CBSE Class 12 Physics Important Questions Atoms and Nuclei

CBSE Class 12 Physics Important Questions Atoms & Nuclei. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Important questions for CBSE Class 12 Physics on atoms and nuclei, covering difficult topics such as the energy of electrons in the Bohr model of the hydrogen atom, decay constant, half-life, average life, beta decay energy, spectral lines of hydrogen, and activity of radioactive material, with answers and teacher's notes.

DIFFICULT TOPICS

1. Energy of Bohr model hydrogen atom.
2. Average life, half life, numerical.

1. Define decay constant (λ) & half-life period (t_h) and hence derive the relationship between decay constant and half-life period.

Answer:
1. Decay constant (λ) is the constant of proportionality in the decay law, that is, the number of atoms decaying per unit time is λ times the number of atoms present.
2. Half-life period (t_h) is the time in which half of the initial number of radioactive atoms in a sample decays.
3. From the decay law, N = N₀e^-λt.
4. At t = \( \frac{1}{\lambda} \), N = \( \frac{N_0}{e} \) = 0.368N₀, so the decay constant is the reciprocal of the time in which the number of atoms falls to 36.8% of its initial value.
5. At t = t_h, N = \( \frac{N_0}{2} \), so \( \frac{N_0}{2} = N_0 e^{-\lambda t_h} \).
6. Therefore \( e^{\lambda t_h} = 2 \), and taking logarithms, λt_h = log_e2 = 2.303 log₁₀2 = 2.303 × 0.3010 = 0.693.
7. Hence, t_h = \( \frac{0.693}{\lambda} \).

Teacher's Note:
a) The decay constant has the unit of inverse time, such as per second.
b) Half-life is always about 0.693 divided by λ, so a larger λ means a shorter half-life.
c) Do not confuse the time for N to fall to half with the time for N to fall to 36.8%.

2. Define average or mean life of the radioactive element and derive its relationship with decay constant and half life time. [3 Marks]

Answer:
1. Average or mean life of a radioactive element is the ratio of the total life time of all the atoms of the element to the total number of atoms initially present in the sample.
2. Let N₀ be the number of atoms at t = 0, and N be the number at time t. The number of atoms decaying in a small time dt is dN, and each of these atoms has an age of about t.
3. Total age of the dN atoms = t dN, so the total life time of all atoms is \( \int_{0}^{N_0} t\,dN \).
4. Average life τ = \( \frac{\int_{0}^{N_0} t\,dN}{N_0} \).
5. Using dN = -λN₀e^-λtdt, the integral becomes τ = \( \frac{1}{N_0}\int_{0}^{\infty} \lambda N_0\, t\, e^{-\lambda t}\,dt \) = λ∫₀^∞ t e^-λt dt.
6. Integrating, τ = \( \frac{1}{\lambda} \).
7. Since t_h = \( \frac{0.693}{\lambda} \), we get τ = \( \frac{t_h}{0.693} \) = 1.44 t_h.

Teacher's Note:
a) Average life is the reciprocal of the decay constant, so it is always longer than the half-life.
b) The average life is about 1.44 times the half-life, which is a common examination result.
c) Students often write the average life as the half-life; keep these two quantities separate.

3. Using Bohr's postulates derive the expression for the total energy of electron in stationery orbits of hydrogen atom. Hence show that the total energy in the stationery orbits are in the ratio 1: 1/4 : 1/9 ……
What is the significance of negative value of energy?

Answer:
1. Electron of charge -e revolves in a circular orbit of radius r around a nucleus of charge +Ze. The Coulomb force provides the centripetal force, so \( \frac{Ze^2}{4\pi\varepsilon_0 r^2} = \frac{mv^2}{r} \).
2. Kinetic energy of the electron: KE = \( \frac{1}{2}mv^2 = \frac{Ze^2}{8\pi\varepsilon_0 r} \).
3. Potential energy of the electron: PE = \( -\frac{Ze^2}{4\pi\varepsilon_0 r} \).
4. Total energy E = KE + PE = \( -\frac{Ze^2}{8\pi\varepsilon_0 r} \).
5. Bohr's first postulate, mvr = n\( \frac{h}{2\pi} \), gives the radius r_n = \( \frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2} \).
6. Substituting r_n gives E_n = \( -\frac{mZ^2e^4}{8\varepsilon_0^2 h^2 n^2} \).
7. Putting the standard values and converting to electron volts, E_n = -13.6 \( \frac{Z^2}{n^2} \) eV.
8. For hydrogen, Z = 1, so E₁ = -13.6 eV, E₂ = -3.4 eV and E₃ = -1.51 eV.
9. Since E_n is proportional to \( \frac{1}{n^2} \), the magnitudes of the energies are in the ratio E₁ : E₂ : E₃ = 1 : \( \frac{1}{4} \) : \( \frac{1}{9} \).
10. Significance of negative energy: the electron is bound to the nucleus and cannot leave the atom unless energy is supplied, and 13.6 eV is needed to remove it from the ground state.

Teacher's Note:
a) The total energy is half of the potential energy, and it is negative because the electron is bound.
b) The ratio 1 : 1/4 : 1/9 comes from the factor \( \frac{1}{n^2} \), so the energy values get closer together for higher orbits.
c) The energy values are in electron volts for hydrogen, and the value -13.6 eV is the ionisation energy from the ground state.
d) Students often write the energy as positive, which is wrong for a bound electron.

4. A nucleus ²³₁₀Ne undergoes β⁻ decay and becomes ²³₁₁Na. Calculate the maximum kinetic energy of electrons emitted assuming that the daughter nucleus and anti − neutrino carry negligible kinetic energy.
mass of ²³₁₀Ne = 22.994466 u
mass of ²³₁₁Na = 22.989770 u
1 u = 931.5 MeV/c²

Answer:
1. The decay is ²³₁₀Ne → ²³₁₁Na + β⁻ + ν̄ + Q.
2. Mass defect Δm = m(Ne) − m(Na) = 22.994466 u − 22.989770 u = 0.004696 u.
3. Energy released Q = Δm × 931.5 MeV = 0.004696 × 931.5 MeV = 4.374 MeV.
4. When the daughter nucleus and the anti-neutrino carry negligible kinetic energy, the β⁻ electron carries the whole Q.
5. Hence, the maximum kinetic energy of the electron is 4.374 MeV.

Teacher's Note:
a) The printed label "Q = kinetic energy of the daughter nucleus" is wrong; Q is the total energy released in the decay, shared among the electron, the antineutrino and the recoiling nucleus.
b) Atomic masses can be used here because the electron masses on both sides cancel out.
c) The electron has the maximum energy only when the antineutrino takes no energy, which is why the spectrum of beta particles is continuous.

5. The energy level diagram of an element is given below. Identify, by doing necessary calculations for each, which transition corresponds to the emission of a spectral line of wavelength 102.7 nm.

[Figure: Energy level diagram with four levels, from the top: -0.85 eV, -1.5 eV, -3.4 eV and -13.6 eV. Four transitions are marked: A from -1.5 eV to -0.85 eV, B from -3.4 eV to -0.85 eV, C from -3.4 eV to -1.5 eV, and D from -13.6 eV to -1.5 eV.]

Answer:
1. Transition A: E₁ = -1.5 eV and E₂ = -0.85 eV, so E = E₂ - E₁ = -0.85 - (-1.5) = 0.65 eV. Wavelength λ = \( \frac{hc}{E} \) = \( \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{0.65 \times 1.6 \times 10^{-19}} \) = 1911.35 nm.
2. Transition B: E₁ = -3.4 eV and E₂ = -0.85 eV, so E = 2.55 eV. Wavelength λ = 487.2 nm.
3. Transition C: E₁ = -3.4 eV and E₂ = -1.5 eV, so E = 1.9 eV. Wavelength λ = 653.88 nm.
4. Transition D: E₁ = -13.6 eV and E₂ = -1.5 eV, so E = -1.5 - (-13.6) = 12.1 eV. Wavelength λ = \( \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{12.1 \times 1.6 \times 10^{-19}} \) = 102.67 nm.
5. Only transition D gives a wavelength of 102.7 nm.
6. Hence, transition D corresponds to the spectral line of wavelength 102.7 nm.

Teacher's Note:
a) The printed value for A, 19311.35 nm, is a slip; the correct wavelength is 1911.35 nm.
b) Wavelength is inversely proportional to energy, so the largest energy gap gives the shortest wavelength.
c) Check the unit conversion from metres to nanometres at the end of each calculation.

6. (i) Define 'activity' of a radioactive material and write its S.I. units.

Answer:
1. Activity of a radioactive material is the rate of decay, that is, the number of disintegrations per second in a sample containing one or more radionuclides.
2. Its S.I. unit is becquerel (Bq), where 1 Bq is one disintegration per second.

Teacher's Note:
a) Activity is written as R or A, and it equals λN.
b) The older unit of activity is the curie, where 1 curie = 3.7 × 10¹⁰ Bq.

(ii) Plot a graph showing variation of activity of a given radioactive sample with time.

Answer:
1. Activity R = λN = λN₀e^-λt, so it decreases exponentially with time.
2. The graph starts at the initial activity R₀ at t = 0 and falls steeply at first.
3. The curve then flattens and approaches the time axis, but never touches it.

Teacher's Note:
a) The graph is an exponential decay curve, not a straight line.
b) The slope of the curve at any point gives the rate at which activity falls.
c) The curve never reaches zero, so it approaches the time axis asymptotically.

(iii) The sequence of stepwise decay of a radioactive nucleus is D →(α) D₁ →(β⁻) D₂. If the atomic number and mass number of D₂ are 71 and 176 respectively, what are their corresponding values of D?

Answer:
1. In β⁻ decay the mass number stays the same and the atomic number increases by 1. So D₁ has atomic number 70 and mass number 176.
2. In α decay the mass number falls by 4 and the atomic number falls by 2. So D has atomic number 70 + 2 = 72 and mass number 176 + 4 = 180.
3. Hence, D is ¹⁸⁰₇₂D, and D₁ is ¹⁷⁶₇₀D₁.

Teacher's Note:
a) The printed working gives atomic number 74 for D; that is wrong, and the correct value is 72, because alpha decay lowers the atomic number by 2.
b) The printed D₁ with atomic number 72 is also wrong, since a beta decay raises the atomic number by 1 and so D₁ must have 70.
c) Always work backwards from the final nucleus when the parent is not given.