JEE Physics Atoms MCQs Set 08

Answer: (d) has a constant value of 7.2 eV Å

Explanation:
1. Binding energy is \( E_n = \dfrac{13.6}{n^2}\ \text{eV} \) and the orbit radius is \( r_n = 0.53\,n^2\ \text{Å} \).
2. Their product is \( 13.6\times0.53 = 7.2\ \text{eV Å} \), independent of n.

Teacher's Note:
1. The n² in the radius cancels the 1/n² in the energy.
2. The product is a constant (for hydrogen, Z = 1).

Answer: (a) Helium ion

Explanation:
1. For a hydrogen-like ion \( \dfrac{1}{\lambda} = RZ^2\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right) \).
2. Wavelengths four times shorter need \( Z^2 = 4 \), so Z = 2.
3. That is He⁺ (the helium ion).

Teacher's Note:
1. For the same transition, wavelength is proportional to \( 1/Z^2 \).
2. Only one-electron species such as He⁺ and Li²⁺ have hydrogen-like spectra.

Answer: (b) E = 3.4 eV, λ = 6.6 × 10⁻¹⁰ m

Explanation:
1. In a Bohr orbit the kinetic energy equals the magnitude of the total energy, so \( E = 3.4\ \text{eV} \).
2. \( \lambda = \dfrac{h}{\sqrt{2mE}} = \dfrac{12.27}{\sqrt{3.4}}\ \text{Å} \approx 6.6\ \text{Å} = 6.6\times10^{-10}\ \text{m} \).

Teacher's Note:
1. For bound electrons \( K = -E_{total} \) and \( U = 2E_{total} \).
2. Use \( \lambda\,(\text{Å}) = 12.27/\sqrt{V} \) when the energy is in eV.

Answer: (c) line 3

Explanation:
1. By Moseley's law \( \sqrt{f} = a(Z-b) \), with b close to 1 for the K series and close to 7.4 for the L series. The pair of lines meeting the Z-axis farther to the right (lines 3 and 4) is therefore the L series.
2. Within a series, the β line has a larger frequency than the α line for the same Z, so its line is steeper.
3. Of lines 3 and 4, line 3 is the steeper, so Lβ is line 3.

Teacher's Note:
1. Moseley's graph of \( \sqrt{f} \) against Z is a straight line for each X-ray line.
2. Within a series, the β line has the greater slope than the α line.

Answer: (c) it emits a 2.55 eV photon

Explanation:
1. \( \Delta E = 13.6\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right) = 13.6\times\dfrac{3}{16} = 2.55\ \text{eV} \).
2. The electron moves to a lower energy level, so this energy is emitted as a photon.

Teacher's Note:
1. Falling to a lower orbit always releases a photon.
2. Its total energy decreases by the photon energy.

Answer: (d) 14.82

Explanation:
1. Spacing \( \Delta E \propto \dfrac{1}{n_1^2}-\dfrac{1}{n_2^2} \) (the factor \( 13.6Z^2 \) cancels).
2. For 3 and 4: \( \dfrac19-\dfrac1{16} = \dfrac{7}{144} \). For 8 and 9: \( \dfrac1{64}-\dfrac1{81} = \dfrac{17}{5184} \).
3. Ratio \( = \dfrac{7}{144}\times\dfrac{5184}{17} = \dfrac{252}{17} \approx 14.82 \).

Teacher's Note:
1. Energy levels crowd together as n increases.
2. The factor Z² is common and cancels in a ratio.

Answer: (d) 7/108

Explanation:
1. The maximum wavelength of a series comes from the transition between its first two levels.
2. Lyman: \( \dfrac{1}{\lambda_L} = R\left(1-\dfrac14\right) = \dfrac{3R}{4} \). Paschen: \( \dfrac{1}{\lambda_P} = R\left(\dfrac19-\dfrac1{16}\right) = \dfrac{7R}{144} \).
3. Ratio \( \dfrac{\lambda_L}{\lambda_P} = \dfrac{7R/144}{3R/4} = \dfrac{7}{108} \).

Teacher's Note:
1. Longest wavelength means smallest energy gap, which is the first line of the series.
2. Lyman ends at n = 1 and Paschen ends at n = 3.

Answer: (c) \( \dfrac{1}{\lambda_1} = \dfrac{1}{\lambda_2} + \dfrac{1}{\lambda_3} \)

Explanation:
1. The shortest wavelength λ₁ comes from the largest energy gap, between the highest and lowest levels.
2. This gap equals the sum of the two smaller gaps, whose wavelengths are λ₂ and λ₃.
3. Since \( E = hc/\lambda \), \( \dfrac{1}{\lambda_1} = \dfrac{1}{\lambda_2} + \dfrac{1}{\lambda_3} \).

Teacher's Note:
1. Energies of transitions add, so reciprocals of wavelengths add.
2. Largest energy means smallest wavelength.

Answer: (a) One photon of energy 10.2 eV and an electron having energy 1.4 eV

Explanation:
1. The 10.2 eV photon excites the atom from n = 1 to n = 2.
2. An excited state lasts only about 10⁻⁸ s, so within a microsecond the atom has returned to the ground state, emitting a 10.2 eV photon.
3. The 15 eV photon then meets a ground-state atom and ionises it (13.6 eV), leaving the electron with \( 15-13.6 = 1.4\ \text{eV} \).

Teacher's Note:
1. A microsecond is much longer than the lifetime of an excited state.
2. So the atom is in the ground state when the second photon arrives.

Answer: (c) 3.4 eV

Explanation:
1. The first excited state is n = 2.
2. Its energy is \( -\dfrac{13.6}{2^2} = -3.4\ \text{eV} \), so 3.4 eV is needed to remove the electron.

Teacher's Note:
1. Ionisation energy from level n is \( 13.6/n^2 \) eV for hydrogen.
2. The ground state is n = 1 and the first excited state is n = 2.

Answer: (d) Doubly ionised lithium

Explanation:
1. The radius of the first Bohr orbit is \( r_1 = \dfrac{0.53}{Z}\ \text{Å} \).
2. Hydrogen and deuterium have Z = 1, He⁺ has Z = 2 and Li²⁺ has Z = 3.
3. The largest Z, that is Li²⁺, gives the smallest radius.

Teacher's Note:
1. Orbit radius is inversely proportional to Z.
2. The small difference in mass between hydrogen and deuterium has a negligible effect.