} = \frac{\text{B.E.}}{\text{A}}\)
Calculation: \(\overline{B} = \frac{508}{58}\)
\( = 8.75862069\)
\(\approx 8.76 \text{ MeV/nucleon}\)
Ans: Binding energy per nucleon of \(^{58}\text{Ni}\) is 8.76 MeV/nucleon.
Calculation using log table:
\(\frac{508}{58}\)
\( = \text{Antilog}_{10}[\text{log}_{10}508 - \text{log}_{10}58]\)
\( = \text{Antilog}_{10}[2.7059-1.7634]\)
\( = \text{Antilog}_{10}[0.9425] = 8.760\)
In simple words: Binding energy per nucleon is calculated by dividing the total binding energy of a nucleus by its mass number (total number of nucleons). For Nickel-58, with a total binding energy of 508 MeV and 58 nucleons, the binding energy per nucleon is 508 MeV / 58 = 8.76 MeV/nucleon.

🎯 Exam Tip: This is a straightforward calculation. Remember that binding energy per nucleon is a measure of nuclear stability, and its unit is typically MeV/nucleon. Always double-check your division.

Question N.Atomic mass of \(_{16}^{32}\text{S}\) is 31.97 u. If masses of neutron and H atom are 1.0087 u and 1.0078 u respectively. What is the mass defect?
Answer:Given: m = 31.97 u, Z = 16, A = 32
\(\text{m}_{\text{n}}\) = 1.0087 u
\(\text{m}_{\text{H}}\) = 1.0078 u
To find: \(\Delta\text{m}\)
Formula: \(\Delta\text{m} = \text{Z}\text{m}_{\text{H}} + (\text{A} - \text{Z})\text{m}_{\text{n}} - \text{m}\)
Calculation: \(\Delta\text{m} = \text{Z}\text{m}_{\text{H}} + (\text{A} - \text{Z})\text{m}_{\text{n}} - \text{m}\)
\( = 16 \times 1.0078 + (16 \times 1.0087) - 31.97\)
\( = [16.1248 + 16.1392] - 31.97\)
\( = 0.294 \text{ u}\)
Ans: The mass defect is 0.294 u.
In simple words: Mass defect is the difference between the actual measured atomic mass of a nucleus and the sum of the masses of its individual constituent protons and neutrons (using the mass of a hydrogen atom for protons). For Sulfur-32, the calculated mass defect is 0.294 atomic mass units.

🎯 Exam Tip: When calculating mass defect, remember to use the mass of a hydrogen atom for protons and account for all protons (Z) and neutrons (A-Z) in the nucleus. The mass defect is always positive, representing the mass converted into binding energy.

Question Q. A produces B by α- emission. If B is in the group 16 of periodic table, what is the group of A ?
Answer: \(_{Z}^{A}\text{A} \xrightarrow{\text{-Emission}} _{Z-2}^{A-4}\text{B} + _{2}^{4}\text{He}\)
When α-emission occurs, atomic number decreases by 2 and atomic mass number by 4.
Thus, if 'B' belongs to group 16 of periodic table, that means outermost orbit will contain 6 electrons.
Thus, 'A' will have 8 electrons in its valence shell and it will belong to group 18 of the periodic table.
In simple words: When an element undergoes alpha decay, its atomic number decreases by 2. If the product element (B) is in group 16, it means the original element (A) had 2 more protons, placing it in group 18.

🎯 Exam Tip: Understanding the change in atomic number and mass number during alpha decay is crucial for determining parent and daughter nuclides and their positions in the periodic table.

Question R. Find the number of α and β- particles emitted in the process \(_{86}^{222}\text{Rn} \longrightarrow _{84}^{214}\text{PO}\)
Answer: The emission of one α-particle decreases the mass number by 4 whereas the emission of β-particles has no effect on mass number.
Net decrease in mass number = 222 - 214 = 8. This decrease is only due to α-particle. Hence, number of α-particle emitted = 8/4 = 2
Now, the emission of one α-particle decreases the atomic number by 2 and one β-particle emission increases it by 1.
The net decrease in atomic number = 86 - 84 = 2
The emission of 2 α-particles causes decrease in atomic number by 4. However, the actual decrease is only 2. It means atomic number increases by 2. This increase is due to emission of 2 β-particles.
Thus, 2 α and 2 β-particles are emitted.
[Note: The above question is modified to include the final decay product so as to determine the number of α-particles and β-particles emitted in the process. Here, the final decay product is assumed to be Po-214.]
In simple words: To find the number of alpha and beta particles, compare the changes in mass and atomic numbers. Each alpha particle reduces mass by 4 and atomic number by 2. Beta particles don't change mass but increase atomic number by 1.

🎯 Exam Tip: Always analyze changes in both mass number (A) and atomic number (Z) systematically to correctly identify the number of alpha and beta emissions in a radioactive decay series. Remember alpha decay reduces A by 4 and Z by 2, while beta decay keeps A same and increases Z by 1.

4. Solve The Problems

Question A. Half life of \(^{18}\text{F}\) is 110 minutes. What fraction of \(^{18}\text{F}\) sample decays in 20 minutes ?
Answer: Given: \(t_{1/2}\) = 110 min
t = 20 min
To find: Fraction of \(^{18}\text{F}\) sample that decays
Formulae: i. \( \lambda = \frac{0.693}{t_{1/2}} \)
ii. \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
Calculation: \( \lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{110} = 0.0063 \text{ min}^{-1} \)
\( \implies \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) = \lambda \)
\( \log_{10} \left( \frac{N_0}{N} \right) = \frac{\lambda \times t}{2.303} = \frac{0.0063 \times 20}{2.303} = 0.0547 \)
\( \frac{N_0}{N} = \text{antilog}(0.0547) = 1.1342 \)
Fraction remaining undecayed = \( \frac{N}{N_0} = \frac{1}{1.1342} = 0.882 \)
Fraction of \(^{18}\text{F}\) sample that decays = 1 – 0.882 = 0.118
Ans: Fraction of \(^{18}\text{F}\) sample that decays in 20 minutes is 0.118.
In simple words: First, calculate the decay constant from the half-life. Then, use the decay constant and time to find the ratio of the initial number of atoms to the remaining atoms. Finally, subtract the remaining fraction from 1 to get the decayed fraction.

🎯 Exam Tip: Remember to use the correct units consistently throughout the calculation. The decay constant (λ) is key and can be derived from the half-life.

Question B. Half life of \(^{35}\text{S}\) is 87.8 d. What percentage of \(^{35}\text{S}\) sample remains after 180 d ?
Answer: Given: \(t_{1/2}\) = 87.8 d,
No = 100,
t = 180 d
To find: % of \(^{35}\text{S}\) that remains after 180 days
Formulae: i. \( \lambda = \frac{0.693}{t_{1/2}} \)
ii. \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
Calculation: i. \( \lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{87.8 \text{ d}} = 7.893 \times 10^{-3} \text{ d}^{-1} \)
ii. Now, \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
\( \log_{10} \left( \frac{N_0}{N} \right) = \frac{\lambda \times t}{2.303} \)
\( = \frac{7.893 \times 10^{-3} \times 180}{2.303} \)
\( = 0.617 \)
Taking antilog on both sides we get,
\( \frac{N_0}{N} = \text{antilog}(0.617) = 4.140 \)
\( N = \frac{100}{4.140} \)
\( = 24.155 \approx 24.2\% \)
Ans: Percentage of \(^{35}\text{S}\) that remains after 180 d is 24.2%
In simple words: First, calculate the decay constant from the given half-life. Then, use the decay constant and the elapsed time to find the ratio of initial to remaining sample, which will give you the percentage remaining.

🎯 Exam Tip: Ensure accurate calculation of the decay constant and correct application of the logarithmic decay formula. Pay attention to significant figures in final answers.

Question C. Half life \(^{67}\text{Ga}\) is 78 h. How long will it take to decay 12% of sample of Ga ?
Answer: Given: \(t_{1/2}\) = 78 h,
No = 100,
N = 100 - 12 = 88
To find: t
Formulae: i. \( \lambda = \frac{0.693}{t_{1/2}} \)
ii. \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
Calculation: i. \( \lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{78} = 8.885 \times 10^{-3} \text{ h}^{-1} \)
ii. \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
\( \implies 8.885 \times 10^{-3} = \frac{2.303}{t} \log_{10} \left( \frac{100}{88} \right) \)
\( 8.885 \times 10^{-3} = \frac{2.303}{t} \log_{10}(1.1364) \)
\( t = \frac{2.303}{8.885 \times 10^{-3}} \times 0.0555 \)
\( = 14.39 \text{ h} \)
Ans: Time taken for decay of 12% of sample of Ga is 14.39 h.
In simple words: First, calculate the decay constant from the half-life. Then, determine the remaining percentage of the sample after 12% decay. Use these values in the decay formula to solve for the time taken.

🎯 Exam Tip: When a percentage decay is given, convert it to the remaining percentage of the sample for use as N in the decay formula. Proper unit conversion (e.g., minutes to hours or vice-versa) is essential.

Question D. 0.5 g Sample of \(^{201}\text{Tl}\) decays to 0.0788 g in 8 days. What is its half life ?
Answer: Given: No = 0.5 g,
N = 0.0788 g,
t = 8 days
To find: \(t_{1/2}\)
Formulae: i. \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
ii. \( t_{1/2} = \frac{0.693}{\lambda} \)
Calculation: i. \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
\( = \frac{2.303}{8} \log_{10} \left( \frac{0.5}{0.0788} \right) \)
\( = \frac{2.303}{8} \log_{10}(6.3452) \)
\( = \frac{2.303}{8} \times 0.8024 \)
\( \lambda = 0.231 \text{ d}^{-1} \)
ii. \( t_{1/2} = \frac{0.693}{\lambda} \)
\( t_{1/2} = \frac{0.693}{0.231} \)
\( t_{1/2} = 3 \text{ d} \)
Ans: The half-life of \(^{201}\text{Tl}\) is 3 d.
In simple words: First, use the initial and final mass of the sample and the given time to calculate the decay constant. Then, use the decay constant to find the half-life.

🎯 Exam Tip: Ensure that the initial (N0) and final (N) amounts are in consistent units (mass or moles). The correct application of logarithmic properties is crucial for accurate calculation.

Question E. 65% of \(^{111}\text{In}\) sample decays in 4.2 d. What is its half life ?
Answer: Given: No = 100,
N = 100 – 65 = 35,
t = 4.2d
To find: \(t_{1/2}\)
Formulae: i. \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
ii. \( t_{1/2} = \frac{0.693}{\lambda} \)
Calculation: i. \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
\( = \frac{2.303}{4.2} \log_{10} \left( \frac{100}{35} \right) \)
\( = 0.548 \times 0.456 = 0.2499 \text{ d}^{-1} \)
ii. \( t_{1/2} = \frac{0.693}{\lambda} \)
\( = \frac{0.693}{0.2499} = 2.773 \text{ d} \)
Ans: Half-life of \(^{111}\text{In}\) sample is 2.773 d
In simple words: Determine the percentage of the sample remaining after 65% decay. Use this value along with the given time to calculate the decay constant, and then use the decay constant to find the half-life.

🎯 Exam Tip: Always convert the decay percentage into the remaining amount (N) before applying the decay formula. Precise calculation of log values and correct inverse operations are vital for accurate results.

Question F. Calculate the binding energy per nucleon of \(_{36}^{84}\text{Kr}\) whose atomic mass is 83.913 u. (Mass of neutron is 1.0087 u and that of H atom is 1.0078 u).
Answer: Given: A = 84, Z = 36,
m = 83.913 u
\(m_n\) = 1.0087 u
\(m_H\) = 1.0078 u
To find: Binding energy per nucleon (B)
Formulae: i. \( \Delta m = Zm_H + (A-Z)m_n - m \)
ii. \( \text{B.E.} = \Delta m \times 931.4 \text{ MeV} \)
iii. \( B = \frac{\text{B.E.}}{A} \)
Calculation: i. \( \Delta m = Zm_H + (A-Z)m_n - m \)
\( = (36 \times 1.0078) + (48 \times 1.0087) - 83.913 \)
\( = 36.2808 + 48.4176 - 83.913 \)
\( = 0.7854 \text{ u} \)
ii. \( \text{B.E.} = \Delta m \times 931.4 \)
\( = 0.7854 \times 931.4 \)
\( = 731.4 \text{ MeV} \) (by using log table)
iii. \( B = \frac{\text{B.E.}}{A} = \frac{731.4}{84} \)
\( = 8.706 \text{ MeV} \) (by using log table)
Ans: Binding energy per nucleon of \(_{36}^{84}\text{Kr}\) = 8.706 MeV
In simple words: First, calculate the mass defect by comparing the mass of the nucleus with the sum of its individual protons and neutrons. Then, convert this mass defect into binding energy using Einstein's mass-energy equivalence. Finally, divide the binding energy by the total number of nucleons (mass number) to get the binding energy per nucleon.

🎯 Exam Tip: Remember that A-Z gives the number of neutrons. Be careful with calculations involving the conversion factor (931.4 MeV/u) and ensure all values are correctly substituted into the formulas.

Question G. Calculate the energy in Mev released in the nuclear reaction
\(_{77}^{174}\text{Ir} \longrightarrow _{75}^{170}\text{Re} + _{2}^{4}\text{He}\)
Atomic masses : Ir = 173.97 u,
Re = 169.96 u and
He = 4.0026 u
Answer: Given: \(m_{\text{Ir}}\) = 173.97 u
\(m_{\text{Re}}\) = 169.96 u
\(m_{\text{He}}\) = 4.0026 u
To find: Energy released
Formulae: i. \( \Delta m = (\text{mass of } ^{174}\text{Ir}) - (\text{mass of } ^{170}\text{Re} + \text{mass of } ^{4}\text{He}) \)
ii. \( E = \Delta m \times 931.4 \text{ MeV} \)
Calculation: i. \( \Delta m = (\text{mass of } ^{174}\text{Ir}) - (\text{mass of } ^{170}\text{Re} + \text{mass of } ^{4}\text{He}) \)
\( = 173.97 - (169.96 + 4.0026) \)
\( = 173.97 - 173.9626 \)
\( = 7.4 \times 10^{-3} \text{ u} \)
ii. \( E = \Delta m \times 931.4 \)
\( = 7.4 \times 10^{-3} \times 931.4 \)
\( = 6.89236 \text{ MeV} \approx 6.892 \text{ MeV} \)
Ans: The energy released in given nuclear reaction is 6.892 MeV.
In simple words: To find the energy released, calculate the mass defect by subtracting the total mass of the products from the mass of the reactant. Then, convert this mass defect into energy using the conversion factor (931.4 MeV/u).

🎯 Exam Tip: Always ensure the masses are summed correctly for reactants and products before calculating the mass defect. A positive mass defect indicates energy release, while a negative one would indicate energy absorption.

Question H. A 3/4 of the original amount of radioisotope decays in 60 minutes. What is its half life ?
Answer: Given: No = 100,
For N, \( 100 \times \frac{3}{4} = 75 \).
\( \implies N = 100 - 75 = 25 \)
t = 60 min
To find: \(t_{1/2}\)
Formulae: i. \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
ii. \( t_{1/2} = \frac{0.693}{\lambda} \)
Calculation: i. \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
\( = \frac{2.303}{60} \log_{10} \left( \frac{100}{25} \right) \)
\( = \frac{2.303}{60} \log_{10} 4 \)
\( = 0.0231 \text{ min}^{-1} \)
ii. \( t_{1/2} = \frac{0.693}{\lambda} \)
\( = \frac{0.693}{0.0231} = 30 \text{ min} \)
Ans: Half-life of the radioisotope is 30 min.
In simple words: If 3/4 of the sample decays, then 1/4 remains. Use the initial amount, the remaining amount, and the time taken in the decay constant formula. Once the decay constant is found, calculate the half-life.

🎯 Exam Tip: Properly converting the fraction decayed into the remaining fraction (N) is a critical first step. Remember that the half-life is inversely proportional to the decay constant.

Question I. How many α-particles are emitted by 0.1 g of \(^{226}\text{Ra}\) in one year?
Answer: Given: t = 1 y,
Amount of sample = 0.1 g
To find: Number of particles emitted
Formulae: i. \( \lambda = \frac{0.693}{t_{1/2}} \)
ii. Activity = Number of α (or β) particles emitted = \( -\frac{dN}{dt} = \lambda N \)
Calculation: Half-life of radium = 1620 y
\( \lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{1620} = 4.28 \times 10^{-4} \text{ y}^{-1} \)
\( N = \frac{0.1 \times 6.022 \times 10^{23}}{226} \)
\( N = 2.665 \times 10^{20} \text{ atoms} \)
Activity = \( -\frac{dN}{dt} = \lambda N \)
\( = 4.28 \times 10^{-4} \times 2.665 \times 10^{20} \text{ atoms} \)
\( = 1.141 \times 10^{17} \text{ particles/year} \)
Ans: Particles emitted by 0.1 g of \(^{226}\text{Ra}\) in one year = 1.141 × 1017 particles/year.
[Note: The half-life of radium is 1620 years. In order to apply appropriate textual concept, we have used this value in calculation.]
In simple words: First, calculate the decay constant from the half-life. Then, find the total number of radium atoms in the given sample. Finally, multiply the decay constant by the number of atoms to determine the activity, which represents the number of particles emitted per year.

🎯 Exam Tip: Ensure consistent units for time (years in this case) for both half-life and activity. Accurately calculate the number of atoms using molar mass and Avogadro's number.

Question J. A sample of \(^{32}\text{P}\) initially shows activity of one Curie. After 303 days the activity falls to 1.5× \(10^4\) dps. What is the half life of \(^{32}\text{P}\) ?
Answer: Given:
\( -\frac{dN_0}{dt} = 1 \text{ Ci} = 3.7 \times 10^{10} \text{ dps} \)
\( -\frac{dN}{dt} = 1.5 \times 10^4 \text{ dps} \)
t = 303 days
To find: \(t_{1/2}\)
Formulae: i. \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
ii. \( t_{1/2} = \frac{0.693}{\lambda} \)
Calculation: i. \( \lambda = \frac{2.303}{t} \log_{10} \left( \frac{N_0}{N} \right) \)
\( \implies \lambda = \frac{2.303}{303} \log_{10} \left( \frac{3.7 \times 10^{10}}{1.5 \times 10^4} \right) \)
\( = \frac{2.303}{303} \log_{10}(2.4666 \times 10^6) \)
\( = \frac{2.303}{303} \times 6.3921 \)
\( = 0.04859 \text{ d}^{-1} \)
ii. \( t_{1/2} = \frac{0.693}{\lambda} \)
\( = \frac{0.693}{0.04859} = 14.27 \text{ d} \) (by using log table)
Ans: Half-life of \(^{32}\text{P}\) is 14.27 days
In simple words: Convert the initial and final activities to disintegrations per second (dps). Use these activities and the elapsed time in the decay formula to find the decay constant, then calculate the half-life from the decay constant.

🎯 Exam Tip: Remember that activity is directly proportional to the number of radioactive nuclei (N), so \(N_0/N\) can be replaced with \(A_0/A\). Precise logarithmic calculations are vital.

Question K. Half life of radon is 3.82 d. By what time would 99.9 % of radon will be decayed.
Answer: Given: \(t_{1/2}\) = 3.82 d,
No = 100
N = 100 - 99.9 = 0.1
To find: t
Formulae: i. \( \lambda = \frac{0.693}{t_{1/2}} \)
ii. \( t = \frac{2.303}{\lambda} \log_{10} \left( \frac{N_0}{N} \right) \)
Calculation: i. \( \lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{3.82} = 0.1814 \text{ d}^{-1} \)
ii. \( t = \frac{2.303}{\lambda} \log_{10} \left( \frac{N_0}{N} \right) \)
\( = \frac{2.303}{0.1814} \log_{10} \left( \frac{100}{0.1} \right) \)
\( = \frac{2.303}{0.1814} \log_{10}(1000) \)
\( = 38.087 \text{ d} \approx 38.1 \text{ d} \)
Ans: Time taken for 99.9% of radon to be decayed is 38.1 d.
In simple words: Calculate the decay constant from the half-life. Since 99.9% decays, 0.1% remains. Use these values in the integrated rate law to find the total time required for the decay.

🎯 Exam Tip: Always correctly calculate the remaining percentage (N) from the decay percentage. Logarithm rules and accurate calculations are crucial for solving for time (t).

Question L. It has been found that the Sun's mass loss is 4.34 × \(10^9\) kg per second. How much energy per second would be radiated into space by the Sun ?
Answer: Given: Sun's mass loss = 4.34 × \(10^9\) kg per second
To find: Energy radiated per second into space by Sun
Calculation: \( \Delta m = 4.34 \times 10^9 \text{ kg per second} \)
Now, \( 1.66 \times 10^{-27} \text{ kg} = 1\text{ u} \)
\( \therefore \quad \Delta m = \frac{4.34 \times 10^9}{1.66 \times 10^{-27}} \text{ u per second} \)
\( = 2.614 \times 10^{36} \text{ u per second} \)
Now, \( 1\text{ u} = 931.4 \text{ MeV} \)
\( 2.614 \times 10^{36} \text{ u per second} = 2.614 \times 10^{36} \times 931.4 \)
\( = 2.435 \times 10^{39} \text{ MeV/s} \)
Now, \( 1\text{ MeV} = 1.6022 \times 10^{-19} \text{ J} \) and \( 1\text{ eV} = 1 \times 10^{-6} \text{ MeV} \)
\( 1\text{ MeV} = 1.6022 \times 10^{-13} \text{ J} \)
\( = 1.6022 \times 10^{-16} \text{ kJ} \)
\( E = 2.435 \times 10^{39} \text{ MeV/s} \times 1.6022 \times 10^{-16} \text{ kJ/MeV} \)
\( = 3.901 \times 10^{23} \text{ kJ/s} \)
Ans: Energy radiated per second into space by Sun is 3.901 × \(10^{23}\) kJ/s.
In simple words: Convert the Sun's mass loss from kilograms to atomic mass units (u). Then, convert this mass in 'u' to energy in MeV using the conversion factor, and finally convert MeV to kilojoules to find the energy radiated per second.

🎯 Exam Tip: Crucial step is the accurate conversion of mass (kg to u) and then (u to MeV) using \(E = mc^2\) relation (represented here by 931.4 MeV/u). Pay attention to unit conversions from MeV to J or kJ.

Question M. A sample of old wood shows 7.0 dps/g. If the fresh sample of tree shows 16.0 dps/g, How old is the given sample of wood ? Half life of \(^{14}\text{C}\) 5730 y.
Answer: Given: \(t_{1/2}\) of \(^{14}\text{C}\) = 5730 y
\( \left( -\frac{dN_0}{dt} \right) = 16 \text{ dps/g} \)
\( \left( -\frac{dN}{dt} \right) = 7 \text{ dps/g} \)
To find: Age of sample of wood
Formulae: i. \( \lambda = \frac{0.693}{t_{1/2}} \)
ii. \( t = \frac{2.303}{\lambda} \log_{10} \left( \frac{N_0}{N} \right) \)
Calculation: \( \lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} = 1.209 \times 10^{-4} \text{ y}^{-1} \)
\( t = \frac{2.303}{\lambda} \log_{10} \left( \frac{N_0}{N} \right) \)
\( t = \frac{2.303}{1.209 \times 10^{-4}} \log_{10} \left( \frac{16.0}{7.0} \right) \)
\( = 6839 \text{ y} \)
Ans: The age of sample of old wood is 6839 y.
In simple words: This problem uses radiocarbon dating. First, find the decay constant for carbon-14 from its half-life. Then, use the ratio of activities (initial and current) and the decay constant in the radioactive decay formula to calculate the age of the wood sample.

🎯 Exam Tip: For carbon dating, the ratio of activities \( (A_0/A) \) can be directly substituted for the ratio of nuclei \( (N_0/N) \). Ensure the half-life and time are in consistent units (usually years for carbon dating).

Activity :

1. Discuss five applications of radioactivity for peaceful purpose.
Answer:

• Development in earth sciences: Like to understand various geographical changes occurring on earth.
• Development in space technology: To study nuclear reactions in stars which may lead to new discoveries.
• Development in medical sciences: Diagnosis and treatment of various diseases.
• Development in industries: As a potent source of electricity or a power generator.
• Development in agriculture: To study or monitor changes in soil like uptake of nutrients from the soil etc.

🎯 Exam Tip: When discussing applications, categorize them by field (e.g., medical, industrial, agricultural) and provide specific examples within each category to demonstrate comprehensive understanding.

2. Organize a trip to Bhabha Atomic Reasearch Centre, Mumbai to learn about nuclear reactor. This will have to be organized through your college.
Answer: Students are expected to visit the place to understand more about nuclear reactors.
In simple words: This is an experiential learning task, encouraging students to visit a nuclear research center to gain practical knowledge about nuclear reactors.

🎯 Exam Tip: This question assesses your initiative and ability to connect theoretical knowledge with real-world applications. While not a direct answer, mentioning the educational benefits of such a visit would be a strong point.

11th Chemistry Digest Chapter 13 Nuclear Chemistry And Radioactivity Intext Questions And Answers

Do You Know? (Textbook Page No. 190)

Question 1. How small is the nucleus in comparison to the rest of the atom?
Answer: The radius of nucleus is of the order of \(10^{-15}\) m whereas that of the outer sphere is of the order of \(10^{-10}\) m. The size of outer sphere, is \(10^5\) times larger than the nucleus i.e., if we consider the atom of size of football stadium then its nucleus will be the size of a pea.
In simple words: The nucleus is incredibly tiny compared to the atom, roughly 100,000 times smaller in diameter, like a pea in the middle of a football stadium.

🎯 Exam Tip: Remember the orders of magnitude for atomic and nuclear radii (\(10^{-10}\) m for atom, \(10^{-15}\) m for nucleus) and be able to articulate the relative size difference using an analogy.

(Textbook Page No. 191)

Question 1. Identify the following nuclides as: isotopes, isobars and isotones.
\(_{1}^{3}\text{H, } _{2}^{3}\text{He, } _{6}^{14}\text{C, } _{7}^{14}\text{N, } _{12}^{24}\text{Mg, } _{17}^{35}\text{Cl, } _{17}^{38}\text{Cl, } _{16}^{32}\text{S, } _{15}^{31}\text{P}\)
Answer: Isotopes: \(_{17}^{35}\text{Cl}\) and \(_{17}^{38}\text{Cl}\)
Isobars: \(_{6}^{14}\text{C}\) and \(_{7}^{14}\text{N}\)
Isotones: i. \(_{1}^{3}\text{H}\) and \(_{2}^{4}\text{He}\)
ii. \(_{16}^{32}\text{S}\) and \(_{15}^{31}\text{P}\)
In simple words: Isotopes have the same number of protons (atomic number Z) but different numbers of neutrons (mass number A). Isobars have the same mass number (A) but different atomic numbers (Z). Isotones have the same number of neutrons (A-Z) but different atomic numbers (Z).

🎯 Exam Tip: Focus on the definitions: Isotopes (same Z, different A), Isobars (same A, different Z), Isotones (same A-Z, different Z). Clearly identify Z and A for each nuclide given.

(Textbook Page No. 194)

Question 1.
(i) What do you understand by the term rate of decay and give its mathematical expression.
(ii) Why is minus sign required in the expression of decay rate?
Answer:
(i) Rate of decay of a radioelement denotes the number of nuclei of its atoms which decay in unit time. It is also called activity of radioelement.
Rate of decay at any time t can be expressed as follows:
Rate of decay (activity) = \(-\frac{dN}{dt}\)
where, dN is the number of nuclei that decay within time interval dt.
(ii) Minus sign in the expression indicates that the number of nuclei decreases with time. Therefore, dN is a negative quantity. But, the rate of decay is a positive quantity. The negative sign is introduced in the rate expression to make the rate positive.
In simple words: The rate of decay is how many radioactive nuclei break down per unit time. Its mathematical expression is \(-\frac{dN}{dt}\), where the negative sign ensures the decay rate is positive because the number of nuclei (N) is decreasing.

🎯 Exam Tip: Understand that rate is always a positive quantity. The negative sign in \(-\frac{dN}{dt}\) is a convention to make the decay rate positive, reflecting a decrease in the number of radioactive nuclei over time.

Try This. (Textbook Page No. 194)

Question 1. Prepare a chart of comparative properties of the above three types of radiations.
Answer:

🎯 Exam Tip: Memorize the key properties for each type of radiation, especially their identity, charge, mass, and relative penetration power, as these are frequently tested comparative points.

Just Think (Textbook Page No. 195)

Question 1. Does half-life increase, decrease or remain constant? Explain.
Answer: Half-life of a particular radioelement remains constant at a given instant. A radioactive half-life refers to the amount of time it takes for half of the original isotope to decay. It is related to decay constant by the expression: \( t_{1/2} = \frac{0.693}{\lambda} \)
From the expression, it is evident that half-life of a radio isotope is dependent only on the decay constant and is independent of the initial amount of the radio isotope. Each successive half-life in which the amount of radio isotope decreases to its half value is the same.
\( 100\% \xrightarrow{t_{1/2}} 50\% \text{ (50% complete)} \xrightarrow{t_{1/2}} 25\% \text{ (75% complete)} \)
Thus, half-life remains constant.
In simple words: Half-life remains constant because it depends only on the decay constant, which is a fixed property of a specific radioisotope, not on the initial amount of the substance.

🎯 Exam Tip: Emphasize that half-life is a characteristic constant for a given radioisotope and is independent of external conditions like temperature, pressure, or the initial amount of the sample.

Try This (Textbook Page No. 198)

Question 1. \(_{12}^{24}\text{Mg}\) and \(_{13}^{27}\text{Al}\), both undergo (\(\alpha, \text{ n}\)) reactions and the products are radioactive. These emit \(\beta^+\) particles having positive charge (called positrons). Write balanced nuclear reactions in both.
Answer:
(i) \(_{12}^{24}\text{Mg} + _{2}^{4}\text{He} \longrightarrow _{14}^{27}\text{Si} + _{0}^{1}\text{n}\)
\(_{14}^{27}\text{Si} \longrightarrow _{13}^{27}\text{Al} + _{+1}^{0}\text{e} (\beta^+)\) (Positron)
(ii) \(_{13}^{27}\text{Al} + _{2}^{4}\text{He} \longrightarrow _{15}^{30}\text{P} + _{0}^{1}\text{n}\)
\(_{15}^{30}\text{P} \longrightarrow _{14}^{30}\text{Si} + _{+1}^{0}\text{e}\) (Positron)
In simple words: In these reactions, magnesium and aluminum absorb an alpha particle and emit a neutron to form a new, unstable nuclide. This unstable nuclide then undergoes positron (beta-plus) decay to become a different stable element.

🎯 Exam Tip: Ensure that both mass number (A) and atomic number (Z) are conserved on both sides of each nuclear reaction. Pay careful attention to the charges and masses of emitted particles like alpha, neutron, and positron.

Do You Know? (Textbook Page No. 198)

Question 1. What is the critical mass of \(_{92}^{235}\text{U}\)?
Answer:
(i) The critical mass is the minimum mass of uranium-235 required to achieve a self-sustaining fission chain reaction under stated conditions.
(ii) The chain reaction in fission of U-235 becomes self-sustaining when the critical mass of uranium-235 is about 50 kilograms.
In simple words: Critical mass is the smallest amount of a fissile material like uranium-235 needed for a nuclear chain reaction to sustain itself. For uranium-235, this is approximately 50 kilograms.

🎯 Exam Tip: Understand that critical mass is essential for sustaining a chain reaction in nuclear fission. Factors like shape, density, and purity can influence its exact value, although for U-235, roughly 50 kg is a standard reference.

Activity (Textbook Page No. 200)

Question 1. You have learnt in Std. 9th, medical, industrial and agricultural applications of radioisotopes. Write at least two applications each.
Answer:
(i) The uses of radioactive isotopes in the field of medicine:
a. Polycythaemia: The red blood cell count increases in the disease polycythaemia. Phosphorus-32 is used in its treatment.
b. Bone cancer: Strontium-89, strontium-90, samarium-153 and radium-223 are used in the treatment of bone cancer.
(ii) The uses of radioactive isotopes in the industrial field:
a. Luminescent paint and radioluminescence: The radioactive substances radium, promethium, tritium with some phosphorus are used to make certain objects visible in the dark.
e.g. Hands of a clock, krypton-85 is used in HID (High Intensity Discharge) lamps.
b. Use in ceramic articles:
1. Luminous colours are used to decorate ceramic tiles, utensils, plates, etc.
2. Uranium oxide was earlier used to colour ceramics.
(iii) The uses of radioactive isotopes in the agriculture field:
a. The genes and chromosomes that give seeds its properties like fast growth, higher productivity, etc., can be modified by means of radiation.
b. Onions and potatoes are irradiated with gamma rays from cobalt-60 to prevent their sprouting.
In simple words: Radioisotopes are used in medicine to treat blood and bone cancers, in industry for glowing paints and ceramic coloring, and in agriculture to improve crop properties and prevent spoilage.

🎯 Exam Tip: When listing applications, provide specific examples of radioisotopes and their direct uses. Categorizing applications helps demonstrate a structured understanding of the topic.

Question 1. Discuss five applications of radioactivity for peaceful purpose.
Answer:
• Development in earth sciences: Like to understand various geographical changes occurring on earth.
• Development in space technology: To study nuclear reactions in stars which may lead to new discoveries.
• Development in medical sciences: Diagnosis and treatment of various diseases.
• Development in industries: As a potent source of electricity or a power generator.
• Development in agriculture: To study or monitor changes in soil like uptake of nutrients from the soil etc.
In simple words: Radioactivity can be used for various beneficial purposes such as understanding geological changes, studying stellar nuclear reactions, diagnosing and treating diseases, generating electricity, and monitoring soil nutrient uptake in agriculture.

🎯 Exam Tip: Listing specific applications clearly and concisely is key to scoring points in questions about the peaceful uses of radioactivity.

Question 2. Organize a trip to Bhabha Atomic Reasearch Centre, Mumbai to learn about nuclear reactor. This will have to be organized through your college.
Answer: Students are expected to visit the place to understand more about nuclear reactors.
In simple words: Students should arrange an educational visit to the Bhabha Atomic Research Centre to learn about nuclear reactors firsthand, as part of their college curriculum.

🎯 Exam Tip: For practical-oriented questions, demonstrating initiative and hands-on learning is often valued.

11th Chemistry Digest Chapter 13 Nuclear Chemistry And Radioactivity Intext Questions And Answers

Do You Know?

Question 1. How small is the nucleus in comparison to the rest of the atom?
Answer: The radius of nucleus is of the order of \( 10^{-15} \) m whereas that of the outer sphere is of the order of \( 10^{-10} \) m. The size of outer sphere, is \( 10^5 \) times larger than the nucleus i.e., if we consider the atom of size of football stadium then its nucleus will be the size of a pea.
In simple words: The nucleus is extremely small compared to the entire atom, with the atom being about \( 10^5 \) times larger than its nucleus, much like a football stadium compared to a pea.

🎯 Exam Tip: Remember the approximate orders of magnitude for atomic and nuclear radii (\( 10^{-10} \) m and \( 10^{-15} \) m respectively) to illustrate the vast size difference effectively.