CBSE Class 12 Physics Atoms and Nuclei Assignment Set 02

(a) Hydrogen atom
(b) Singly ionised helium atom (He+)
(c) Deuteron atom
(d) Singly ionised neon atom (Ne+) 

Answer :   D

Question. The total energy of an electron in an atom in an orbit is – 3.4 eV. Its kinetic and potential energies are, respectively

(a) 3.4 eV, 3.4 eV
(b) – 3.4 eV, – 3.4 eV
(c) – 3.4 eV, – 6.8 eV
(d) 3.4 eV, – 6.8 eV

Answer :   D

Question. The radius of the first permitted Bohr orbit for the electron, in a hydrogen atom equals 0.51 Å and its ground state energy equals –13.6 eV. If the electron in the hydrogen atom is replaced by muon (m–) [charge same as electron and mass 207 me], the first Bohr radius and ground state energy will be
(a) 0.53 × 10^–13 m, –3.6 eV
(b) 25.6 × 10^–13 m, –2.8 eV
(c) 2.56 × 10^–13 m, –2.8 keV
(d) 2.56 × 10^–13 m, –13.6 eV 

Answer :   C

Question. The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is 
(a) 1 : 1
(b) 1 : –1
(c) 2 : –1
(d) 1 : –2 

Answer :   B

Question. Consider 3rd orbit of He+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = 9 × 109 constant, Z = 2 and h (Planck’s constant) = 6.6 × 10–34 J s]
(a) 0.73 × 10⁶ m/s
(b) 3.0 × 10⁸ m/s
(c) 2.92 × 10⁶ m/s
(d) 1.46 × 10⁶ m/s

Answer :   D

Question. An electron in hydrogen atom makes a transition n1 → n2 where n1 and n2 are principal quantum numbers of the two states. Assuming Bohr’s model to be valid, the time period of the electron in the initial state is eight times that in the final state. The possible values of n1 and n2 are
(a) n1 = 6 and n2 = 2
(b) n1 = 8 and n2 = 1
(c) n1 = 8 and n2 = 2
(d) n1 = 4 and n2 = 2

Answer :   D

Question. Which of the following statements are true regarding Bohr’s model of hydrogen atom? 
(I) Orbiting speed of electron decreases as it shifts to discrete orbits away from the nucleus
(II) Radii of allowed orbits of electron are proportional to the principal quantum number
(III) Frequency with which electrons orbit around the nucleus in discrete orbits is inversely proportional to the cube of principal quantum number
(IV) Binding force with which the electron is bound to the nucleus increases as it shifts to outer orbits
Select correct answer using the codes given below.
Codes :
(a) I and II
(b) II and IV
(c) I, II and III
(d) II, III and IV

Answer :   A

Question. Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material.
The stopping potential is measured to be 3.57 V. The threshold frequency of the material is
(a) 4 × 10¹⁵ Hz
(b) 5 × 10¹⁵ Hz
(c) 1.6 × 10¹⁵ Hz
(d) 2.5 × 10¹⁵ Hz 

Answer :   C

Question. What is the energy of He+ electron in first orbit?
(a) 40.8 eV
(b) –27.2 eV 
(c) –54.4 eV
(d) –13.6 eV

Answer :   C

Question. The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is[
(a) 3
(b) 4
(c) 1
(d) 2

Answer :   D

Important Questions for NCERT Class 12 Physics Nuclei 

VERY SHORT ANSWER QUESTIONS (1 Mark)

Question 1. Define atomic mass unit. Write its energy equivalent in MeV.
Answer: One atomic mass unit (u) is defined as \(\frac{1}{12}\)th of the mass of an unbound neutral carbon-12 atom in its nuclear and electronic ground state.
\(1\text{ u} = 1.6605 \times 10^{-27}\text{ kg}$
Energy equivalent of 1 u = 931.5 MeV.
In simple words: Atomic mass unit is a tiny unit used to measure the mass of atoms and subatomic particles.

Exam Tip: Remember the exact value 931.5 MeV for numerical problems involving mass-energy conversion.

Question 2. What was the drawback of Rutherford model of atom?
Answer: According to classical electromagnetic theory, an accelerated charged particle (like an orbiting electron) must continuously radiate energy. As a result, the electron would spiral into the nucleus, making the atom unstable. Also, it could not explain the discrete line spectrum of atoms.
In simple words: Rutherford's model could not explain why atoms are stable and why they emit light of only specific frequencies.

Exam Tip: Mention both atomic instability due to continuous energy radiation and inability to explain line spectra.

Question 3. What are the number of electrons and neutrons in singly ionised \(^{236}_{\) 92}\text{U}\) atom?
Answer: For a neutral uranium atom, atomic number \(Z = 92\) and mass number \(A = 236\).
Number of protons = 92
Since the atom is singly ionized, it has lost one electron. Number of electrons = \(92 - 1 = 91\).
Number of neutrons = \(A - Z = 236 - 92 = 144\).
In simple words: Neutrons are found by subtracting atomic number from mass number, and electrons are found by adjusting the atomic number for the charge.

Exam Tip: Always account for the ionic charge when calculating the number of electrons.

Question 4. Name the series of hydrogen spectrum which has least wavelength.
Answer: Lyman series.
In simple words: The Lyman series falls in the ultraviolet region and corresponds to transitions to the ground state, yielding the highest energy and shortest wavelength.

Exam Tip: Wavelength is inversely proportional to energy. Maximum energy transition gives minimum wavelength.

Question 5. Any two protons repel each other, then how is this possible for them to remain together in a nucleus.
Answer: They remain together due to strong nuclear forces, which are attractive and much stronger than the electrostatic repulsive force between protons at nuclear distances.
In simple words: The powerful nuclear glue (strong nuclear force) overpowers the electric repulsion between protons.

Exam Tip: Note that strong nuclear force is charge-independent and operates over very short ranges (\(\sim 1\text{ fm}\)).

Question 6. Define radioactive decay constant.
Answer: Radioactive decay constant is defined as the reciprocal of the time during which the number of radioactive atoms decreases to \(\frac{1}{e}\) (or about \(36.8\%\)) of their initial value.
Alternatively, it is the probability of decay of a nucleus per unit time.
In simple words: Decay constant tells us how fast a radioactive substance decays.

Question 7. You are given reaction: \(_1^2\text{H} + _1^2\text{H} \rightarrow _2^4\text{He} + 24\text{ MeV}\). What type of nuclear reaction is this?
Answer: Nuclear fusion reaction.
In simple words: Two light nuclei combine to form a heavier nucleus with the release of energy.

Exam Tip: Distinguish fusion (combining light nuclei) from fission (splitting heavy nuclei).

Question 8. After losing two electrons, to which particle does a helium atom get transformed into?
Answer: An alpha particle (\(_2^4\text{He}^{2+}\)).
In simple words: A helium nucleus stripped of its two electrons is an alpha particle.

Exam Tip: An alpha particle is identical to a doubly ionized helium nucleus.

Question 9. Write two important inferences drawn from Geiger-Marsden's \(\alpha\)-particle scattering experiment.
Answer: (i) The entire positive charge and most of the mass of the atom are concentrated in a very small volume called the nucleus.
(ii) The size of the nucleus is extremely small compared to the total size of the atom.
In simple words: Atoms are mostly empty space with a tiny, dense, positively charged center.

Question 10. What will be effect on the ratio of the radii of the nuclei of mass number \(A_1\) and \(A_2\)?
Answer: Using the relation \(R = R_0 A^{1/3}\), the ratio of radii is:
\(\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3}\)
In simple words: Nuclear radius is proportional to the cube root of the mass number.

Exam Tip: Always write the formula \(R = R_0 A^{1/3}\) before deriving ratios.

Question 11. In nuclear reaction \(_1^1\text{H} \rightarrow _0^1\text{n} + _Q^P\text{X}\) find P, Q and hence identify X.
Answer: Using conservation of mass number (A) and atomic number (Z):
Mass number: \(1 = 1 + P \implies P = 0\)
Atomic number: \(1 = 0 + Q \implies Q = 1$
Particle X is \(_1^0\text{X}\), which is a positron (\(_{+1}^{}\text{e}\)).
In simple words: Balance the top numbers (mass) and bottom numbers (charge) on both sides of the reaction equation.

Exam Tip: Conservation laws of charge and mass number must always hold in nuclear reactions.

Question 12. Binding energies of neutron (\(_1^2\text{H}\)) and \(\alpha\)-particle (\(_2^4\text{He}\)) are 1.25 MeV/nucleon and 7.2 MeV/nucleon respectively. Which nucleus is more stable?
Answer: \(_2^4\text{He}\) is more stable because it has a higher binding energy per nucleon (7.2 MeV/nucleon > 1.25 MeV/nucleon).
In simple words: Higher binding energy per nucleon means the nucleus is held together more tightly, making it more stable.

Question 13. \(\alpha\)-particles are incident on a thin gold foil. For what angle of deviation will the number of deflected \(\alpha\)-particles be minimum?
Answer: For \(180^\circ\) (or very large angles), the number of deflected particles is minimum.
In simple words: Very few alpha particles bounce straight back; most pass right through with little or no deflection.

Question 14. A and B are two isotopes having mass numbers 14 and 16 respectively. If the number of electrons in A is 7, then give the number of neutrons in B.
Answer: Since A and B are isotopes, they have the same atomic number \(Z\).
Number of electrons in A = 7, so \(Z = 7\).
For isotope B, mass number \(A_B = 16\) and \(Z = 7\).
Number of neutrons in B = \(A_B - Z = 16 - 7 = 9\).
In simple words: Isotopes share the same atomic number, so use that to find the neutron count for the second isotope.

Exam Tip: Isotopes have the same number of protons but different numbers of neutrons.

Question 15. If the amount of a radioactive substance is increased four times then how many times will the number of atoms disintegrating per unit time be increased?
Answer: Activity (\(-\frac{dN}{dt} = \lambda N\)) is directly proportional to the number of atoms \(N\). If the amount is increased four times, the disintegration rate also increases four times.
In simple words: Four times more atoms mean four times more disintegrations per second.

Question 16. An electron jumps from fourth to first orbit in an atom. How many maximum number of spectral lines can be emitted by the atom?
Answer: Using the formula for maximum spectral lines: \(\frac{n(n-1)}{2}\) where \(n = 4\):
\(\text{Number of lines} = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6\)
In simple words: An electron falling from level 4 can take 6 different transition paths to reach level 1.

Exam Tip: Use \(\frac{n(n-1)}{2}\) for transitions from level \(n\) to ground state.

Question 17. Under what conditions of electronic transition will the emitted light be monochromatic?
Answer: When electrons make transitions from a single higher energy level to a single lower energy level (\(\Delta E\) is fixed, so frequency \(\nu\) is fixed).
In simple words: A single frequency of light is emitted only when transitions occur between the exact same pair of energy levels.

Question 18. Why does only a slow neutron (\(0.03\text{ eV}\) energy) cause the fission in the uranium nucleus and not the fast one?
Answer: A slow neutron stays in the vicinity of the uranium nucleus for a longer duration, allowing sufficient time for the nucleus to capture it and destabilize into fission fragments.
In simple words: Slow neutrons have more time to interact and be captured by the heavy nucleus.

Exam Tip: Slow neutrons have a larger capture cross-section for uranium fission.

Question 19. Write the relation for distance of closest approach.
Answer: \(r_0 = \frac{1}{4\pi\varepsilon_0} \frac{2Ze^2}{K}\)
where \(K\) is the initial kinetic energy of the alpha particle and \(Z\) is the atomic number of the target nucleus.
In simple words: Distance of closest approach depends inversely on the kinetic energy of the incoming particle.

Exam Tip: At the distance of closest approach, kinetic energy is completely converted into electrostatic potential energy.

Question 20. In Bohr's atomic model, the potential energy is negative and has a magnitude greater than the kinetic energy, what does this imply?
Answer: It implies that the electron is bound to the nucleus.
In simple words: The negative total energy shows the electron is trapped in an attractive potential well.

Exam Tip: Total energy \(E = K + U = -K\). Since \(U = -2K\), \(\lvert U\rvert > K\).

Question 21. Name the physical quantity whose dimensions are same as Planck's constant.
Answer: Angular momentum.
In simple words: Both Planck's constant and angular momentum have dimensions of \([ML^2T^{-1}]\).

Exam Tip: Recall Bohr's quantization condition: \(L = \frac{nh}{2\pi}\).

Question 22. Define ionisation potential.
Answer: Ionization potential is defined as the accelerating potential difference required to provide an electron with enough energy to completely remove an electron from the ground state of an atom to infinity.
In simple words: The voltage needed to knock an electron completely out of an atom.

Exam Tip: Ionization potential (in volts) numerically equals ionization energy (in eV).

Question 23. The ionisation potential of helium atom is \(24.6\text{ V}\). How much energy will be required to ionise it?
Answer: Energy required = \(24.6\text{ eV}\).
In simple words: Multiply the potential in volts by the electronic charge to get energy in electron-volts.

Exam Tip: 1 V \(\times\) 1 e = 1 eV.

Question 24. What is the energy possessed by an electron whose principal quantum number is infinite?
Answer: Zero energy.
In simple words: At infinity, the electron is completely free from the nucleus, so its energy is zero.

Question 25. Write the value of Rydberg constant?
Answer: \(R = 1.097 \times 10^7\text{ m}^{-1}\).
In simple words: Rydberg constant is a fundamental physical constant used in atomic spectra formulas.

Exam Tip: Memorize the value and units (\(\text{m}^{-1}\)) for spectral calculations.

Question 26. Name the spectral series of hydrogen atom which lie in uv region.
Answer: Lyman Series.
In simple words: Lyman series emits ultraviolet light.

Exam Tip: Lyman = UV, Balmer = Visible, Paschen/Brackett/Pfund = Infrared.

Question 27. Name the series of hydrogen spectrum lying in the infra red region.
Answer: Paschen, Brackett, and Pfund series.
In simple words: Transitions ending in higher orbits (\(n_1 \ge 3\)) produce infrared radiation.

Question 28. What is the order of velocity of electron in a hydrogen atom in ground state?
Answer: \(10^6\text{ ms}^{-1}\).
In simple words: Electrons orbit the nucleus at about one-hundredth the speed of light.

Exam Tip: Useful order-of-magnitude value for atomic physics estimations.

Question 29. Write a relation for Paschen series lines of hydrogen spectrum.
Answer: \(\bar{\nu} = R \left( \frac{1}{3^2} - \frac{1}{n^2} \right)\) where \(n = 4, 5, 6, \dots$
In simple words: Paschen series formula uses \(n_1 = 3\) as the lower energy level.

Exam Tip: General formula is \(\bar{\nu} = R (\frac{1}{n_1^2} - \frac{1}{n_2^2})\); put \(n_1 = 3\) for Paschen.

Question 30. Arrange radioactive radiation in the increasing order of penetrating power.
Answer: \(\alpha\text{-particles} < \beta\text{-particles} < \gamma\text{-rays}\).
In simple words: Alpha rays are stopped by paper, beta by aluminum, and gamma can penetrate thick lead.

Exam Tip: Ionizing power follows the exact reverse order (\(\alpha > \beta > \gamma\)).

Question 31. Write a relation between average life and decay constant.
Answer: \(\tau = \frac{1}{\lambda}\)
where \(\tau\) is the mean life and \(\lambda\) is the decay constant.
In simple words: Mean life is simply the reciprocal of the decay constant.

Exam Tip: Average life is related to half-life by \(T_{1/2} = 0.693 \tau\).

Question 32. Write two units for activity of radioactive element and relate them with number of disintegration per second.
Answer: The two units are Becquerel (Bq) and Curie (Ci).
\(1\text{ Bq} = 1\text{ disintegration per second}\)
\(1\text{ Ci} = 3.7 \times 10^{10}\text{ disintegrations per second}\).
In simple words: Becquerel is the SI unit, while Curie is a larger traditional unit based on radium.

Exam Tip: Memorize the conversion factor for 1 Curie.

Question 33. The half life of a radioactive element A is same as the mean life time of another radioactive element B. Initially, both have same number of atoms. B decay faster than A. Why?
Answer: Given \(T_{A} = \tau_{B} \implies T_{A} = 1.44 T_{B}\). Since \(T_{A} > T_{B}\), the decay constant \(\lambda_A < \lambda_B\). Therefore, B decays faster than A.
In simple words: Element B has a shorter half-life, meaning it decays at a faster rate than A.

Exam Tip: Smaller half-life implies a larger decay constant and faster decay.

Question 34. Draw the graph showing the distribution of Kinetic energy of electrons emitted during \(\beta\) decay.
Answer:
(Description of graph: A continuous bell-shaped curve plotting number of electrons on the y-axis versus kinetic energy on the x-axis, starting from zero up to a maximum kinetic energy \(K_{\max}\)).
In simple words: Beta particles are emitted with a continuous spectrum of kinetic energies up to a maximum limit.

Exam Tip: Note that unlike alpha decay, beta decay energy is shared with antineutrinos, causing a continuous spectrum.

Question 35. Compare radii of two nuclei of mass numbers 1 and 27 respectively.
Answer: \(\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{1}{27}\right)^{1/3} = \frac{1}{3}\). Ratio is \(1:3$.
In simple words: Cube root of 1 is 1 and cube root of 27 is 3, so the radius ratio is 1 to 3.

Exam Tip: Always apply \(R \propto A^{1/3}\).

Question 36. Define atomic mass unit.
Answer: One atomic mass unit (u) is \(\frac{1}{12}\)th of the mass of a carbon-12 atom (\(_6^{12}\text{C}\)).
In simple words: The standard unit for measuring atomic masses.

Exam Tip: Mention carbon-12 explicitly in definitions.

Question 37. Write the energy equivalent of MeV.
Answer: \(1\text{ u} = 931.5\text{ MeV}\).
In simple words: 1 atomic mass unit equals 931.5 million electron-volts of energy.

Question 38. Which element has highest value of Binding Energy per nucleon?
Answer: Iron (\(_{26}^{56}\text{Fe}\)) with approximately \(8.8\text{ MeV/nucleon}\).
In simple words: Iron is the most tightly bound and stable nucleus.

Exam Tip: Peak of the binding energy curve occurs around \(A = 56\).

Question 39. Mention the range of mass number for which the Binding energy curve is almost horizontal.
Answer: From \(A = 30\) to \(A = 170\).
In simple words: For medium-sized nuclei, binding energy per nucleon remains nearly constant at about 8 MeV.

Exam Tip: This constancy explains the saturation property of nuclear forces.

Question 40. What is the ratio of nuclear densities of the two nuclei having mass numbers in the ratio \(1:4\)?
Answer: \(1:1$.
In simple words: Nuclear density is constant and independent of mass number.

Exam Tip: Nuclear density is independent of \(A\); all nuclei have roughly the same density (\(\sim 2.3 \times 10^{17}\text{ kg/m}^3\)).

Question 41. Write two important inferences drawn from Rutherford \(\alpha\) particle scattering experiment.
Answer: (i) Most space inside the atom is empty.
(ii) The entire positive charge and almost all mass are concentrated in a tiny nucleus.
In simple words: Atoms are mostly empty space with a dense central nucleus.

Exam Tip: Keep answers concise for 1-mark questions.

Question 42. Draw a graph of number of undecayed nuclei to the time, for a radioactive nuclei.
Answer:
(Description of graph: An exponential decay curve starting at \(N_0\) on the y-axis at \(t = 0\) and asymptotically approaching zero as time \(t\) increases).
In simple words: The number of radioactive atoms decreases exponentially over time.

Exam Tip: Label axes clearly: \(N\) on y-axis and \(t\) on x-axis.

Question 43. Write an equation to represent \(\alpha\) decay.
Answer: \(_Z^A\text{X} \rightarrow _{Z-2}^{A-4}\text{Y} + _2^4\text{He} + Q\)
In simple words: During alpha decay, a nucleus loses 2 protons and 2 neutrons by emitting an alpha particle.

Exam Tip: Check that mass and atomic numbers balance on both sides of the equation.

SHORT ANSWER QUESTIONS (2 Marks)

Question 1. Define distance of the closest approach. An \(\alpha\)-particle of kinetic energy \(K\) is bombarded on a thin gold foil. The distance of the closest approach is \(r\). What will be the distance of closest approach for an \(\alpha\)-particle of double the kinetic energy?
Answer: Distance of closest approach is the minimum distance up to which an alpha particle approaches the target nucleus before halting and reversing its direction due to electrostatic repulsion.
Formula: \(r = \frac{1}{4\pi\varepsilon_0} \frac{2Ze^2}{K}\)
Since \(r \propto \frac{1}{K}\), when kinetic energy is doubled (\(2K\)), the distance of closest approach becomes halved (\(\frac{r}{2}\)).
In simple words: Faster particles get closer to the nucleus before being repelled back.

Exam Tip: Use proportionality \(r \propto 1/K\) for quick two-mark conceptual problems.

Question 2. Show that nuclear density is independent of the mass number.
Answer: Nuclear density \(\rho = \frac{\text{Mass of nucleus}}{\text{Volume of nucleus}} = \frac{m A}{\frac{4}{3}\pi R^3}\)
Substituting \(R = R_0 A^{1/3}\):
\(\rho = \frac{m A}{\frac{4}{3}\pi (R_0 A^{1/3})^3} = \frac{m A}{\frac{4}{3}\pi R_0^3 A} = \frac{m}{\frac{4}{3}\pi R_0^3}\)
Since \(\rho\) does not contain mass number \(A\), nuclear density is independent of \(A\).
In simple words: Mass and volume both scale linearly with mass number \(A\), so the \(A\) terms cancel out, leaving density constant.

Exam Tip: This is a very common derivation. Practice substituting \(R = R_0 A^{1/3}\) cleanly.

Question 3. Which of the following radiations \(\alpha\), \(\beta\) and \(\gamma\) :
(i) similar to x-rays?
(ii) easily absorbed by matter
(iii) travel with greatest speed?
(iv) similar to the nature of cathode rays?
Answer:
(i) \(\gamma\)-rays (both are electromagnetic waves)
(ii) \(\alpha\)-particles (due to high charge and mass)
(iii) \(\gamma\)-rays (travel at the speed of light)
(iv) \(\beta\)-particles (both are streams of high-speed electrons)
In simple words: Match each property to the correct type of radiation based on its physical nature.

Exam Tip: Memorize the nature of alpha (helium nuclei), beta (electrons), and gamma (photons).

Question 4. Some scientist have predicted that a global nuclear war on earth would be followed by 'Nuclear winter'. What could cause nuclear winter?
Answer: Massive nuclear explosions would ignite widespread firestorms, sending enormous quantities of black soot and dust high into the stratosphere. This aerosol layer would block sunlight for months, causing temperatures to plummet drastically.
In simple words: Thick smoke and dust from nuclear explosions would block sunlight and freeze the Earth.

Exam Tip: Focus on atmospheric absorption of solar radiation by soot particles.

Question 5. If the total number of neutrons and protons in a nuclear reaction is conserved how then is the energy absorbed or evolved in the reaction?
Answer: Energy is absorbed or evolved due to the "mass defect" (\(\Delta m\)). Although nucleon number is conserved, the total rest mass of the products differs slightly from that of the reactants, and this mass difference is converted into energy via \(E = \Delta m c^2\).
In simple words: Even though nucleon count stays the same, tiny changes in total mass release or absorb energy.

Exam Tip: Distinguish between nucleon conservation and mass-energy equivalence.

Question 6. In the ground state of hydrogen atom orbital radius is \(5.3 \times 10^{-11}\text{ m}\). The atom is excited such that atomic radius becomes \(21.2 \times 10^{-11}\text{ m}\). What is the principal quantum number of the excited state of atom?
Answer: Since radius \(r_n \propto n^2\):
\(\frac{r_2}{r_1} = \frac{n_2^2}{n_1^2}\)
\(\frac{21.2 \times 10^{-11}}{5.3 \times 10^{-11}} = \frac{n_2^2}{1^2}\)
\(4 = n_2^2 \implies n_2 = 2\)
In simple words: Radius quadruples, so the quantum number squared is 4, making \(n = 2\).

Exam Tip: Use \(r_n = r_1 n^2\) for hydrogen atom orbit radii.

Question 7. Calculate the percentage of any radioactive substance left undecayed after half of half life.
Answer: Half of half-life means time \(t = \frac{T}{2}\).
Using \(\frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/T} = \left(\frac{1}{2}\right)^{1/2} = \frac{1}{\sqrt{2}}\)
Percentage left = \(\frac{1}{\sqrt{2}} \times 100\% = \frac{100}{1.414}\% \approx 70.9\%\)
In simple words: Calculate \((1/2)^{0.5}\) to find the remaining fraction.

Exam Tip: Be careful with fractional exponents when time is not an integer multiple of \(T\).

Question 8. Why is the density of the nucleus more than that of atom?
Answer: Almost the entire mass of an atom is concentrated in its tiny nucleus, whereas the volume of the atom is enormously large due to empty space occupied by orbiting electrons. Since density equals mass divided by volume, the tiny volume of the nucleus results in a huge nuclear density.
In simple words: The nucleus packs almost all the atom's mass into a tiny fraction of its volume.

Exam Tip: Mention that atomic volume is about \(10^{15}\) times larger than nuclear volume.

Question 9. The atom \(_8\text{O}^{16}\) has 8 protons, 8 neutrons and 8 electrons while atom \(_4\text{Be}^8\) has 4 proton, 4 neutrons and 4 electrons, yet the ratio of their atomic masses is not exactly 2. Why?
Answer: Due to mass defect. Different nuclei have different total binding energies, meaning varying amounts of mass are converted into binding energy when nucleons bind together.
In simple words: Mass is not simply additive because some mass is converted into binding energy to hold the nucleus together.

Exam Tip: Always cite mass defect and binding energy for deviations from simple mass addition.

Question 10. What is the effect on neutron to proton ratio in a nucleus when \(\beta^-\) particle is emitted? Explain your answer with the help of a suitable nuclear reaction.
Answer: During \(\beta^-\) decay, a neutron converts into a proton: \(_0^1\text{n} \rightarrow _1^1\text{p} + _{-1}^0\text{e} + \bar{\nu}\).
Since a neutron is lost and a proton is gained, the neutron-to-proton (\(N/P\)) ratio decreases.
In simple words: Neutron turns into a proton, so neutrons decrease and protons increase, lowering the ratio.

Exam Tip: \(\beta^-\) emission decreases \(N/P\) ratio, while \(\beta^+\) emission increases it.

Question 11. Why must heavy stable nucleus contain more neutrons than protons?
Answer: Protons repel each other via electrostatic forces which have an infinite range, whereas strong nuclear forces have a very short range. In heavy nuclei, additional neutrons are needed to provide extra attractive strong nuclear forces to counterbalance the cumulative electrostatic repulsion among protons.
In simple words: Extra neutrons act as neutral glue to stabilize the growing electrical repulsion among protons in large nuclei.

Exam Tip: Mention short-range strong force vs. long-range Coulomb force.

Question 12. Show that the decay rate R of a sample of radio nuclide at some instant is related to the number of radio active nuclei N at the same instant by the expression \(R = -\lambda N\).
Answer: According to the radioactive decay law, the rate of disintegration is proportional to the number of undecayed nuclei present:
\(-\frac{dN}{dt} \propto N\)
\(-\frac{dN}{dt} = \lambda N\)
Since decay rate \(R = -\frac{dN}{dt}\), we get:
\(R = \lambda N \text{ (magnitude)} \text{ or } R = -\frac{dN}{dt}\)
In simple words: The rate at which nuclei decay is simply the decay constant times the number of remaining nuclei.

Exam Tip: State the radioactive decay law clearly before deriving the relation.

Question 13. What is a nuclear fusion reaction? Why is nuclear fusion difficult to carry out for peaceful purpose?
Answer: Nuclear fusion is the process in which two light nuclei combine to form a heavier nucleus, releasing energy.
It is difficult because positively charged nuclei repel each other. Extremely high temperatures (\(\sim 10^8\text{ K}\)) and pressures are required to give nuclei enough kinetic energy to overcome Coulomb repulsion, making containment extremely difficult.
In simple words: Fusion requires extreme heat and pressure to force repellent nuclei together, making it hard to control for power generation.

Question 14. Write two characteristic features of nuclear forces which distinguish them from coulomb force.
Answer: (i) Nuclear forces are short-range forces (effective up to a few femtometers), whereas Coulomb forces are long-range forces (\(1/r^2\)).
(ii) Nuclear forces are strongly attractive and independent of charge (act equally between proton-proton, neutron-neutron, and proton-neutron), whereas Coulomb forces act only between charged particles and are repulsive between like charges.
In simple words: Nuclear forces are powerful short-range attractive forces that do not care about electrical charge.

Exam Tip: Highlight range and charge independence.

Question 15. Half life of certain radioactive nuclei is 3 days and its activity is 8 times the 'safe limit'. After how much time will the activity of the radioactive sample reach the 'safe limit'?
Answer: Let initial activity be \(R_0 = 8 R_{\text{safe}}\). Final activity \(R = R_{\text{safe}}\).\(\frac{R}{R_0} = \frac{1}{8}\).
Using \(\frac{R}{R_0} = \left(\frac{1}{2}\right)^{t/T}\):
\(\frac{1}{8} = \left(\frac{1}{2}\right)^{t/3}\)
\(\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^{t/3}\)
\(3 = \frac{t}{3} \implies t = 9\text{ days}\)
In simple words: Activity drops by a factor of 8 in 3 half-lives (\(3 \times 3 = 9\) days).

Exam Tip: Relate activity ratios directly to powers of 2 for quick half-life calculations.

Question 16. Derive \(mvr = \frac{nh}{2\pi}\) using de Broglie equation.
Answer: According to de Broglie hypothesis, a wave is associated with a moving electron of momentum \(p = mv\):
\(\lambda = \frac{h}{mv}\)
For a standing wave to form along a circular orbit of radius \(r$, the circumference must be an integral multiple of the de Broglie wavelength:
\(2\pi r = n\lambda\)
Substitute \(\lambda = \frac{h}{mv}\):
\(2\pi r = \frac{nh}{mv}\)
Rearranging gives:
\(mvr = \frac{nh}{2\pi}\)
In simple words: By requiring electron waves to fit neatly around a circular orbit, we derive Bohr's angular momentum condition.

Exam Tip: Combine standing wave condition (\(2\pi r = n\lambda\)) with de Broglie wavelength (\(\lambda = h/p\)).

Question 17. Draw graph of number of scattered particles to scattering angle in Rutherford's experiment.
Answer:
(Description of graph: A steeply decreasing curve plotting number of scattered particles \(N(\theta)\) on the logarithmic or sharp vertical y-axis against scattering angle \(\theta\) on the x-axis, showing high counts at small angles and very near-zero counts at large angles like \(180^\circ\)).
In simple words: Most alpha particles pass straight through (\(\theta \approx 0^\circ\)), while very few bounce back at large angles.

Exam Tip: The graph follows an inverse fourth power dependence on \(\sin(\theta/2)\).

Question 18. Show that nuclear density is same for all the nuclei.
Answer: Nuclear density \(\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{mA}{\frac{4}{3}\pi R_0^3 A} = \frac{m}{\frac{4}{3}\pi R_0^3}\). Since this expression is constant and independent of mass number \(A\), nuclear density is the same for all nuclei.
In simple words: Mass and volume both scale with \(A\), so the density remains constant across all elements.

Exam Tip: This is identical to Short Answer Q2; write out the steps clearly.

Question 19. What is the shortest wavelength present in the (i) Paschen series (ii) Balmer series of spectral lines?
Answer:
(i) For Paschen series (\(n_1 = 3\)), shortest wavelength corresponds to \(n_2 = \infty\):
\(\frac{1}{\lambda} = R \left(\frac{1}{3^2} - \frac{1}{\infty}\right) = \frac{R}{9} \implies \lambda = \frac{9}{R} \approx 820\text{ nm}\)(\(8200\text{ \AA}\))
(ii) For Balmer series (\(n_1 = 2\)), shortest wavelength corresponds to \(n_2 = \infty\):
\(\frac{1}{\lambda} = R \left(\frac{1}{2^2} - \frac{1}{\infty}\right) = \frac{R}{4} \implies \lambda = \frac{4}{R} \approx 365\text{ nm}\)(\(3650\text{ \AA}\))
In simple words: Shortest wavelength occurs at the series limit where the electron transitions from infinity.

Exam Tip: Series limit corresponds to \(n_2 = \infty\).

Question 20. The radius of the inner most electron orbit of a hydrogen atom \(0.53\text{ \AA}\). What are the radii of the \(n = 2\) and \(n = 3\) orbits?
Answer: Given \(r_1 = 0.53\text{ \AA}\).
For \(n = 2\): \(r_2 = r_1 \times 2^2 = 0.53 \times 4 = 2.12\text{ \AA}\).
For \(n = 3\): \(r_3 = r_1 \times 3^2 = 0.53 \times 9 = 4.77\text{ \AA}\).
In simple words: Orbit radii scale as the square of the principal quantum number (\(n^2\)).

Exam Tip: Remember \(r_n \propto n^2\).

Question 21. The ground state energy of hydrogen atom is \(-13.6\text{ eV}\). What are the Kinetic and potential energies of the electron in this state?
Answer: Total energy \(E = -13.6\text{ eV}\).
Kinetic energy \(K = -E = +13.6\text{ eV}\).
Potential energy \(U = 2E = 2 \times (-13.6) = -27.2\text{ eV}\).
In simple words: Kinetic energy is positive and half the magnitude of potential energy, while potential energy is negative and double the total energy.

Exam Tip: Memorize relationships: \(K = -E\), \(U = 2E\), and \(U = -2K\).

Question 22. Write formula of frequency to represent (i) Lyman series (ii) Balmer series.
Answer:
(i) Lyman series frequency: \(\nu = c R \left(\frac{1}{1^2} - \frac{1}{n^2}\right)\) where \(n = 2, 3, 4, \dots$
(ii) Balmer series frequency: \(\nu = c R \left(\frac{1}{2^2} - \frac{1}{n^2}\right)\) where \(n = 3, 4, 5, \dots$
In simple words: Multiply the wave number formula by the speed of light \(c\) to get frequency.

Question 23. From the relation \(R = R_0 A^{1/3}\) where \(R_0\) is a constant and A is the mass number of a nucleus, show that nuclear matter density is nearly constant.
Answer:
Nuclear density \(\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{mA}{\frac{4}{3}\pi R^3} = \frac{mA}{\frac{4}{3}\pi (R_0 A^{1/3})^3} = \frac{m}{\frac{4}{3}\pi R_0^3}\)
Since all parameters in the final expression are constants, nuclear matter density is constant.
In simple words: Substituting \(R\) in terms of \(A\) eliminates \(A\), proving density does not change across nuclei.

Exam Tip: Standard derivation for nuclear density constancy.

Question 24. Find the energy equivalent of one atomic mass unit in joules and then in MeV.
Answer:
\(E = \Delta m c^2 = 1.6605 \times 10^{-27}\text{ kg} \times (3 \times 10^8\text{ m/s})^2 = 1.4924 \times 10^{-10}\text{ J}\)
Converting to eV: \(\frac{1.4924 \times 10^{-10}}{1.6 \times 10^{-19}} = 9.328 \times 10^8\text{ eV} = 931.5\text{ MeV}\)
In simple words: Multiply 1 u in kg by \(c^2\) and divide by the charge of an electron to convert Joules to MeV.

Exam Tip: Practice this conversion as it frequently appears in board exams.

Question 25. Write four properties of nuclear force.
Answer:
(i) They are the strongest forces in nature.
(ii) They are short-range attractive forces.
(iii) They are charge-independent.
(iv) They are non-central forces.
In simple words: Nuclear forces are super-strong, short-range, independent of electric charge, and depend on nucleon orientation.

Exam Tip: List bullet points clearly to secure full credit.

SHORT ANSWER QUESTIONS (3 Marks)

Question 1. Give one example of a nuclear reaction. Also define the Q-value of the reaction. What does \(Q > 0\) signify?
Answer: Example: \(_7^{14}\text{N} + _2^4\text{He} \rightarrow _8^{17}\text{O} + _1^1\text{H} + Q\)
Q-value is defined as the total energy released or absorbed during a nuclear reaction, equal to the difference in rest mass energy between reactants and products: \(Q = (\sum m_{\text{reactants}} - \sum m_{\text{products}}) c^2\).
\(Q > 0\) signifies an exothermic reaction where energy is released.
In simple words: Q-value measures energy change in a nuclear reaction; positive Q means energy is released.

Exam Tip: Remember \(Q > 0\) is exothermic (energy released) and \(Q < 0\) is endothermic (energy absorbed).

Question 2. Explain how radio-active nucleus can emit \(\beta\)-particles even though nuclei do not contain these particles. Hence explain why the mass number of radioactive nuclide does not change during \(\beta\)-decay.
Answer: Beta particles are created at the instant of decay when a neutron inside the nucleus converts into a proton, an electron (\(\beta^-\)), and an antineutrino (\(\bar{\nu}\)): \(_0^1\text{n} \rightarrow _1^1\text{p} + _{-1}^0\text{e} + \bar{\nu}\).
The mass number does not change because the total number of nucleons (\(A = N + Z\)) remains constant (one neutron is lost while one proton is gained).
In simple words: Electrons are born inside the nucleus when a neutron transforms into a proton, keeping the total nucleon count unchanged.

Exam Tip: Emphasize neutron-to-proton conversion inside the nucleus.

Question 3. Define the term half life period and decay constant. Derive the relation between these terms.
Answer:
Half-life (\(T_{1/2}\)): Time in which half of the radioactive atoms disintegrate.
Decay constant (\(\lambda\)): Probability of decay per unit time.
Derivation:
\(N = N_0 e^{-\lambda t}\)
At \(t = T_{1/2}\), \(N = \frac{N_0}{2}\):
\(\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}}\)
\(\frac{1}{2} = e^{-\lambda T_{1/2}} \implies 2 = e^{\lambda T_{1/2}}\)
Taking natural logarithm on both sides:
\(\ln 2 = \lambda T_{1/2} \implies T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}\)
In simple words: Half-life is linked to the decay constant through the natural logarithm of 2 (\(0.693\)).

Exam Tip: Practice this standard 3-mark derivation thoroughly.

Question 4. State the law of radioactive decay. Deduce the relation \(N = N_0 e^{-\lambda t}\), where symbols have their usual meaning.
Answer: Law: The rate of disintegration of radioactive atoms is directly proportional to the number of undecayed atoms present at that instant.
Derivation:
\(-\frac{dN}{dt} \propto N \implies \frac{dN}{dt} = -\lambda N\)
\(\frac{dN}{N} = -\lambda dt\)
Integrating both sides from \(t = 0\) (where \(N = N_0\)) to time \(t\) (where \(N = N\)):
\(\int_{N_0}^{N} \frac{dN}{N} = -\lambda \int_{0}^{t} dt\)
\(\ln\left(\frac{N}{N_0}\right) = -\lambda t\)
\(\frac{N}{N_0} = e^{-\lambda t} \implies N = N_0 e^{-\lambda t}\)
In simple words: Integrating the decay rate equation yields exponential decay over time.

Exam Tip: Show integration limits clearly to secure full marks.

Question 5. Give the properties of \(\alpha\)-particles, \(\beta\)-particles and \(\gamma\)-rays.
Answer:
- \(\alpha\)-particles: Helium nuclei (\(_2^4\text{He}^{2+}\)), positive charge, high ionizing power, low penetration, deflected by electric and magnetic fields.
- \(\beta\)-particles: Fast-moving electrons (\(_{-1}^0\text{e}\)), negative charge, moderate ionization and penetration power, deflected by fields.
- \(\gamma\)-rays: High-frequency electromagnetic radiation, neutral, very high penetration power, low ionization, not deflected by fields.
In simple words: Compare them based on charge, nature, ionizing power, and penetrating ability.

Exam Tip: Use a comparative table format if time permits.

Question 6. With the help of one example, explain how the neutron to proton ratio changes during alpha decay of a nucleus.
Answer: Example: \(_92^{238}\text{U} \rightarrow _{90}^{234}\text{Th} + _2^4\text{He}\).
For Uranium-238: Protons = 92, Neutrons = \(238 - 92 = 146\), \(N/P = 146/92 \approx 1.587\).
For Thorium-234: Protons = 90, Neutrons = \(234 - 90 = 144\), \(N/P = 144/90 = 1.600\).
Thus, the neutron-to-proton ratio increases during alpha decay.
In simple words: Alpha decay removes 2 protons and 2 neutrons, which slightly increases the neutron-to-proton ratio.

Exam Tip: Always calculate numerical \(N/P\) values before and after to prove the change.

Question 7. Distinguish between nuclear fusion and fission. Give an example of each.
Answer:
- Nuclear Fission: Splitting of a heavy nucleus into two lighter nuclei. Example: \(_92^{235}\text{U} + _0^1\text{n} \rightarrow _{56}^{141}\text{Ba} + _{36}^{92}\text{Kr} + 3_0^1\text{n} + \text{Energy}\).
- Nuclear Fusion: Combining two light nuclei to form a heavy nucleus. Example: \(_1^2\text{H} + _1^2\text{H} \rightarrow _2^4\text{He} + \text{Energy}\).
In simple words: Fission breaks heavy nuclei apart, while fusion combines light nuclei together.

Exam Tip: Provide balanced chemical-nuclear equations for both examples.

Question 8. A radioactive isotope decays in the following sequence: \(A \xrightarrow{\text{neutron}} A_1 \xrightarrow{\alpha} A_2\). If the mass and atomic numbers of \(A_2\) are 171 and 76 respectively, find mass and atomic number of A and \(A_1\). Which of the three elements are isobars?
Answer:
Given \(A_2\) has mass \(A' = 171\) and atomic number \(Z' = 76\).
Step 1 (\(A_1 \xrightarrow{\alpha} A_2\)): Alpha decay reduces mass number by 4 and atomic number by 2.
So for \(A_1\): Mass \(= 171 + 4 = 175\), Atomic number \(= 76 + 2 = 78\).
Step 2 (\(A \xrightarrow{\text{neutron}} A_1\)): Neutron capture adds 1 to mass number and 0 to atomic number.
So for \(A\): Mass \(= 175 - 1 = 174\), Atomic number \(= 78 - 0 = 78\).
Isobars are elements with the same mass number: \(A\) (mass 174) and \(A_1\) (mass 175)... wait, let's check: \(A\) has mass 174, \(A_1\) has mass 175, \(A_2\) has mass 171. None share the same mass number here, or wait—let's re-verify: Neutron addition to \(A\) gives \(A_1\), so mass of \(A\) is 174, mass of \(A_1\) is 175. Let's check atomic numbers: \(Z_A = 78\), \(Z_{A1} = 78\). Thus \(A\) and \(A_1\) are isotopes (same \(Z\)), not isobars. Wait, is there any isobar pair? Let's re-read standard problem variants; here \(A\) and \(A_1\) have same \(Z = 78\), making them isotopes.
In simple words: Work backwards from \(A_2\) by reversing alpha decay and neutron capture rules.

Exam Tip: Track changes in mass number (\(A\)) and atomic number (\(Z\)) step-by-step.

Question 9. Obtain a relation for total energy of the electron in terms of orbital radius. Show that total energy is negative of K.E. and half of potential energy.
Answer:
Electrostatic force provides the necessary centripetal force:
\(\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} = \frac{mv^2}{r} \implies mv^2 = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}\)
Kinetic energy \(K = \frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{2r}\)
Potential energy \(U = -\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r}\)
Total energy \(E = K + U = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{2r} - \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r} = -\frac{1}{4\pi\varepsilon_0} \frac{e^2}{2r}\)
Thus, \(E = -K = \frac{U}{2}\).
In simple words: Total energy is negative, equal in magnitude to kinetic energy, and half of potential energy.

Exam Tip: Essential 3-mark derivation; practice combining Coulomb force and centripetal force.

Question 10. Draw energy level diagram for hydrogen atom and show the various line spectra originating due to transition between energy levels.
Answer:
(Description of diagram: Horizontal lines representing energy levels \(n = 1, 2, 3, 4, 5, \infty\) with vertical arrows pointing down to \(n = 1\) for Lyman, \(n = 2\) for Balmer, \(n = 3\) for Paschen, etc.).
In simple words: Energy level diagrams show electron transitions producing different spectral series.

Exam Tip: Clearly label Lyman, Balmer, Paschen, and Brackett series transitions.

Question 11. The total energy of an electron in the first excited state of the hydrogen atom is about \(-3.4\text{ eV}\). What is (a) the kinetic energy, (b) the potential energy of the electron? (c) Which of the answers above would change if the choice of the zero of potential energy is changed to (i) \(+0.5\text{ eV}\) (ii) \(-0.5\text{ eV}\)?
Answer:
(a) Kinetic energy \(K = -E = -(-3.4\text{ eV}) = +3.4\text{ eV}\).
(b) Potential energy \(U = 2E = 2 \times (-3.4\text{ eV}) = -6.8\text{ eV}\).
(c) (i) If potential energy zero is shifted to \(+0.5\text{ eV}\), P.E. becomes \(-6.8 - 0.5 = -7.3\text{ eV}\), and Total Energy becomes \(-3.4 - 0.5 = -3.9\text{ eV}\). Kinetic energy remains unchanged (\(+3.4\text{ eV}\)).
(ii) If shifted to \(-0.5\text{ eV}\), P.E. becomes \(-6.8 - (-0.5) = -6.3\text{ eV}\), and Total Energy becomes \(-3.4 - (-0.5) = -2.9\text{ eV}\). Kinetic energy remains unchanged (\(+3.4\text{ eV}\)).
In simple words: Kinetic energy is independent of the choice of zero potential energy, but potential and total energies shift by the chosen value.

Exam Tip: Kinetic energy depends on relative velocity and is invariant to potential energy reference shifts.

Question 12. What is beta decay? Write an equation to represent \(\beta^-\) and \(\beta^+\) decay. Explain the energy distribution curve in \(\beta\) decay.
Answer: Beta decay is a radioactive decay process where a nucleus emits an electron or positron.
\(\beta^-\) decay: \(_Z^A\text{X} \rightarrow _{Z+1}^A\text{Y} + _{-1}^0\text{e} + \bar{\nu}\)
\(\beta^+\) decay: \(_Z^A\text{X} \rightarrow _{Z-1}^A\text{Y} + _{+1}^0\text{e} + \nu\)
The energy distribution curve is continuous because the decay energy is shared dynamically between the beta particle and the neutrino/antineutrino.
In simple words: Beta decay emits particles with a continuous range of energies because energy is shared with a neutrino.

Exam Tip: Mention neutrinos and antineutrinos to explain the continuous energy spectrum.

Question 13. Using energy level diagram show emission \(\gamma\) rays by \(_{27}^{60}\text{Co}\) nucleus and subsequent \(\beta\) decay to obtain \(_{28}^{60}\text{Ni}\).
Answer:
(Description: Cobalt-60 undergoes beta decay to an excited state of Nickel-60, which then transitions to its ground state by emitting one or more gamma-ray photons).
In simple words: Cobalt decays into excited nickel, which then sheds excess energy by emitting gamma rays.

Exam Tip: Refer to NCERT textbook diagram for exact energy level values.

LONG ANSWER QUESTIONS (5 Marks)

Question 1. State Bohr's postulates. Using these postulates, derive an expression for total energy of an electron in the \(n^{\text{th}}\) orbit of an atom. What does negative of this energy signify?
Answer:
Postulates:
1. Electrons revolve around the nucleus in certain stable circular orbits called stationary orbits without radiating energy.
2. Angular momentum of an electron is quantized: \(mvr = \frac{nh}{2\pi}\).
3. Energy is radiated only when an electron jumps from a higher to a lower orbit: \(h\nu = E_2 - E_1\).
Derivation for Total Energy:
Electrostatic force = centripetal force: \(\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} = \frac{mv^2}{r} \implies mv^2 = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r}\).
From Bohr's quantization: \(v = \frac{nh}{2\pi mr}\).
Substituting \(v\) into force equation yields radius \(r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2}\).
Total energy \(E_n = -\frac{me^4}{8n^2 h^2 \varepsilon_0^2} = -\frac{13.6}{n^2}\text{ eV}\).
Significance of negative sign: It signifies that the electron is bound to the nucleus by attractive forces and requires outside energy to escape.
In simple words: Combine Coulomb's law and Bohr's angular momentum condition to derive quantized energy levels.

Exam Tip: This is a major 5-mark derivation. Practice all algebraic steps clearly.

Question 2. Define binding energy of a nucleus. Draw a curve between mass number and average binding energy per nucleon. On the basis of this curve, explain fusion and fission reactions.
Answer:
Binding energy is the energy required to break a nucleus into its constituent protons and neutrons.
Curve features: Binding energy per nucleon increases sharply for light nuclei, reaches a peak around \(A = 56\) (iron) at \(\sim 8.8\text{ MeV}\), and slowly decreases for heavier nuclei.
Explanation of Fission & Fusion:
- Fission: Heavy nuclei (\(A > 200\)) split into medium nuclei to increase binding energy per nucleon, releasing energy.
- Fusion: Very light nuclei (\(A < 20\)) combine to form heavier nuclei, also increasing binding energy per nucleon and releasing energy.
In simple words: Nuclei try to reach the peak of stability at iron by splitting (fission) or combining (fusion).

Question 3. State the law of radioactive disintegration. Hence define disintegration constant and half life period. Establish relation between these terms.
Answer:
Law: Rate of decay is proportional to the number of undecayed atoms: \(-\frac{dN}{dt} = \lambda N\).
Disintegration constant (\(\lambda\)): Reciprocal of time in which atoms reduce to \(1/e\) of initial value.
Half-life (\(T_{1/2}\)): Time taken for half the radioactive atoms to decay.
Relation derivation: Starting from \(N = N_0 e^{-\lambda t}\), substitute \(N = N_0/2\) at \(t = T_{1/2}\) to obtain \(T_{1/2} = \frac{0.693}{\lambda}\).
In simple words: Combine radioactive decay law with half-life definition to link half-life and decay constant.

Exam Tip: Standard 5-mark question combining law statement, definitions, and derivation.

Question 4. What is meant by nuclear fission and nuclear chain reaction? Outline the conditions necessary for nuclear chain reaction.
Answer:
Nuclear fission: Splitting of a heavy nucleus into smaller fragments with release of enormous energy.
Nuclear chain reaction: A self-sustaining series of fission reactions where neutrons released in one fission trigger further fissions.
Conditions necessary:
1. Critical mass: A minimum amount of fissile material is required so neutrons are not lost without causing further fissions.
2. Moderators: Slow neutrons are needed to sustain fission in Uranium-235 effectively.
In simple words: Chain reactions require enough fissile material (critical mass) and slow neutrons to keep the reaction going.

Exam Tip: Mention critical mass, neutron moderation, and multiplication factor \(k\).

Question 5. Briefly explain Rutherford's experiment for scattering of \(\alpha\) particle with the help of a diagram. Write the conclusion made and draw the model suggested.
Answer:
Experiment setup: Alpha particles from a radioactive source are collimated and directed at a thin gold foil. Scattered particles are detected using a zinc sulfide screen and microscope.
Conclusions:
1. Most particles pass undeflected \(\implies\) atom is mostly empty space.
2. Few particles suffer large deflections \(\implies\) positive charge is concentrated in a tiny nucleus.
3. Very few bounce back (\(180^\circ\)) \(\implies\) nucleus contains almost all the mass.
Model suggested: Planetary model where electrons orbit a central positive nucleus.
In simple words: Firing alpha particles at gold foil proved atoms have a tiny, dense, positive center.

Question 6. State law of radioactive decay obtain relation (i) \(N = N_0 e^{-\lambda t}\) (ii) \(R = R_0 e^{-\lambda t}\).
Answer:
Law: \(-\frac{dN}{dt} \propto N \implies \frac{dN}{dt} = -\lambda N\).
(i) Integrating from \(0\) to \(t$:
\(\int_{N_0}^{N} \frac{dN}{N} = -\lambda \int_{0}^{t} dt \implies \ln\left(\frac{N}{N_0}\right) = -\lambda t \implies N = N_0 e^{-\lambda t}\)
(ii) Multiplying both sides by decay constant \(\lambda\):
\(\lambda N = \lambda N_0 e^{-\lambda t} \implies R = R_0 e^{-\lambda t}\)
In simple words: Integrating the decay law gives the exponential formula for nuclei, which multiplies by \(\lambda\) to give the decay rate formula.

Exam Tip: Show both parts clearly to secure full marks.

NUMERICALS

Question 1. The radius of innermost orbit of Hydrogen atom is \(5.3 \times 10^{-1}\text{ m}\) (Note: typical value is \(0.53\text{ \AA} = 5.3 \times 10^{-11}\text{ m}\)). What are the radii of \(n = 2\) and \(n = 3\) orbits?
Answer: Given \(r_1 = 5.3 \times 10^{-11}\text{ m}\).
For \(n = 2\): \(r_2 = r_1 \times 2^2 = 5.3 \times 10^{-11} \times 4 = 2.12 \times 10^{-10}\text{ m}\).
For \(n = 3\): \(r_3 = r_1 \times 3^2 = 5.3 \times 10^{-11} \times 9 = 4.77 \times 10^{-10}\text{ m}\).
In simple words: Multiply the ground state radius by \(n^2\).

Exam Tip: Watch out for powers of 10 in atomic radius calculations.

Question 2. Calculate the radius of the third Bohr orbit of hydrogen atom and energy of electron in that orbit.
Answer:
Radius \(r_3 = 0.53 \times 3^2 = 0.53 \times 9 = 4.77\text{ \AA} = 4.77 \times 10^{-10}\text{ m}\).
Energy \(E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51\text{ eV} = -1.51 \times 1.6 \times 10^{-19}\text{ J} = -2.43 \times 10^{-19}\text{ J}\).
In simple words: Use \(r_n = 0.53 n^2\) \(\text{\AA}\) and \(E_n = -13.6/n^2\) eV.

Exam Tip: Convert eV to Joules correctly using \(1\text{ eV} = 1.6 \times 10^{-19}\text{ J}\).

Question 3. Calculate the longest and shortest wavelength in the Balmer series of Hydrogen atom. Rydberg constant \(= 1.0987 \times 10^7\text{ m}^{-1}\).
Answer: Balmer series formula: \(\frac{1}{\lambda} = R (\frac{1}{2^2} - \frac{1}{n^2})\) where \(n = 3, 4, 5, \dots$
- Longest wavelength (\(n = 3\)):
\(\frac{1}{\lambda_L} = 1.0987 \times 10^7 \left(\frac{1}{4} - \frac{1}{9}\right) = 1.0987 \times 10^7 \times \frac{5}{36} \implies \lambda_L = 6563\text{ \AA}\)
- Shortest wavelength (\(n = \infty\)):
\(\frac{1}{\lambda_S} = 1.0987 \times 10^7 \left(\frac{1}{4} - 0\right) \implies \lambda_S = 3646\text{ \AA}\)
In simple words: Longest wavelength corresponds to the adjacent transition (\(n = 3\)), while shortest corresponds to \(n = \infty\).

Question 4. What will be the distance of closest approach of a \(5\text{ MeV}\) proton as it approaches a gold nucleus?
Answer: Energy \(K = 5\text{ MeV} = 5 \times 10^6 \times 1.6 \times 10^{-19}\text{ J} = 8 \times 10^{-13}\text{ J}\).
For a proton, charge \(q_1 = e\), gold nucleus charge \(q_2 = Ze\) (\(Z = 79\)).
Distance of closest approach \(r_0 = \frac{1}{4\pi\varepsilon_0} \frac{Ze^2}{K}\)
\(r_0 = \frac{9 \times 10^9 \times 79 \times (1.6 \times 10^{-19})^2}{8 \times 10^{-13}} = 4.55 \times 10^{-14}\text{ m}\)
In simple words: Set electrostatic potential energy equal to initial kinetic energy and solve for distance.

Exam Tip: For alpha particles use \(2Ze^2\); for protons use \(Ze^2\).

Question 5. A \(12.5\text{ MeV}\) alpha-particle approaching a gold nucleus is deflected by \(180^\circ\). What is the closest distance to which it approaches the nucleus?
Answer: \(K = 12.5\text{ MeV} = 12.5 \times 10^6 \times 1.6 \times 10^{-19}\text{ J} = 2 \times 10^{-12}\text{ J}\).
\(r_0 = \frac{1}{4\pi\varepsilon_0} \frac{2Ze^2}{K} = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{2 \times 10^{-12}} = 1.82 \times 10^{-14}\text{ m}\)
In simple words: Use the closest approach formula with \(2Ze^2\) for alpha particles.

Exam Tip: Remember factor of 2 for alpha particle charge (\(2e\)).

Question 6. Determine the speed of the electron in \(n = 3\) orbit of hydrogen atom.
Answer: Speed in \(n^{\text{th}}\) orbit is given by \(v_n = \frac{v_1}{n}\) where \(v_1 = 2.18 \times 10^6\text{ m/s}\).
\(v_3 = \frac{2.18 \times 10^6}{3} = 7.29 \times 10^5\text{ m/s}\)
In simple words: Electron speed decreases inversely with principal quantum number \(n\).

Exam Tip: Memorize ground state orbital speed \(2.18 \times 10^6\text{ m/s}\).

Question 7. There are \(4\sqrt{2} \times 10^6\) radioactive nuclei in a given radioactive element. If half-life is 20 seconds, how many nuclei will remain after 10 seconds?
Answer: Given \(N_0 = 4\sqrt{2} \times 10^6\), \(T = 20\text{ s}$, \(t = 10\text{ s}\).
\(N = N_0 \left(\frac{1}{2}\right)^{t/T} = 4\sqrt{2} \times 10^6 \times \left(\frac{1}{2}\right)^{10/20} = 4\sqrt{2} \times 10^6 \times \left(\frac{1}{2}\right)^{1/2}\)
\(N = 4\sqrt{2} \times 10^6 \times \frac{1}{\sqrt{2}} = 4 \times 10^6\)
In simple words: After half a half-life, the remaining nuclei amount to \(N_0 / \sqrt{2}\).

Question 8. The half life of a radioactive substance is 5 hours. In how much time will \(15/16\) of the material decay?
Answer: If \(\frac{15}{16}\) decays, the fraction remaining is \(1 - \frac{15}{16} = \frac{1}{16}\).
Using \(\frac{N}{N_0} = (\frac{1}{2})^{t/T}\):
\(\frac{1}{16} = \left(\frac{1}{2}\right)^{t/5} \implies \left(\frac{1}{2}\right)^4 = \left(\frac{1}{2}\right)^{t/5}\)
\(4 = \frac{t}{5} \implies t = 20\text{ hours}\)
In simple words: Remaining fraction \(\frac{1}{16}\) corresponds to 4 half-lives (\(4 \times 5 = 20\) hours).

Exam Tip: Always find the remaining fraction, not the decayed fraction, for the decay formula.

Question 9. At a given instant, there are \(25\%\) undecayed radioactive nuclei in a sample. After 10 seconds, the number of undecayed nuclei reduces to \(12.5\%\). Calculate the mean life of nuclei.
Exam Tip: Remember \(\tau = \frac{T_{1/2}}{0.693}\).

Question 10. Binding energy of \(_2^4\text{He}\) and \(_3^7\text{Li}\) nuclei are \(27.37\text{ MeV}\) and \(39.4\text{ MeV}\) respectively. Which of the two nuclei is more stable? Why?
Answer:
- For \(_2^4\text{He}\): Binding energy per nucleon \(= \frac{27.37}{4} = 6.84\text{ MeV/nucleon}\).
- For \(_3^7\text{Li}\): Binding energy per nucleon \(= \frac{39.4}{7} = 5.63\text{ MeV/nucleon}\).
\(_2^4\text{He}\) is more stable because it has a higher binding energy per nucleon.
In simple words: Compare binding energy per nucleon; higher value means greater stability.

Exam Tip: Stability depends strictly on binding energy *per nucleon*.

Question 11. Find the binding energy and binding energy per nucleon of nucleus \(_{83}\text{Bi}^{209}\). Given: mass of proton \(= 1.0078254\text{ u}\), mass of neutron \(= 1.008665\text{ u}\). Mass of \(_{83}\text{Bi}^{209} = 208.980388\text{ u}\).
Answer:
Number of protons \(Z = 83\), Number of neutrons \(N = 209 - 83 = 126\).
Total mass of nucleons \(= 83 \times 1.0078254 + 126 \times 1.008665 = 83.6495 + 127.0918 = 210.7413\text{ u}\).
Mass defect \(\Delta m = 210.7413 - 208.980388 = 1.7609\text{ u}\).
Binding energy \(\text{BE} = 1.7609 \times 931.5\text{ MeV} \approx 1639.38\text{ MeV}\).
Binding energy per nucleon \(= \frac{1639.38}{209} \approx 7.84\text{ MeV/nucleon}\).
In simple words: Subtract actual nuclear mass from total nucleon mass to find mass defect, then multiply by 931.5 MeV.

Question 12. Is the fission of iron (\(_{26}\text{Fe}^{56}\)) into (\(_13\text{Al}^{28}\)) as given below possible? \(_26\text{Fe}^{56} \rightarrow _13\text{Al}^{28} + _13\text{Al}^{28} + Q\). Given mass of \(_{26}\text{Fe}^{56} = 55.934940\text{ u}\) and \(_{13}\text{Al}^{28} = 27.98191\text{ u}\).
Answer:
Mass of reactants \(= 55.934940\text{ u}\).
Mass of products \(= 2 \times 27.98191 = 55.96382\text{ u}\).
Mass defect \(\Delta m = 55.934940 - 55.96382 = -0.02888\text{ u}\).
Since Q-value is negative (\(Q = -0.02888 \times 931.5\text{ MeV} < 0\)), this fission reaction is not possible.
In simple words: Negative mass defect means energy must be supplied rather than released, making spontaneous fission impossible.

Exam Tip: Negative Q-value indicates an endothermic reaction that cannot occur spontaneously.

Question 13. Find the maximum energy that \(\beta\)-particle may have in the following decay: \(_{8}\text{O}^{19} \rightarrow _{9}\text{F}^{19} + _{-1}e^0 + \bar{\nu}\). Given \(m(_8\text{O}^{19}) = 19.003576\text{ a.m.u.}\), \(m(_9\text{F}^{19}) = 18.998403\text{ a.m.u.}\), \(m(_{-1}e^0) = 0.000549\text{ a.m.u.}\).
Answer:
Mass of reactants \(= 19.003576\text{ u}\).
Mass of products \(= 18.998403 + 0.000549 = 18.998952\text{ u}\).
Mass difference \(\Delta m = 19.003576 - 18.998952 = 0.004624\text{ u}\).
Maximum energy \(E = 0.004624 \times 931.5\text{ MeV} \approx 4.3049\text{ MeV}\).
In simple words: Convert the mass difference into energy using 931.5 MeV/u.

Exam Tip: Maximum beta energy equals the total Q-value of the decay.

Question 14. The value of wavelength in the lyman series is given as \(\lambda = \frac{93.4 n_i^2}{n_i^2 - 1}\text{ \AA}\). Calculate the wavelength corresponding to transition from energy level 2, 3 and 4. Does wavelength decrease or increase?
Answer:
- For \(n_i = 2\): \(\lambda_{21} = \frac{93.4 \times 2^2}{2^2 - 1} = \frac{93.4 \times 4}{3} = 1218\text{ \AA}\).
- For \(n_i = 3\): \(\lambda_{31} = \frac{93.4 \times 3^2}{3^2 - 1} = \frac{93.4 \times 9}{8} = 1028\text{ \AA}\).
- For \(n_i = 4\): \(\lambda_{41} = \frac{93.4 \times 4^2}{4^2 - 1} = \frac{93.4 \times 16}{15} = 974.3\text{ \AA}\).
Wavelength decreases as \(n_i\) increases (\(\lambda_{41} < \lambda_{31} < \lambda_{21}\)).
In simple words: Higher initial levels mean higher energy transitions, which produce shorter wavelengths.

Question 15. The half life of \(_{92}\text{U}^{238}\) undergoing \(\alpha\) decay is \(4.5 \times 10^9\) years what is the activity of 1g. sample of \(_{92}\text{U}^{238}\).
Answer:
Half-life \(T_{1/2} = 4.5 \times 10^9\text{ years} = 4.5 \times 10^9 \times 3.15 \times 10^7\text{ s} = 1.42 \times 10^{17}\text{ s}\).
Decay constant \(\lambda = \frac{0.693}{T_{1/2}}\).
Number of atoms in 1g of \(\text{U}^{238}\):
\(N = \frac{1}{238} \times 6.023 \times 10^{23} \approx 2.53 \times 10^{21}\text{ atoms}\)
Activity \(R = \lambda N = \frac{0.693}{1.42 \times 10^{17}} \times 2.53 \times 10^{21} = 1.23 \times 10^4\text{ Bq}\).
In simple words: Calculate decay constant and total atoms in 1 gram, then multiply them to find activity.

Exam Tip: Convert years to seconds carefully when calculating activity.