CBSE Class 12 Physics Atoms Nuclei Frequently Asked Questions

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UNIT VIII

ATOMS & NUCLEI

2 MARKS

1. Define disintegration constant and mean life of a radioactive substance. Give the unit of each. [2 Marks]

Answer:
1. Disintegration constant is defined as the reciprocal of the time interval in which the number of undecayed atoms of a radioactive substance falls to 1/e times of its initial value. Its S.I. unit is s^-1.
2. Mean life is defined as the ratio of the total life time of all radioactive atoms to the total number of atoms present initially. Its S.I. unit is second (s).

Teacher's Note:
a) Disintegration constant (\(\lambda\)) represents the probability of decay per unit time.
b) Mean life (\(\tau\)) is related to the disintegration constant by the relation \(\tau = \frac{1}{\lambda}\).
c) Always include the correct dimensional units for both terms to avoid losing half marks.

2. What is impact parameter? What is the value of impact parameter for a head on collision? The sequence of the stepwise decays of radioactive nucleus is:
\(D \xrightarrow{\alpha} D_1 \xrightarrow{\beta} D_2 \xrightarrow{\alpha} D_3 \xrightarrow{\alpha} D_4\)
If the nucleon number and atomic number for \(D_2\) are respectively 176 & 71, what are the corresponding values for D and \(D_4\) nuclei? Justify your answer. [2 Marks]

Answer:
1. Impact parameter is defined as the perpendicular distance of the velocity vector of the alpha particle from the central line of the nucleus when it is far away.
2. For a head-on collision, the value of the impact parameter is zero.
3. For \(D_2\), Mass number A = 176, Atomic number Z = 71.
4. Decay sequence: \(D \xrightarrow{\alpha} D_1 \xrightarrow{\beta} D_2 \xrightarrow{\alpha} D_3 \xrightarrow{\alpha} D_4\).
5. Moving backward from \(D_2\) to D: \(D_2\) to \(D_1\) involves beta decay (\(\Delta A = 0, \Delta Z = +1\)), so for \(D_1\) (A = 176, Z = 70); \(D_1\) to D involves alpha decay (\(\Delta A = -4, \Delta Z = -2\)), so for D (A = 180, Z = 68).
6. Moving forward from \(D_2\) to \(D_4\): \(D_2\) to \(D_3\) involves alpha decay (\(\Delta A = -4, \Delta Z = -2\)), giving \(D_3\) (A = 172, Z = 69); \(D_3\) to \(D_4\) involves alpha decay (\(\Delta A = -4, \Delta Z = -2\)), giving \(D_4\) (A = 168, Z = 67).

Teacher's Note:
a) An alpha particle emission decreases mass number by 4 and atomic number by 2.
b) A beta particle emission increases atomic number by 1 while keeping mass number unchanged.
c) Trace the chain carefully step by step in both directions from the given intermediate nucleus.

3. Draw a diagram to show the variation of binding energy per nucleon with mass number for different nuclei. Explain with the help of this plot the release of energy in the processes of nuclear fission and fusion? [2 Marks]

Answer:
1. In nuclear fission, a heavy nucleus splits into two medium-sized nuclei, resulting in an increase in total binding energy per nucleon, which releases energy.
2. In nuclear fusion, two light nuclei combine to form a heavier nucleus, also resulting in an increase in binding energy per nucleon and release of energy.

Teacher's Note:
a) The binding energy curve peaks around mass number 56 corresponding to iron (Fe).
b) Systems naturally move towards higher stability and lower potential energy, releasing the difference as nuclear energy.

4. The value of ground state energy of hydrogen atom is: -13.6 eV; (i) what does the negative sign signify? & (ii) How much energy is required to take an electron in this atom from the ground state to the first excited state? [2 Marks]

Answer:
1. The negative sign signifies that the electron is bound to the nucleus by an attractive electrostatic force.
2. Ground state energy \(E_1 = -13.6\) eV, First excited state energy \(E_2 = \frac{-13.6}{2^2} = -3.4\) eV.
3. Energy required = \(E_2 - E_1 = -3.4 - (-13.6) = 10.2\) eV.

Teacher's Note:
a) Bound systems always possess negative total mechanical energy.
b) The first excited state corresponds to the principal quantum number n = 2.
c) Always subtract ground state energy correctly considering sign conventions.

5. Give one point of difference between 'nuclear fission' & 'nuclear fusion'. Will neutron to proton ratio increase or decrease in a nucleus when: (i) an electron, (ii) a positron is emitted? [2 Marks]

Answer:
1. Difference: Nuclear fission involves splitting a heavy nucleus into lighter nuclei, whereas nuclear fusion involves combining light nuclei to form a heavier nucleus.
2. (i) When an electron (\(\beta^-\) decay) is emitted, a neutron converts into a proton (\(n \rightarrow p + e^- + \bar{\nu}\)), so the neutron-to-proton ratio decreases.
3. (ii) When a positron (\(\beta^+\) decay) is emitted, a proton converts into a neutron (\(p \rightarrow n + e^+ + \nu\)), so the neutron-to-proton ratio increases.

Teacher's Note:
a) Fission is common in extremely heavy elements, while fusion powers stellar bodies.
b) Keep track of nucleon transformation equations to determine neutron-to-proton ratios accurately.

6. Sketch the graph showing the variation of potential energy of a pair of nucleons as a function of their separation. Write three characteristic properties of nuclear force which distinguish it from the electrostatic force. [2 Marks]

Answer:
1. Nuclear forces are strongly attractive at distances greater than 0.8 fm and become repulsive at distances less than 0.8 fm.
2. Three distinguishing properties: (a) Nuclear force is charge-independent, unlike electrostatic force acting only between charges; (b) Nuclear force is short-ranged (effective up to a few fm), whereas electrostatic force is long-ranged; (c) Nuclear force is much stronger than electrostatic force.

Teacher's Note:
a) The potential energy curve reaches a minimum at about 0.8 fm denoting stable equilibrium.
b) Electrostatic forces obey the inverse square law, but nuclear forces do not follow a simple central-force mathematical expression.

7. State two characteristics of nuclear force. Why does the binding energy per nucleon decrease with increase in mass number for heavy nuclei like \(^{235}\text{U}\)? [2 Marks]

Answer:
1. Two characteristics: (a) Nuclear forces are saturated forces; (b) They are non-central forces.
2. For heavy nuclei, the nuclear size becomes large compared to the short range of nuclear forces. Each nucleon interacts only with its immediate neighbors, whereas the repulsive electrostatic force between protons is long-ranged and acts cumulatively, lowering the net binding energy per nucleon.

Teacher's Note:
a) Saturation property means a nucleon interacts only with a limited number of closest neighbors.
b) Coulomb repulsion among a large number of protons in heavy nuclei destabilizes the nucleus.

8. Define activity of a substance. State its S.I unit. Derive an expression for activity of a substance. [2 Marks]

Answer:
1. Activity of a radioactive substance is defined as the rate of disintegration of the radioactive substance. S.I. unit is Becquerel (Bq) or decay per second.
2. According to radioactive decay law, the rate of disintegration is proportional to the number of undecayed atoms present: \(-\frac{dN}{dt} = \lambda N\).
3. Integrating the law gives \(N = N_0 e^{-\lambda t}\).
4. Substituting N into the decay rate gives activity \(R = -\frac{dN}{dt} = \lambda N_0 e^{-\lambda t} = R_0 e^{-\lambda t}\).

Teacher's Note:
a) Activity decreases exponentially with time following the radioactive decay law.
b) \(R_0 = \lambda N_0\) represents the initial activity at time t = 0.

SECTION A

3 MARKS

9. State the basic postulates of Bohr's atomic model & derive an expression for the energy of an electron in any orbit of hydrogen atom. [3 Marks]

Answer:
1. Postulates of Bohr's atomic model:
    (a) An electron can revolve in certain stable, non-radiating circular orbits called stationary orbits.
    (b) The angular momentum of the revolving electron is an integral multiple of \(\frac{h}{2\pi}\), so that \(mvr = \frac{nh}{2\pi}\).
    (c) Energy is radiated only when an electron jumps from a higher stationary orbit to a lower one.
2. Total energy \(E = K + U = -\frac{ke^2}{2r}\) where \(k = \frac{1}{4\pi\varepsilon_0}\).
3. From radius expression \(r = \frac{n^2h^2}{4\pi^2mke^2}\), substituting \(r\) into energy gives \(E_n = -\frac{2\pi^2mk^2e^4}{n^2h^2}\).

Teacher's Note:
a) Emphasize the quantization condition of angular momentum as proposed by Bohr.
b) Watch out for sign conventions in potential energy and total energy expressions.
c) The negative sign in total energy signifies that the electron is bound to the nucleus.
d) Verify that total energy is inversely proportional to \(n^2\).

10. Derive an expression for the radius of stationary orbit. Prove that the various stationary orbits are not equally spaced. [3 Marks]

Answer:
1. Equating electrostatic force to centripetal force: \(\frac{ke^2}{r^2} = \frac{mv^2}{r}\).
2. Using Bohr's quantization condition \(v = \frac{nh}{2\pi mr}\) and substituting it yields \(r_n = \frac{n^2h^2}{4\pi^2mke^2}\).
3. Since radius \(r_n \propto n^2\), the spacing between successive orbits \(r_{n+1} - r_n \propto (2n + 1)\) increases with \(n\), proving they are not equally spaced.

Teacher's Note:
a) Combine Coulomb's law and Newton's second law with angular momentum quantization.
b) A common mistake is forgetting that radius depends quadratically on the principal quantum number.
c) The non-linear spacing is a fundamental result distinguishing quantum orbits from planetary orbits.

11. Derive mathematical expressions for: (i) kinetic energy, & (ii) potential energy of an electron revolving in an orbit of radius 'r'; how does the potential energy change with increase in principal quantum number (n) for the electron and why? [3 Marks]

Answer:
1. Kinetic energy \(K = \frac{1}{2}mv^2 = \frac{ke^2}{2r}\) using centripetal force balance.
2. Potential energy \(U = -\frac{ke^2}{r}\) due to electrostatic attraction between nucleus and electron.
3. Since \(U = -\frac{ke^2}{r}\) and \(r \propto n^2\), \(U\) becomes less negative and increases (increases algebraically) as \(n\) increases because the electron moves farther from the positive nucleus.

Teacher's Note:
a) Remember the crucial relation \(U = -2K\) and total energy \(E = -K\).
b) Students often confuse algebraic increase with magnitude decrease when dealing with negative potential energy.
c) Connect the variation of potential energy directly to the increasing radius of higher stationary orbits.

12. Define the decay constant for a radioactive sample. Which of the following radiations \(\alpha\), \(\beta\), & \(\gamma\) rays are: (i) similar to X-rays? (ii) easily absorbed by matter? & (iii) similar in nature to cathode rays? [3 Marks]

Answer:
1. Decay constant is defined as the reciprocal of time during which the number of undecayed atoms falls to \(\frac{1}{e}\) times its initial value, or the probability of decay per unit time.
2. (i) \(\gamma\)-rays are similar to X-rays as both are high-frequency electromagnetic radiation.
3. (ii) \(\alpha\)-rays are easily absorbed by matter due to their large mass and high ionizing power.
4. (iii) \(\beta\)-rays are similar in nature to cathode rays as both consist of high-speed fast-moving electrons.

Teacher's Note:
a) Clearly state the definition of decay constant using the radioactive decay law equation.
b) Ensure all three sub-parts (i), (ii), and (iii) are distinctly matched to the correct radiation type.
c) Note that \(\beta\)-particles are nuclear electrons whereas cathode rays are extranuclear, but their physical nature as electrons is identical.

13. Define the terms: half life period and decay constant of a radioactive sample. Derive the relation between these terms. [3 Marks]

Answer:
1. Half-life period (\(T_{1/2}\)) is the time in which half of the active nuclei in a radioactive sample decay.
2. Decay constant (\(\lambda\)) is the relative rate of disintegration per unit time.
3. Using decay law \(N = N_0 e^{-\lambda t}\), at \(t = T_{1/2}\), \(N = \frac{N_0}{2}\), giving \(\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}}\), which leads to \(T_{1/2} = \frac{\ln(2)}{\lambda} = \frac{0.693}{\lambda}\).

Teacher's Note:
a) Use the exponential decay law as the starting point for the derivation.
b) Take natural logarithms correctly on both sides to isolate the half-life expression.
c) Memorize the numerical value \(0.693\) for quick calculation in numerical problems.

14. In Rutherford's scattering experiment, mention two important conclusions which can be drawn by studying the scattering of \(\alpha\) particles by an atom. Draw the schematic arrangement of Geiger and Marsden experiment showing the scattering of \(\alpha\) particle by a thin foil of gold. How does one get the information regarding the size of the nucleus in this experiment? [3 Marks]

Answer:
1. Conclusions: (i) Most of the space inside an atom is empty. (ii) The entire positive charge and almost all the mass of the atom are concentrated in a very small central core called the nucleus.
2. [Figure: Schematic diagram showing a radioactive source in a lead block, collimating slits, a thin gold foil, a movable zinc sulphide screen, and a microscope to detect scattered alpha particles]
3. The distance of closest approach (\(r_0\)) gives an upper limit to the size of the nucleus, obtained by equating kinetic energy of the alpha particle to electrostatic potential energy at the turning point.

Teacher's Note:
a) List precise conclusions derived from the scattering angle distribution.
b) Describe the experimental setup clearly including source, gold foil, and detector screen.
c) Explain the energy conservation principle used to determine the nuclear size estimate.

15. Sketch the energy level diagram for hydrogen atom. Mark the transitions corresponding to Lyman and Balmer series. [3 Marks]

Answer:
1. [Figure: Energy level diagram of hydrogen showing horizontal lines for principal quantum numbers n = 1, 2, 3, 4, 5, infinity with corresponding energy values -13.6 eV, -3.4 eV, -1.51 eV, -0.85 eV, 0 eV, and vertical arrows indicating downward transitions to n = 1 for Lyman series and to n = 2 for Balmer series]
2. Lyman series transitions originate from higher levels (\(n \ge 2\)) terminating at \(n = 1\) in the ultraviolet region.
3. Balmer series transitions originate from higher levels (\(n \ge 3\)) terminating at \(n = 2\) in the visible region.

Teacher's Note:
a) Ensure energy levels are drawn closer together as \(n\) increases.
b) Clearly label the ground state and excited states with correct energy values in eV.
c) Differentiate between the spectral regions (UV for Lyman, Visible for Balmer) to earn full credit.

16. Prove that the instantaneous rate of change of the activity of a radioactive substance is inversely proportional to the square of its half life. [3 Marks]

Answer:
1. Activity \(R = \lambda N = \lambda N_0 e^{-\lambda t}\).
2. Rate of change of activity is \(\frac{dR}{dt} = \frac{d}{dt} (\lambda N_0 e^{-\lambda t}) = -\lambda^2 N_0 e^{-\lambda t} = -\lambda^2 R\).(magnitude is \(\lambda^2 R\))
3. Since decay constant \(\lambda = \frac{0.693}{T_{1/2}}\), substituting \(\lambda\) gives \(\left|\frac{dR}{dt}\right| = \left(\frac{0.693}{T_{1/2}}\right)^2 R\), proving that the rate of change of activity is inversely proportional to \((T_{1/2})^2\).

Teacher's Note:
a) Differentiate the activity equation with respect to time using the chain rule.
b) Substitute the standard relationship linking decay constant and half-life.
c) State the final inverse square proportionality clearly to complete the proof.