Maharashtra Board Class 12 Physics Chapter 15 Structure of Atoms and Nuclei Exercise Solutions

In solving problems, use \(m_e = 0.00055\) u \(= 0.5110\) MeV/c\(^2\), \(m_p = 1.00728\) u, \(m_n = 1.00866\) u, \(m_H = 1.007825\) u, u \(= 931.5\) MeV, \(e = 1.602 \times 10^{-19}\) C, \(h = 6.626 \times 10^{-34}\) Js, \(\varepsilon_0 = 8.854 \times 10^{-12}\) SI units and \(m_e = 9.109 \times 10^{-31}\) kg.

1. Choose the Correct Option.

Question 1. (i) In which of the following systems will the radius of the first orbit of the electron be the smallest?
(a) hydrogen
(b) singly ionized helium
(c) deuteron
(d) tritium
Answer: (b) Singly ionized helium
In simple words: The radius of the first Bohr orbit is inversely proportional to the atomic number Z. Singly ionized helium has Z = 2 (the highest among these options), so its first orbit radius is the smallest.

Teacher's Note: Students often confuse deuteron and tritium with hydrogen isotopes having different Z values - they don't. Deuteron and tritium are hydrogen isotopes with Z = 1. Only singly ionized helium has Z = 2. Use the formula \(r_n \propto \frac{n^2}{Z}\) to make this clear.

Exam Tip: Remember: smaller orbit = larger Z. Singly ionized helium (He+) has Z = 2, giving half the radius of hydrogen for the same orbit number.

Question 1. (ii) The radius of the 4th orbit of the electron will be smaller than its 8th orbit by a factor of
(a) 2
(b) 4
(c) 8
(d) 16
Answer: (b) 4
In simple words: The radius of the nth orbit is proportional to n\(^2\). So \(r_4 \propto 16\) and \(r_8 \propto 64\). The ratio \(r_8/r_4 = 64/16 = 4\). The 8th orbit is 4 times larger than the 4th orbit.

Teacher's Note: Reinforce the formula \(r_n = n^2 r_1\) where \(r_1 = 0.529\) Å for hydrogen. Ask students to calculate actual radii for n=4 and n=8 to verify the factor of 4.

Exam Tip: Use the ratio method: \(\frac{r_8}{r_4} = \frac{8^2}{4^2} = \frac{64}{16} = 4\). Write this ratio clearly in your answer.

Question 1. (iii) In the spectrum of the hydrogen atom which transition will yield the longest wavelength?
(a) n = 2 to n = 1
(b) n = 5 to n = 4
(c) n = 7 to n = 6
(d) n = 8 to n = 7
Answer: (d) n = 8 to n = 7
In simple words: Longest wavelength means smallest energy difference. The energy levels get closer together as n increases, so the transition between the highest consecutive levels (n=8 to n=7) releases the least energy and gives the longest wavelength.

Teacher's Note: Use the energy level diagram to show students that energy gaps decrease as n increases. The formula \(\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\) shows that larger n values give smaller \(\frac{1}{\lambda}\), hence larger \(\lambda\).

Exam Tip: For longest wavelength, pick the transition with highest quantum numbers that are consecutive (smallest energy gap). For shortest wavelength, pick transition to n=1 from the highest level.

Question 1. (iv) Which of the following properties of a nucleus does not depend on its mass number?
(a) radius
(b) mass
(c) volume
(d) density
Answer: (d) density
In simple words: Nuclear density is approximately the same for all nuclei - about \(2.3 \times 10^{17}\) kg/m\(^3\). Although both mass and volume increase with mass number A, their ratio (density) stays nearly constant.

Teacher's Note: This is a key conceptual point - nuclear density is uniform across all elements. Compare this with atomic density which varies widely. The formula \(\rho = \frac{3m_p}{4\pi r_0^3}\) is independent of A.

Exam Tip: Remember: radius \(\propto A^{1/3}\), volume \(\propto A\), mass \(\propto A\), so density = mass/volume \(\propto A/A\) = constant. Density is the odd one out.

Question 1. (v) If the number of nuclei in a radioactive sample at a given time is N, what will be the number at the end of two half-lives?
(a) N/2
(b) N/4
(c) 3N/4
(d) N/8
Answer: (b) N/4
In simple words: After one half-life, half the nuclei remain: N/2. After two half-lives, half of that remains: N/4. Each half-life halves the number of remaining nuclei.

Teacher's Note: Use a simple table: Start = N, after 1 half-life = N/2, after 2 half-lives = N/4, after 3 = N/8. This pattern (\(N/2^n\) after n half-lives) should be memorized.

Exam Tip: The formula is \(N = N_0 \left(\frac{1}{2}\right)^n\) where n is the number of half-lives. For n=2: \(N = N_0/4\).

2. Answer in Brief.

Question 2. (i) State the postulates of Bohr's atomic model.
Answer: The postulates of Bohr's atomic model (for the hydrogen atom):
(1) The electron revolves with a constant speed in a circular orbit around the nucleus. The necessary centripetal force is the Coulomb force of attraction of the positive nuclear charge on the negatively charged electron.
(2) The electron can revolve without radiating energy only in certain orbits, called allowed or stable orbits, in which the angular momentum of the electron is equal to an integral multiple of h/2\(\pi\), where h is Planck's constant. That is, \(L = \frac{nh}{2\pi} = n\hbar\), where n = 1, 2, 3, ...
(3) Energy is radiated by the electron only when it jumps from one of its orbits to another orbit having lower energy. The energy of the quantum of electromagnetic radiation (photon) emitted is equal to the energy difference of the two states: \(h\nu = E_m - E_n\).
In simple words: Electrons move in fixed circular orbits without losing energy. Only specific orbits are allowed where angular momentum is a multiple of h/2\(\pi\). Energy is released only when an electron jumps to a lower orbit.

Teacher's Note: Emphasize that Bohr's model solved two major failures of Rutherford's model - atomic stability and line spectra. Students often forget to mention the quantization condition (postulate 2) in exams. Make them write all three postulates separately.

Exam Tip: Write all three postulates numbered clearly. Include the mathematical form \(mvr = \frac{nh}{2\pi}\) for postulate 2 and \(h\nu = E_2 - E_1\) for postulate 3 to score full marks.

Question 2. (ii) State the difficulties faced by Rutherford's atomic model.
Answer:
(1) According to Rutherford, electrons revolve in circular orbits around the atomic nucleus. Circular motion is an accelerated motion. According to classical electromagnetic theory, an accelerated charge continuously radiates energy. Therefore, an electron during its orbital motion should go on radiating energy. Due to the loss of energy, the radius of its orbit should go on decreasing. The electron should move along a spiral path and finally fall into the nucleus in a very short time (of the order of \(10^{-16}\) s for hydrogen). Thus, the atom should be unstable - but atoms are stable.
(2) If the electron moves along such a spiral path, the radius of its orbit would continuously decrease. As a result, the speed and frequency of revolution would continuously increase. The electron would emit radiation of continuously changing frequency, giving rise to a continuous spectrum. However, the atomic spectrum is a line spectrum, not a continuous spectrum.
In simple words: Rutherford's model predicts atoms should collapse in a tiny fraction of a second, and should emit all colors of light continuously. But atoms are stable and emit only specific colors (lines). This is why his model failed.

Teacher's Note: Ask students: "If Rutherford's model were correct, would you exist?" The answer is no - all atoms would have collapsed. This makes the problem concrete and memorable. Connect difficulty 2 to the observation of line spectra in the lab.

Exam Tip: State two difficulties clearly and separately. For each: state what the model predicts, then state what is actually observed. Always mention the time scale \(10^{-16}\) s for the collapse - examiners look for this detail.

Question 2. (iii) What are alpha, beta and gamma decays?
Answer:
(a) Alpha decay: A radioactive transformation in which an \(\alpha\)-particle (helium nucleus, \(_2^4\text{He}\)) is emitted is called \(\alpha\)-decay. The atomic number decreases by 2 and mass number decreases by 4.
Example: \(_{92}^{238}\text{U} \rightarrow _{90}^{234}\text{Th} + _2^4\alpha\)
Energy released: \(Q = [m_U - m_{Th} - m_\alpha]c^2\)

(b) Beta-minus decay (\(\beta^-\)): A radioactive transformation in which a \(\beta^-\)-particle (electron) is emitted. The atomic number increases by 1; mass number unchanged.
Example: \(_{90}^{234}\text{Th} \rightarrow _{91}^{234}\text{Pa} + _{-1}^0e + \bar{\nu}_e\)
where \(\bar{\nu}_e\) is the antineutrino emitted to conserve momentum, energy and spin.
\(Q = [m_{Th} - m_{Pa} - m_e]c^2\)

Beta-plus decay (\(\beta^+\)): The atomic number decreases by 1; mass number unchanged.
Example: \(_{15}^{30}\text{P} \rightarrow _{14}^{30}\text{Si} + _{+1}^0e + \nu_e\)
\(Q = [m_P - m_{Si} - m_e]c^2\)

(c) Gamma decay: A given nucleus does not emit \(\alpha\) and \(\beta\)-particles simultaneously. However, on emission of \(\alpha\) or \(\beta\)-particles, most nuclei are left in an excited state. A nucleus in an excited state emits a \(\gamma\)-ray photon in a transition to the lower energy state. Hence, \(\alpha\) and \(\beta\)-particle emissions are often accompanied by \(\gamma\)-rays.
In simple words: Alpha decay releases a helium nucleus (shrinks the atom). Beta decay converts a neutron to a proton or vice versa (changes the element). Gamma decay releases pure energy (X-ray-like) after alpha or beta decay - the nucleus just settles down to a lower energy state.

Teacher's Note: Use the analogy of a ball rolling down steps: alpha and beta are like the ball jumping down big steps (changing identity), while gamma is like the vibration that follows the jump settling down. Make sure students know that gamma decay does NOT change Z or A.

Exam Tip: For each decay, state: what particle is emitted, how Z changes, how A changes, and write the example equation. For \(\beta\) decay, always mention the neutrino/antineutrino - many students forget this.

Question 2. (iv) Define excitation energy, binding energy and ionization energy of an electron in an atom.
Answer:
(1) Excitation energy: The energy required to transfer an electron from the ground state to an excited state (a state of higher energy) is called the excitation energy of the electron in that state.
(2) Binding energy of an electron in an atom: The minimum energy that should be provided to an orbital electron to remove it from the atom such that its total energy is zero.
(3) Ionization energy of an electron in an atom: The minimum energy required to remove the least strongly bound electron from a neutral atom such that its total energy is zero.
In simple words: Excitation energy lifts an electron to a higher orbit without removing it. Binding energy is the energy needed to free one specific electron. Ionization energy is the energy to remove the outermost (easiest to remove) electron from a neutral atom.

Teacher's Note: Students often confuse binding energy and ionization energy. Clarify: binding energy applies to any electron in any orbit; ionization energy specifically refers to the least bound (outermost) electron in a neutral atom. For hydrogen, they are equal.

Exam Tip: Each definition should include the key phrase "such that its total energy is zero" for binding energy and ionization energy. Examiners look for this condition.

Question 2. (v) Show that the frequency of the first line in Lyman series is equal to the difference between the limiting frequencies of Lyman and Balmer series.
Answer:
For the first line in the Lyman series (n = 1 to m = 2):
\[\frac{1}{\lambda_{L1}} = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}\]
\[\therefore \nu_{L1} = \frac{c}{\lambda_{L1}} = \frac{3Rc}{4}\]

For the limit of the Lyman series (n = 1, m = \(\infty\)):
\[\frac{1}{\lambda_{L\infty}} = R\left(\frac{1}{1^2} - \frac{1}{\infty}\right) = R\]
\[\therefore \nu_{L\infty} = \frac{c}{\lambda_{L\infty}} = Rc\]

For the limit of the Balmer series (n = 2, m = \(\infty\)):
\[\frac{1}{\lambda_{B\infty}} = R\left(\frac{1}{2^2} - \frac{1}{\infty}\right) = \frac{R}{4}\]
\[\therefore \nu_{B\infty} = \frac{c}{\lambda_{B\infty}} = \frac{Rc}{4}\]

Now: \(\nu_{L\infty} - \nu_{B\infty} = Rc - \frac{Rc}{4} = \frac{3Rc}{4} = \nu_{L1}\)

Hence, the frequency of the first line in the Lyman series equals the difference between the limiting frequencies of the Lyman and Balmer series.
In simple words: The first Lyman line frequency (3Rc/4) equals the Lyman limit frequency (Rc) minus the Balmer limit frequency (Rc/4). This is a beautiful mathematical relationship in hydrogen's spectrum.

Teacher's Note: This proof tests students' ability to apply the Rydberg formula systematically. Encourage them to write each step clearly, substituting the appropriate values of n and m. The key insight is that both series share the same Rydberg constant R.

Exam Tip: Start by writing the Rydberg formula for each case. Compute \(\nu_{L1}\), \(\nu_{L\infty}\) and \(\nu_{B\infty}\) separately, then show the difference equals \(\nu_{L1}\). Present as a logical sequence - don't skip steps.

Question 3. State the postulates of Bohr's atomic model and derive the expression for the energy of an electron in the atom.
Answer:
Postulates of Bohr's atomic model:
(1) The electron revolves with constant speed in a circular orbit around the nucleus. The centripetal force equals the Coulomb force of attraction.
(2) Only orbits where angular momentum \(L = mvr = \frac{nh}{2\pi}\) (n = 1, 2, 3, ...) are allowed.
(3) Energy is radiated only when an electron jumps from a higher to a lower orbit. The photon energy equals the energy difference: \(h\nu = E_m - E_n\).

Derivation of energy expression:
Consider an electron of mass m and charge e revolving in the nth orbit of radius r around a nucleus of atomic number Z.

By Bohr's first postulate (centripetal force = Coulomb force):
\[\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{Ze^2}{r^2} \quad \text{...(1)}\]

Kinetic energy (KE):
\[KE = \frac{1}{2}mv^2 = \frac{Ze^2}{8\pi\varepsilon_0 r} \quad \text{...(2)}\]

Potential energy (PE) of electron:
\[PE = -e \times \frac{1}{4\pi\varepsilon_0}\cdot\frac{Ze}{r} = -\frac{Ze^2}{4\pi\varepsilon_0 r} \quad \text{...(3)}\]

Total energy:
\[E = KE + PE = \frac{Ze^2}{8\pi\varepsilon_0 r} - \frac{Ze^2}{4\pi\varepsilon_0 r} = -\frac{Ze^2}{8\pi\varepsilon_0 r} \quad \text{...(4)}\]

The radius of the nth orbit (from quantization condition):
\[r_n = \frac{\varepsilon_0 h^2 n^2}{\pi m Z e^2} \quad \text{...(5)}\]

Substituting (5) in (4):
\[E_n = -\frac{mZ^2e^4}{8\varepsilon_0^2 h^2 n^2} \quad \text{...(6)}\]

For hydrogen (Z = 1):
\[E_n \approx -\frac{13.6}{n^2} \text{ eV}\]

The negative sign shows that the electron is bound to the nucleus. Energy is discrete and inversely proportional to \(n^2\).
In simple words: The energy of an electron in orbit n is \(-13.6/n^2\) eV for hydrogen. The negative sign means the electron is trapped (bound). Higher orbits (larger n) have less negative energy - the electron is less tightly held.

Teacher's Note: The derivation has four key steps: (1) equate centripetal and Coulomb forces, (2) find KE, (3) find PE, (4) add to get total E, then substitute the orbit radius. Many students make sign errors with PE - emphasize PE is negative because opposite charges attract.

Exam Tip: For 8 marks, write: postulates (3 marks), then the derivation in four clear steps with equations numbered (5 marks). Always substitute Z=1 at the end and state \(E_n = -13.6/n^2\) eV for hydrogen.

Question 4. Starting from the formula for energy of an electron in the nth orbit of hydrogen atom, derive the formula for the wavelengths of Lyman and Balmer series spectral lines and determine the shortest wavelengths of lines in both these series.
Answer:
The energy of an electron in the nth orbit of hydrogen:
\[E_n = -\frac{me^4}{8\varepsilon_0^2 h^2 n^2}\]

When an electron jumps from orbit m to orbit n (m > n), energy released:
\[E_m - E_n = \frac{me^4}{8\varepsilon_0^2 h^2}\left(\frac{1}{n^2} - \frac{1}{m^2}\right) = h\nu\]

Wave number:
\[\bar{\nu} = \frac{1}{\lambda} = \frac{me^4}{8\varepsilon_0^2 h^3 c}\left(\frac{1}{n^2} - \frac{1}{m^2}\right) = R\left(\frac{1}{n^2} - \frac{1}{m^2}\right)\]
where \(R = \frac{me^4}{8\varepsilon_0^2 h^3 c}\) is the Rydberg constant.

Lyman series (n = 1, m = 2, 3, 4, ... \(\infty\)):
\[\frac{1}{\lambda_L} = R\left(\frac{1}{1^2} - \frac{1}{m^2}\right)\]
Shortest wavelength (m = \(\infty\)):
\[\frac{1}{\lambda_{Ls}} = R\left(\frac{1}{1^2}\right) = R \quad \Rightarrow \quad \lambda_{Ls} = \frac{1}{R} = 912 \text{ Å}\]

Balmer series (n = 2, m = 3, 4, 5, ... \(\infty\)):
\[\frac{1}{\lambda_B} = R\left(\frac{1}{4} - \frac{1}{m^2}\right)\]
Shortest wavelength (m = \(\infty\)):
\[\frac{1}{\lambda_{Bs}} = \frac{R}{4} \quad \Rightarrow \quad \lambda_{Bs} = \frac{4}{R} = 4 \times 912 = 3648 \text{ Å}\]
In simple words: Lyman series (UV light) has n=1 as the lower level. Balmer series (visible light) has n=2. The shortest wavelength in each series comes when the electron falls from very far away (m = infinity), giving maximum energy.

Teacher's Note: Draw the energy level diagram and mark both series. Students should understand that shortest wavelength = largest energy = electron falling from infinity. The Balmer series is in visible light - this is what Balmer originally observed with a prism.

Exam Tip: State the Rydberg formula, substitute n=1 for Lyman and n=2 for Balmer. For shortest wavelength, set m = \(\infty\) so \(1/m^2 = 0\). Always express the final answer in Ångströms.

Question 5. Determine the maximum angular speed of an electron moving in a stable orbit around the nucleus of hydrogen atom.
Answer:
The radius of the nth Bohr orbit:
\[r_n = \frac{\varepsilon_0 h^2 n^2}{\pi m Z e^2} \quad \text{...(1)}\]

The linear speed of an electron:
\[v = \frac{Ze^2}{2\varepsilon_0 n h} \quad \text{...(2)}\]

Angular speed \(\omega = v/r\):
\[\omega = \frac{v}{r} = \frac{Ze^2}{2\varepsilon_0 nh} \cdot \frac{\pi m Ze^2}{\varepsilon_0 h^2 n^2} = \frac{\pi m Z^2 e^4}{2\varepsilon_0^2 h^3 n^3}\]

Angular speed is maximum when n is minimum, i.e., n = 1. For hydrogen (Z = 1):
\[\omega_{max} = \frac{\pi m e^4}{2\varepsilon_0^2 h^3}\]

Substituting values:
\[\omega_{max} = \frac{(3.142)(9.1 \times 10^{-31})(1.6 \times 10^{-19})^4}{(2)(8.85 \times 10^{-12})^2(6.63 \times 10^{-34})^3}\]
\[\omega_{max} = 4.105 \times 10^{16} \text{ rad/s}\]
In simple words: The electron moves fastest in the innermost orbit (n=1). Maximum angular speed occurs at n=1 in hydrogen because \(\omega \propto 1/n^3\).

Teacher's Note: Guide students through the substitution step by step - the power of 10 arithmetic is where most errors occur. Note that frequency of revolution \(f = \omega/2\pi \approx 6.6 \times 10^{15}\) Hz - faster than visible light oscillations!

Exam Tip: Derive the expression for \(\omega\) from the formulas for v and r. Then substitute n=1 and Z=1. Show the numerical substitution clearly with all powers of 10. State the answer as \(4.105 \times 10^{16}\) rad/s.

Question 6. Determine the series limit of Balmer, Paschen and Bracket series, given the limit for Lyman series is 912 Å.
Answer:
Data: \(\lambda_{L\infty} = 912\) Å

For hydrogen spectrum: \(\frac{1}{\lambda} = R_H\left(\frac{1}{n^2} - \frac{1}{m^2}\right)\)

Lyman limit (n=1, m=\(\infty\)): \(\frac{1}{\lambda_{L\infty}} = R_H\) ...(1)

Balmer limit (n=2, m=\(\infty\)): \(\frac{1}{\lambda_{B\infty}} = \frac{R_H}{4}\) ...(2)

From (1) and (2): \(\frac{\lambda_{B\infty}}{\lambda_{L\infty}} = 4\)
\[\lambda_{B\infty} = 4 \times 912 = 3648 \text{ Å}\]

Paschen limit (n=3, m=\(\infty\)): \(\frac{1}{\lambda_{Pa\infty}} = \frac{R_H}{9}\)
\[\frac{\lambda_{Pa\infty}}{\lambda_{L\infty}} = 9 \quad \Rightarrow \quad \lambda_{Pa\infty} = 9 \times 912 = 8208 \text{ Å}\]

Brackett limit (n=4, m=\(\infty\)): \(\frac{1}{\lambda_{Br\infty}} = \frac{R_H}{16}\)
\[\lambda_{Br\infty} = 16 \times 912 = 14592 \text{ Å}\]
In simple words: The series limit wavelength is proportional to \(n^2\). So Balmer limit = 4 × Lyman limit, Paschen = 9 × Lyman limit, Brackett = 16 × Lyman limit. All series limits are just multiples of 912 Å.

Teacher's Note: The pattern \(\lambda_{series} = n^2 \times \lambda_{Lyman}\) is a powerful shortcut. Make students derive it once, then use the pattern. Also note that higher series (Paschen, Brackett) are in infrared - not visible.

Exam Tip: Write the general formula for each series limit, take the ratio with the Lyman limit, and multiply by 912 Å. State the result for each series clearly: Balmer = 3648 Å, Paschen = 8208 Å, Brackett = 14592 Å.

Question 7. Describe alpha, beta and gamma decays and write down the formulae for the energies generated in each of these decays.
Answer:
(a) Alpha decay: A radioactive transformation in which an \(\alpha\)-particle \((_2^4\text{He})\) is emitted. Atomic number decreases by 2, mass number decreases by 4.
General form: \(_Z^A\text{X} \rightarrow _{Z-2}^{A-4}\text{Y} + _2^4\alpha + \text{energy released}\)
Example: \(_{92}^{238}\text{U} \rightarrow _{90}^{234}\text{Th} + _2^4\alpha\)
Energy: \(Q = [m_U - m_{Th} - m_\alpha]c^2\)

(b) Beta-minus decay (\(\beta^-\)): Atomic number increases by 1, mass number unchanged. An electron and antineutrino are emitted.
General form: \(_Z^A\text{X} \rightarrow _{Z+1}^A\text{Y} + _{-1}^0\beta + \bar{\nu}_e + \text{energy released}\)
Example: \(_{90}^{234}\text{Th} \rightarrow _{91}^{234}\text{Pa} + _{-1}^0e + \bar{\nu}_e\)
Energy: \(Q = [m_{Th} - m_{Pa} - m_e]c^2\)

Beta-plus decay (\(\beta^+\)): Atomic number decreases by 1, mass number unchanged.
General form: \(_Z^A\text{X} \rightarrow _{Z-1}^A\text{Y} + _{+1}^0\beta + \nu_e + \text{energy released}\)
Example: \(_{15}^{30}\text{P} \rightarrow _{14}^{30}\text{Si} + _{+1}^0e + \nu_e\)
Energy: \(Q = [m_P - m_{Si} - m_e]c^2\)

(c) Gamma decay: On emission of \(\alpha\) or \(\beta\)-particles, the nucleus is often left in an excited state. It emits a \(\gamma\)-ray photon to reach the ground state. Neither Z nor A changes.
General form: \(_Z^A\text{X}^* \rightarrow _Z^A\text{X} + \gamma\)
Energy: \(Q = E_{excited} - E_{ground}\)
In simple words: Think of nuclear decay like a person changing identity (alpha/beta) or just calming down (gamma). Alpha: loses 2 protons + 2 neutrons. Beta: converts neutron↔proton. Gamma: releases leftover excitement energy as a high-energy photon.

Teacher's Note: Connect each decay to conservation laws: alpha conserves Z+N (baryon number), beta conserves lepton number (hence the neutrino). Gamma conserves everything - just releases energy. Use the general equations on the board with variables before specific examples.

Exam Tip: For each decay: (i) state which particle is emitted, (ii) state changes in Z and A, (iii) write the general equation, (iv) give the specific example, (v) write the Q-value formula. This structured approach ensures full marks.

Question 8. Explain what are nuclear fission and fusion giving an example of each. Write down the formulae for energy generated in each of these processes.
Answer:
Nuclear Fission:
Nuclear fission is a nuclear reaction in which a heavy nucleus splits into two or more fragments of comparable size, either spontaneously or due to neutron bombardment (induced fission), followed by emission of 2-3 neutrons.

The mass of the original nucleus is greater than the sum of masses of fragments. This mass defect is released as enormous energy.

Typical fission equations:
(1) \(_{92}^{235}\text{U} + _0^1\text{n} \rightarrow _{92}^{236}\text{U} \rightarrow _{54}^{140}\text{Xe} + _{38}^{94}\text{Sr} + 2_0^1\text{n} + 200 \text{ MeV}\)
(2) \(_{92}^{235}\text{U} + _0^1\text{n} \rightarrow _{56}^{144}\text{Ba} + _{36}^{90}\text{Kr} + 2_0^1\text{n} + 200 \text{ MeV}\)
(3) \(_{92}^{235}\text{U} + _0^1\text{n} \rightarrow _{57}^{148}\text{La} + _{35}^{85}\text{Br} + 3_0^1\text{n} + 200 \text{ MeV}\)

Energy: \(Q = [m_{U-235} + m_n - m_{fragments} - n \cdot m_n]c^2\)

Nuclear Fusion:
Nuclear fusion is a reaction in which lighter nuclei (low atomic number) fuse to form a heavier nucleus, releasing enormous energy. Very high temperatures (10\(^7\) to 10\(^8\) K) are required - hence also called a thermonuclear reaction.

Example (D-T reaction):
\[_1^2\text{D} + _1^3\text{T} \rightarrow _2^4\text{He} + _0^1\text{n} + 17.6 \text{ MeV}\]
(deuteron + triton → helium nucleus + neutron)

Energy: \(Q = [m_D + m_T - m_{He} - m_n]c^2\)
In simple words: Fission splits heavy atoms like uranium - like breaking a large rock into pieces. Fusion joins light atoms like hydrogen - like pressing small magnets together. Both release huge energy from mass converted to energy (E=mc²). The sun uses fusion!

Teacher's Note: Compare fission and fusion using a table: type of reaction (splitting vs. joining), type of fuel (heavy vs. light nuclei), temperature needed, energy per reaction, and applications (nuclear power vs. H-bomb, future reactors). Note that fusion gives more energy per kg of fuel but is much harder to control.

Exam Tip: Define each process, give one specific equation with masses balancing (check proton and neutron numbers), write the Q-value formula, and state the energy released. For fission: ~200 MeV per reaction. For D-T fusion: 17.6 MeV.

Question 9. Describe the principles of a nuclear reactor. What is the difference between a nuclear reactor and a nuclear bomb?
Answer:
Principles of a Nuclear Reactor:
A nuclear reactor uses controlled nuclear fission chain reactions to generate heat, which produces electricity.

Key components:
(1) Fuel rods: Provide fissionable material such as \(_{92}^{235}\text{U}\) or \(_{92}^{238}\text{U}\) enriched with \(^{235}\text{U}\).
(2) Control rods: Made of boron or cadmium; absorb neutrons to control or stop the chain reaction.
(3) Moderator: Slows down fast neutrons to thermal (slow) neutrons needed for fission. Common moderators: heavy water (D\(_2\)O), graphite, ordinary water.
(4) Coolant: Removes heat generated by fission (water, heavy water, CO\(_2\)) and transfers it to produce steam for electricity generation.

Difference between Nuclear Reactor and Nuclear Bomb:
In a nuclear reactor, the fission chain reaction is controlled - the number of neutrons is regulated to maintain a steady rate of fission.
In a nuclear bomb, the fission chain reaction is uncontrolled - the number of neutrons multiplies explosively, releasing tremendous energy in a fraction of a second.
In simple words: A nuclear reactor is like a controlled campfire - you regulate the flames for steady heat. A nuclear bomb is like an explosion - uncontrolled and instantaneous. The key difference is control rods: reactors have them, bombs don't.

Teacher's Note: The concept of critical mass is important here - in a reactor, the fission is kept just at or below criticality. In a bomb, supercriticality is achieved instantly. Connect to current events: India's nuclear power plants (Tarapur, Kudankulam) use reactor principles.

Exam Tip: Name and explain all four components of a reactor. For the comparison, use a two-column table or clearly state: reactor = controlled chain reaction; bomb = uncontrolled chain reaction. Mention the role of control rods explicitly.

Question 10. Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.
Answer:
Data: \(M_\alpha = 4.00151\) u, \(m_p = 1.00728\) u, \(m_n = 1.00866\) u, 1 u = 931.5 MeV/c\(^2\)

Alpha particle has Z = 2 (protons) and N = 2 (neutrons).

Binding energy:
\[E_B = (Zm_p + Nm_n - M_\alpha)c^2\]
\[= (2 \times 1.00728 + 2 \times 1.00866 - 4.00151) \times 931.5 \text{ MeV}\]
\[= (2.01456 + 2.01732 - 4.00151) \times 931.5\]
\[= 0.03037 \times 931.5 = 28.29 \text{ MeV}\]

Converting: \(28.29 \times 10^6 \times 1.602 \times 10^{-19} = 4.531 \times 10^{-12}\) J
In simple words: The alpha particle is lighter than 2 protons + 2 neutrons added separately. The "missing" mass (0.03037 u) has been converted into binding energy - the glue holding the nucleus together. This energy is 28.29 MeV.

Teacher's Note: This is a standard numerical - practice it until students can do it confidently. The mass defect \(\Delta m = 0.03037\) u is key. Multiply by 931.5 MeV/u for energy in MeV, or by \(1.66 \times 10^{-27} \times (3\times10^8)^2\) for energy in joules.

Exam Tip: Write: (1) Z and N of alpha, (2) the binding energy formula, (3) substitute masses, (4) calculate mass defect, (5) multiply by 931.5 MeV/u, (6) convert to joules if asked. Show all intermediate steps.

Question 11. An electron in hydrogen atom stays in its second orbit for \(10^{-8}\) s. How many revolutions will it make around the nucleus in that time?
Answer:
Data: Z = 1, n = 2, t = \(10^{-8}\) s
\(m = 9.1 \times 10^{-31}\) kg, \(e = 1.6 \times 10^{-19}\) C
\(\varepsilon_0 = 8.85 \times 10^{-12}\) C\(^2\)/N·m\(^2\), \(h = 6.63 \times 10^{-34}\) J·s

Period of revolution in nth orbit:
\[T = \frac{4\varepsilon_0^2 h^3 n^3}{\pi m e^4}\]

For n = 2:
\[T = \frac{(4)(8.85 \times 10^{-12})^2(6.63 \times 10^{-34})^3(8)}{\pi(9.1 \times 10^{-31})(1.6 \times 10^{-19})^4}\]
\[T = 3.898 \times 10^{-16} \text{ s}\]

Number of revolutions:
\[N = \frac{t}{T} = \frac{10^{-8}}{3.898 \times 10^{-16}} = 2.565 \times 10^7\]
In simple words: The electron orbits about 25 million times per microsecond! Even in the "slow" second orbit, the electron completes over 25 million revolutions in just 10 nanoseconds - showing how incredibly fast atomic-scale motion is.

Teacher's Note: The key formula \(T = 4\varepsilon_0^2 h^3 n^3 / (\pi m e^4)\) should be derived once in class (from v and r formulas), then used directly. Emphasize that T \(\propto n^3\), so higher orbits have much longer periods.

Exam Tip: Write the formula for period T, substitute n=2, calculate T carefully (show powers of 10 separately), then divide t by T. Final answer: \(N = 2.565 \times 10^7\) revolutions.

Question 12. Determine the binding energy per nucleon of the americium isotope \(_{95}^{244}\text{Am}\), given the mass of \(_{95}^{244}\text{Am}\) to be 244.06428 u.
Answer:
Data: Z = 95, A = 244, N = 244 - 95 = 149
\(m_p = 1.00728\) u, \(m_n = 1.00866\) u
\(M = 244.06428\) u, 1 u = 931.5 MeV/c\(^2\)

\[\frac{E_B}{A} = \frac{(Zm_p + Nm_n - M)c^2}{A}\]
\[= \frac{[95(1.00728) + 149(1.00866) - 244.06428] \times 931.5}{244}\]
\[= \frac{[95.6916 + 150.2901 - 244.06428] \times 931.5}{244}\]
\[= \frac{1.91742 \times 931.5}{244} = \frac{1786.1}{244}\]
\[= 7.3209 \text{ MeV/nucleon}\]
In simple words: Each nucleon in americium-244 is held in the nucleus with about 7.32 MeV of energy. This is typical for heavy nuclei - most stable nuclei have binding energy per nucleon around 7-9 MeV, with iron-56 having the highest at ~8.8 MeV.

Teacher's Note: This calculation is identical in structure to Question 10 but for a heavy nucleus. Guide students to identify Z, N, and A correctly first. Americium is a synthetic element - its existence illustrates that the periodic table extends beyond uranium through nuclear reactions.

Exam Tip: Step 1: Find Z, N, A. Step 2: Calculate \(Zm_p + Nm_n\). Step 3: Subtract M. Step 4: Multiply by 931.5 MeV. Step 5: Divide by A (244). Answer: 7.3209 MeV/nucleon.

Question 13. Calculate the energy released in the nuclear reaction \(_3^7\text{Li} + p \rightarrow 2\alpha\) given mass of \(_3^7\text{Li}\) atom and of helium atom to be 7.016 u and 4.0026 u respectively.
Answer:
Data: \(M_1\)(Li atom) = 7.016 u, \(M_2\)(He atom) = 4.0026 u
\(m_p = 1.00728\) u, 1 u = 931.5 MeV/c\(^2\)

Mass defect:
\[\Delta M = M_1 + m_p - 2M_2\]
\[= [7.016 + 1.00728 - 2(4.0026)] \text{ u}\]
\[= [8.02328 - 8.0052] = 0.01808 \text{ u}\]

Energy released:
\[Q = \Delta M \times c^2 = 0.01808 \times 931.5 = 16.84 \text{ MeV}\]
In simple words: When lithium-7 captures a proton and breaks into two alpha particles, 16.84 MeV of energy is released. This is the reaction Cockroft and Walton used in 1932 to first artificially split the atom!

Teacher's Note: This is a historically important reaction - the first artificially induced nuclear reaction. Note that the products (2 alpha particles) are more stable than the reactants, which is why energy is released. Mass defect = reactant mass − product mass.

Exam Tip: Write the balanced nuclear equation first. Then: \(\Delta M\) = (total reactant mass) - (total product mass). Multiply by 931.5 MeV/u. If \(\Delta M > 0\), energy is released (exothermic).

Question 14. Complete the following equations describing nuclear decays.
Answer:
(a) \(_{86}^{226}\text{Ra} \rightarrow _2^4\alpha + _{86}^{222}\text{Em}\)
(Em = Emanation = Rn, Radon)
Here, an \(\alpha\)-particle is emitted and radon is formed.

(b) \(_8^{19}\text{O} \rightarrow _{-1}^0e^- + _9^{19}\text{F}\)
Here, \(e^- \equiv _{-1}^0\beta\) is emitted and fluorine is formed. (Beta-minus decay)

(c) \(_{90}^{228}\text{Th} \rightarrow _2^4\alpha + _{88}^{224}\text{Ra}\)
Here, an \(\alpha\)-particle is emitted and radium is formed.

(d) \(_7^{12}\text{N} \rightarrow _6^{12}\text{C} + _{+1}^0\beta\)
\(_{+1}^0\beta\) is \(e^+\) (positron). Here, \(\beta^+\) is emitted and carbon is formed.
In simple words: For alpha decay: Z decreases by 2, A decreases by 4. For beta-minus: Z increases by 1, A unchanged. For beta-plus: Z decreases by 1, A unchanged. Always check that both mass numbers and atomic numbers balance on both sides.

Teacher's Note: Teach students to approach these systematically: (1) identify decay type from the missing particle or product, (2) apply conservation of mass number and atomic number, (3) find the missing element using periodic table. Conservation is the key skill here.

Exam Tip: Always verify: sum of mass numbers on left = sum on right; sum of atomic numbers on left = sum on right. Write conservation equations explicitly if you are unsure. Name the product element - don't just write numbers.

Question 15. Calculate the energy released in the following reactions.
Answer:
Given masses: \(_{88}^{223}\text{Ra}\): 223.0185 u, \(_{82}^{209}\text{Pb}\): 208.9811 u, \(_6^{14}\text{C}\): 14.00324 u, \(_{92}^{236}\text{U}\): 236.0456 u, \(_{56}^{140}\text{Ba}\): 139.9106 u, \(_{36}^{94}\text{Kr}\): 93.9341 u, \(_6^{11}\text{C}\): 11.01143 u, \(_5^{11}\text{B}\): 11.0093 u, \(m_n = 1.00866\) u, \(m(e^+) = 0.00055\) u, 1 u = 931.5 MeV/c\(^2\).

(a) \(_{88}^{223}\text{Ra} \rightarrow _{82}^{209}\text{Pb} + _6^{14}\text{C}\)
\[\Delta M = 223.0185 - (208.9811 + 14.00324) = 0.03416 \text{ u}\]
\[Q = 0.03416 \times 931.5 = 31.82 \text{ MeV}\]

(b) \(_{92}^{236}\text{U} \rightarrow _{56}^{140}\text{Ba} + _{36}^{94}\text{Kr} + 2_0^1\text{n}\)
\[\Delta M = 236.0456 - (139.9106 + 93.9341 + 2 \times 1.00866) = 0.18368 \text{ u}\]
\[Q = 0.18368 \times 931.5 = 171.1 \text{ MeV}\]

(c) \(_6^{11}\text{C} \rightarrow _5^{11}\text{B} + e^+ + \nu_e\)
\[\Delta M = 11.01143 - (11.0093 + 0.00055) = 0.00158 \text{ u}\]
\[Q = 0.00158 \times 931.5 = 1.472 \text{ MeV}\]
In simple words: In each reaction, the products are lighter than the reactants. The "lost" mass converts to energy via \(E = \Delta mc^2\). Reaction (b) releases the most energy (171 MeV) - this is typical nuclear fission.

Teacher's Note: Reaction (a) is an unusual cluster decay (Ra emitting C-14). Reaction (b) is standard uranium fission. Reaction (c) is beta-plus decay. Using the same formula \(Q = \Delta M \times 931.5\) for all three shows the universality of mass-energy equivalence.

Exam Tip: For each part: write \(\Delta M\) = (reactant masses) - (product masses), then \(Q = \Delta M \times 931.5\) MeV. Show the subtraction step explicitly. Be careful to include all neutrons in (b).

Question 16. Sample of carbon obtained from any living organism has a decay rate of 15.3 decays per gram per minute. A sample of carbon obtained from very old charcoal shows a disintegration rate of 12.3 disintegrations per gram per minute. Determine the age of the old sample given the decay constant of carbon to be \(3.839 \times 10^{-12}\) per second.
Answer:
Data: \(A_0 = 15.3\) dis/g/min, \(A(t) = 12.3\) dis/g/min
\(\lambda = 3.839 \times 10^{-12}\) per second

\[A(t) = A_0 e^{-\lambda t}\]
\[\therefore e^{\lambda t} = \frac{A_0}{A(t)} = \frac{15.3}{12.3}\]
\[\therefore \lambda t = \log_e\left(\frac{15.3}{12.3}\right) = \frac{2.303}{\lambda}\log_{10}\left(\frac{15.3}{12.3}\right)\]
\[t = \frac{2.303}{3.839 \times 10^{-12}} \times \log_{10}(1.2439)\]
\[= \frac{2.303 \times 0.0948}{3.839 \times 10^{-12}} = \frac{0.2183}{3.839 \times 10^{-12}}\]
\[= 5.687 \times 10^{10} \text{ s} = \frac{5.687 \times 10^{10}}{3.156 \times 10^7} \approx 1802 \text{ years}\]
In simple words: Carbon-14 dating works because living organisms absorb C-14 from the air. When they die, C-14 decays without replacement. By comparing the decay rate of old charcoal (12.3) to living wood (15.3), we can calculate the sample is about 1802 years old.

Teacher's Note: This is radiocarbon dating - one of the most powerful archaeological tools. Explain that \(A_0\) is the activity when the organism was alive (same as modern living organisms since C-14 atmospheric ratio is approximately constant). Willard Libby won the Nobel Prize for this method in 1960.

Exam Tip: The key formula is \(t = \frac{2.303}{\lambda}\log_{10}\left(\frac{A_0}{A}\right)\). Substitute and calculate \(\log_{10}(15.3/12.3) = \log_{10}(1.2439) = 0.0948\). Convert seconds to years by dividing by \(3.156 \times 10^7\).

Question 17. The half-life of \(_{38}^{90}\text{Sr}\) is 28 years. Determine the disintegration rate of its 5 mg sample.
Answer:
Data: \(T_{1/2} = 28\) years \(= 28 \times 3.156 \times 10^7 = 8.837 \times 10^8\) s
Mass M = 5 mg = \(5 \times 10^{-3}\) g

Number of atoms N:
\[N = \frac{6.02 \times 10^{23} \times 5 \times 10^{-3}}{90} = 3.344 \times 10^{19} \text{ atoms}\]

Decay constant:
\[\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{8.837 \times 10^8} = 7.843 \times 10^{-10} \text{ s}^{-1}\]

Disintegration rate:
\[A = N\lambda = 3.344 \times 10^{19} \times 7.843 \times 10^{-10} = 2.622 \times 10^{10} \text{ disintegrations/s}\]
In simple words: Even a tiny 5 mg sample of strontium-90 undergoes over 26 billion nuclear decays every second! This is why radioactive contamination from nuclear fallout is so dangerous - strontium-90 mimics calcium and gets deposited in bones.

Teacher's Note: Strontium-90 is a major concern in nuclear fallout (like Chernobyl) because it replaces calcium in bones and irradiates bone marrow. This problem illustrates why even small amounts of long-lived radioisotopes are dangerous.

Exam Tip: Three steps: (1) find N using Avogadro's number, (2) find \(\lambda = 0.693/T_{1/2}\) (convert years to seconds), (3) multiply \(A = N\lambda\). Answer: \(2.622 \times 10^{10}\) dis/s.

Question 18. What is the amount of \(_{27}^{60}\text{Co}\) necessary to provide a radioactive source of strength 10.0 mCi, its half-life being 5.3 years?
Answer:
Data: Activity = 10.0 mCi = \(10 \times 10^{-3} \times 3.7 \times 10^{10} = 3.7 \times 10^8\) dis/s
\(T_{1/2} = 5.3 \times 3.156 \times 10^7 = 1.673 \times 10^8\) s

\[\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{1.673 \times 10^8} = 4.142 \times 10^{-9} \text{ s}^{-1}\]

Number of atoms required:
\[N = \frac{A}{\lambda} = \frac{3.7 \times 10^8}{4.142 \times 10^{-9}} = 8.933 \times 10^{16} \text{ atoms}\]

Mass:
\[m = \frac{8.933 \times 10^{16}}{6.02 \times 10^{23}} \times 60 = 8.903 \times 10^{-6} \text{ g} = 8.903 \text{ μg}\]
In simple words: Only 8.9 micrograms (less than a speck of dust) of cobalt-60 provides 10 mCi of radioactivity - enough for cancer radiation therapy. This shows how intensely radioactive Co-60 is due to its relatively short half-life.

Teacher's Note: Cobalt-60 is widely used in cancer radiotherapy (gamma knife surgery) and food irradiation. The tiny mass needed (8.9 μg) for a clinically useful source illustrates the intense radioactivity of short-lived isotopes. Connect to the inverse relationship: shorter half-life = higher activity for same amount.

Exam Tip: Convert mCi to dis/s (1 Ci = \(3.7 \times 10^{10}\) dis/s). Find \(\lambda\), then \(N = A/\lambda\), then mass = \(N \times 60 / (6.02 \times 10^{23})\). Answer: 8.903 μg.

Question 19. Disintegration rate of a sample is \(10^{10}\) per hour at 20 hrs from the start. It reduces to \(6.3 \times 10^9\) per hour after 30 hours. Calculate its half life and the initial number of radioactive atoms in the sample.
Answer:
Data: \(A(t_1) = 10^{10}\)/hr at \(t_1 = 20\) h; \(A(t_2) = 6.3 \times 10^9\)/hr at \(t_2 = 30\) h

\[\frac{A(t_1)}{A(t_2)} = e^{\lambda(t_2-t_1)} = e^{10\lambda}\]
\[\frac{10^{10}}{6.3 \times 10^9} = 1.587 = e^{10\lambda}\]
\[10\lambda = \ln(1.587) = 2.303\log_{10}(1.587) = 2.303 \times 0.2007 = 0.4622\]
\[\lambda = 0.04622 \text{ per hour}\]

Half-life:
\[T_{1/2} = \frac{0.693}{\lambda} = \frac{0.693}{0.04622} = 14.99 \approx 15 \text{ hours}\]

Initial activity \(A_0\) (at t = 0):
\[A_0 = A(t_1) \times e^{\lambda t_1} = 10^{10} \times e^{0.04622 \times 20} = 10^{10} \times e^{0.9244}\]
\[e^{0.9244} = \text{antilog}(0.9244/2.303) = \text{antilog}(0.4014) = 2.52\]
\[A_0 = 2.52 \times 10^{10} \text{ per hour}\]

Initial number of atoms:
\[N_0 = \frac{A_0}{\lambda} = \frac{2.52 \times 10^{10}}{0.04622} = 5.452 \times 10^{11}\]
In simple words: By measuring how quickly the activity drops between two time points (from \(10^{10}\) to \(6.3 \times 10^9\) in 10 hours), we can calculate the half-life is 15 hours. Then working backwards, we find there were about 545 billion atoms at the start.

Teacher's Note: This problem requires working with two activity measurements to find \(\lambda\) first, then \(T_{1/2}\), then back-calculate to t=0. Students often get confused about direction of time - emphasize that \(A_0 > A(t_1) > A(t_2)\) since activity decreases with time.

Exam Tip: Key step: take the ratio \(A(t_1)/A(t_2) = e^{\lambda(t_2-t_1)}\). This eliminates \(A_0\) and gives \(\lambda\) directly. Then find \(T_{1/2} = 0.693/\lambda\), then \(N_0 = A_0/\lambda\).

Question 20. The isotope \(_{27}^{57}\text{Co}\) decays by electron capture to \(_{26}^{57}\text{Fe}\) with a half-life of 272 d. The Fe nucleus is produced in an excited state, and it almost instantaneously emits gamma rays.
Answer:
Data: \(T_{1/2} = 272\) d = \(272 \times 24 \times 3600 = 2.35 \times 10^7\) s
Activity \(A_0 = 2.0\) μCi = \(2.0 \times 10^{-6} \times 3.7 \times 10^{10} = 7.4 \times 10^4\) dis/s
t = 1 year = \(3.156 \times 10^7\) s

(a) Mean lifetime and decay constant:
\[\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{2.35 \times 10^7} = 2.949 \times 10^{-8} \text{ s}^{-1}\]
\[\tau = \frac{1}{\lambda} = \frac{T_{1/2}}{0.693} = \frac{2.35 \times 10^7}{0.693} = 3.391 \times 10^7 \text{ s}\]

(b) Number of \(^{57}\text{Co}\) nuclei:
\[N_0 = \frac{A_0}{\lambda} = A_0 \times \tau = 7.4 \times 10^4 \times 3.391 \times 10^7 = 2.509 \times 10^{12} \text{ nuclei}\]

(c) Activity after one year:
\[A(t) = A_0 e^{-\lambda t} = 2.0 \times e^{-(2.949 \times 10^{-8})(3.156 \times 10^7)}\]
\[= 2.0 \times e^{-0.9307} = \frac{2.0}{e^{0.9307}} = \frac{2.0}{2.536} = 0.7886 \text{ μCi}\]
In simple words: Co-57 is used in medical imaging (Mössbauer spectroscopy). After one year (about 1.3 half-lives), the activity drops from 2.0 μCi to about 0.79 μCi - reduced to about 39% of the original. The mean lifetime (39 million seconds ≈ 1.25 years) is longer than the half-life because some nuclei survive much longer than average.

Teacher's Note: Clarify the difference between \(T_{1/2}\) and mean lifetime \(\tau\): \(\tau = T_{1/2}/0.693 = 1.44 T_{1/2}\). Mean lifetime is always longer than half-life. Also note that \(N_0 = A_0 \tau\) is a useful shortcut.

Exam Tip: For part (a): \(\lambda = 0.693/T_{1/2}\), \(\tau = 1/\lambda\). For (b): \(N = A/\lambda\). For (c): use \(A = A_0 e^{-\lambda t}\), converting t to seconds. Express final answer in μCi.

Question 21. A source contains two species of phosphorous nuclei, \(_{15}^{32}\text{P}\) (\(T_{1/2} = 14.3\) d) and \(_{15}^{33}\text{P}\) (\(T_{1/2} = 25.3\) d). At time t = 0, 90% of the decays are from \(_{15}^{32}\text{P}\). How much time has to elapse for only 15% of the decays to be from \(_{15}^{32}\text{P}\)?
Answer:
Decay constants:
\[\lambda_1 = \frac{0.693}{14.3} = 0.04846 \text{ d}^{-1} \quad (\text{for }^{32}\text{P})\]
\[\lambda_2 = \frac{0.693}{25.3} = 0.02739 \text{ d}^{-1} \quad (\text{for }^{33}\text{P})\]

At t = 0, 90% from \(^{32}\text{P}\), 10% from \(^{33}\text{P}\):
\[\frac{N_{01}\lambda_1}{N_{02}\lambda_2} = \frac{90\%}{10\%} = 9 \quad \text{...(1)}\]

At time t, 15% from \(^{32}\text{P}\), 85% from \(^{33}\text{P}\):
\[\frac{N_{01}\lambda_1 e^{-\lambda_1 t}}{N_{02}\lambda_2 e^{-\lambda_2 t}} = \frac{15\%}{85\%} = \frac{3}{17} \quad \text{...(2)}\]

Dividing (1) by (2):
\[\frac{e^{(\lambda_1 - \lambda_2)t}}{9} = \frac{17}{3} \quad \Rightarrow \quad e^{(\lambda_1 - \lambda_2)t} = \frac{153}{3} = 51\]

Wait - let me redo: divide (2) by (1):
\[e^{-(\lambda_1 - \lambda_2)t} = \frac{3/17}{9} = \frac{3}{153} = \frac{1}{51}\]
\[\therefore (\lambda_1 - \lambda_2)t = \ln(51) = 2.303\log_{10}(51) = 2.303 \times 1.7076\]
\[(0.04846 - 0.02739)t = 2.303 \times 1.7076\]
\[0.02107 \times t = 3.931\]
\[t = \frac{3.931}{0.02107} = 186.6 \text{ days}\]
In simple words: Since \(^{32}\text{P}\) decays faster (shorter half-life), its fraction of the total activity decreases over time. After about 187 days, \(^{32}\text{P}\) contributes only 15% of the total decays instead of the original 90%.

Teacher's Note: The key insight is setting up the ratio equation at two different times, then dividing to eliminate the initial amounts. This technique works whenever two competing processes have different rates - common in nuclear physics, chemistry kinetics, and biology.

Exam Tip: Set up ratio equations at t=0 and t, divide to eliminate \(N_0\) terms, then solve for t using logarithms. The answer t = 186.6 days is approximately 6.6 months.

Question 22. Before the year 1900 the activity per unit mass of atmospheric carbon due to the presence of \(^{14}\text{C}\) averaged about 0.255 Bq per gram of carbon. (a) What fraction of carbon atoms were \(^{14}\text{C}\)? (b) An archaeological specimen containing 500 mg of carbon shows 174 decays in one hour. What is the age of the specimen? Half-life of \(^{14}\text{C}\) is 5730 years.
Answer:
Data: \(T_{1/2} = 5730\) y = \(5730 \times 3.156 \times 10^7 = 1.808 \times 10^{11}\) s
\[\lambda = \frac{0.693}{1.808 \times 10^{11}} = 3.832 \times 10^{-12} \text{ s}^{-1}\]

(a) Activity \(A = 0.255\) Bq/g = 0.255 dis/s/g
\[N = \frac{A}{\lambda} = \frac{0.255}{3.832 \times 10^{-12}} = 6.654 \times 10^{10} \text{ atoms/g of }^{14}\text{C}\]
\[\text{Total C atoms per gram} = \frac{6.02 \times 10^{23}}{12} = 5.017 \times 10^{22}\]
\[\text{Fraction} = \frac{6.654 \times 10^{10}}{5.017 \times 10^{22}} = 1.326 \times 10^{-12}\]
About 1 \(^{14}\text{C}\) atom per \(7.54 \times 10^{11}\) carbon atoms.

(b) Activity of specimen:
\[A = \frac{174 \text{ decays}}{3600 \text{ s} \times 0.5 \text{ g}} = 0.09667 \text{ Bq/g}\]
\[A_0 = 0.255 \text{ Bq/g (when alive)}\]
\[t = \frac{2.303}{\lambda}\log_{10}\left(\frac{A_0}{A}\right) = \frac{2.303}{3.832 \times 10^{-12}}\log_{10}\left(\frac{0.255}{0.09667}\right)\]
\[= \frac{2.303}{3.832 \times 10^{-12}} \times \log_{10}(2.638)\]
\[= \frac{2.303 \times 0.4213}{3.832 \times 10^{-12}} = \frac{0.9703}{3.832 \times 10^{-12}} = 2.532 \times 10^{11} \text{ s}\]
\[= \frac{2.532 \times 10^{11}}{3.156 \times 10^7} \approx 8023 \text{ years}\]
In simple words: Part (a) shows C-14 is extremely rare - only 1 in a trillion carbon atoms. Part (b) shows this specimen died about 8000 years ago - possibly from a neolithic civilization! Radiocarbon dating has revolutionized archaeology.

Teacher's Note: For part (b), carefully convert 174 decays/hour/500mg to decays/second/gram. Many students forget to divide by mass (500 mg = 0.5 g) and by 3600 s. The specimen at 8023 years is consistent with early human settlements.

Exam Tip: Part (a): \(N = A/\lambda\), then divide by total atoms per gram. Part (b): calculate present activity per gram, apply \(t = (2.303/\lambda)\log(A_0/A)\). Convert seconds to years at the end.

Question 23. How much mass of \(^{235}\text{U}\) is required to undergo fission each day to provide 3000 MW of thermal power? Average energy per fission is 202.79 MeV.
Answer:
Data: Power = 3000 MW = \(3 \times 10^9\) J/s
Energy per fission = 202.79 MeV = \(202.79 \times 10^6 \times 1.6 \times 10^{-19} = 3.245 \times 10^{-11}\) J

Energy needed per day:
\[E = 3 \times 10^9 \times 86400 = 2.592 \times 10^{14} \text{ J}\]

Number of fissions per day:
\[n = \frac{E}{\text{energy per fission}} = \frac{2.592 \times 10^{14}}{3.245 \times 10^{-11}} = 7.988 \times 10^{24}\]

Mass of \(^{235}\text{U}\) (0.235 kg contains \(6.02 \times 10^{23}\) atoms):
\[m = \frac{7.988 \times 10^{24}}{6.02 \times 10^{23}} \times 0.235 = 13.27 \times 0.235 = 3.118 \text{ kg}\]
In simple words: A 3000 MW nuclear power station (similar to Tarapur in India) needs only 3.118 kg of uranium-235 per day. A coal power plant of the same capacity would need thousands of tonnes of coal - showing the enormous energy density of nuclear fuel.

Teacher's Note: This problem beautifully illustrates the practical advantage of nuclear energy. Compare: 3 kg of U-235 vs. ~6000 tonnes of coal for the same daily energy output. Connect to energy policy discussions and India's nuclear energy program.

Exam Tip: Step 1: Energy per day = Power × 86400 s. Step 2: Fissions per day = Total energy / energy per fission (convert MeV to J). Step 3: Mass = (fissions / Avogadro) × 0.235 kg. Answer: 3.118 kg.

Question 24. In a periodic table the average atomic mass of magnesium is given as 24.312 u. The three isotopes and their masses are \(_{12}^{24}\text{Mg}\) (23.98504 u), \(_{12}^{25}\text{Mg}\) (24.98584 u) and \(_{12}^{26}\text{Mg}\) (25.98259 u). The natural abundance of \(_{12}^{24}\text{Mg}\) is 78.99% by mass. Calculate the abundances of other two isotopes.
Answer:
Let abundance of \(^{25}\text{Mg}\) = x% and \(^{26}\text{Mg}\) = y%
Then: \(78.99 + x + y = 100 \Rightarrow x + y = 21.01\) ...(1)

Average atomic mass equation:
\[24.312 = \frac{(23.98504)(78.99) + (24.98584)x + (25.98259)(100 - 78.99 - x)}{100}\]
\[2431.2 = 1894.5 + 24.98584x + 25.98259(21.01 - x)\]
\[2431.2 = 1894.5 + 24.98584x + 546.1 - 25.98259x\]
\[2431.2 = 2440.6 - 0.99675x\]
\[0.99675x = 9.4 \Rightarrow x = 9.30\%\]
\[y = 21.01 - 9.30 = 11.71\%\]

\(\therefore\) Abundance of \(_{12}^{25}\text{Mg}\) = 9.30% and \(_{12}^{26}\text{Mg}\) = 11.71%
In simple words: Most magnesium (79%) is the lightest isotope \(^{24}\text{Mg}\). About 9.3% is \(^{25}\text{Mg}\) and 11.7% is \(^{26}\text{Mg}\). These percentages produce the observed average atomic mass of 24.312 u. Isotopic abundances are fixed by nature and are the same everywhere on Earth.

Teacher's Note: This is a two-variable linear algebra problem disguised as physics. Students should set up two equations: (1) abundances sum to 100%, (2) weighted average equals given atomic mass. Solve simultaneously. This concept explains why atomic masses are not whole numbers.

Exam Tip: Let x = % of \(^{25}\text{Mg}\). Then % of \(^{26}\text{Mg} = (21.01 - x)\). Write the weighted average equation, expand, collect x terms, and solve. Answer: \(^{25}\text{Mg}\) = 9.30%, \(^{26}\text{Mg}\) = 11.71%.

Question 1. Why don't heavy nuclei decay by emitting a single proton or a single neutron?
Answer: According to quantum mechanics, the probability for these emissions (single proton or single neutron) is extremely low. Heavy nuclei preferentially emit alpha particles because the alpha particle (\(_2^4\text{He}\)) is a very stable configuration (doubly magic nucleus with Z=2, N=2). The quantum mechanical tunneling probability for alpha emission is much higher than for single nucleon emission.
In simple words: A single proton or neutron has very low chance of escaping a heavy nucleus because quantum mechanics makes it unlikely. The alpha particle (2 protons + 2 neutrons together) escapes much more easily because it is extremely stable and has a higher tunneling probability through the nuclear barrier.

Teacher's Note: This connects to quantum tunneling - nuclear particles can "tunnel" through the potential barrier, but only with certain probabilities. Alpha particles tunnel because their binding energy makes them stable pre-formed clusters inside the nucleus. This concept bridges nuclear physics with quantum mechanics.

Exam Tip: State: (1) probability of single nucleon emission is extremely low according to quantum mechanics, (2) alpha particle is a stable configuration preferred by nature. Keep the answer concise - this is an intext question worth fewer marks.

This is a preview of the chapter solutions. To get complete study materials for Maharashtra Board Class 12 Physics, visit studiestoday.com.