JEE Physics Atoms MCQs Set 04

Answer: (b) 18750 Å and 973 Å

Explanation:
1. Photon energy = 12400/975 = 12.7 eV, which raises the electron from n = 1 to n = 4 (E₄ − E₁ = 12.75 eV)
2. Longest wavelength is for 4 → 3: 1/λ = R(1/9 − 1/16) = 7R/144, so λ = 144/(7 × 1.097 × 10⁷) ≈ 1.875 × 10⁻⁶ m = 18750 Å
3. Shortest wavelength is for 4 → 1: 1/λ = R(1 − 1/16) = 15R/16, so λ = 16/(15 × 1.097 × 10⁷) ≈ 973 Å.

Teacher's Note:
1. First find the highest level reached by comparing the photon energy with 13.6(1 − 1/n²).
2. Longest wavelength means smallest energy gap, shortest means the largest gap.

Answer: (d) 3

Explanation:
1. The photon takes the atom to n = 4
2. In absorption the atom starts in the ground state, so only transitions 1 → 2, 1 → 3 and 1 → 4 are possible
3. Number of absorption lines = 3 (the 6 lines belong to the emission spectrum).

Teacher's Note:
1. Absorption lines start from the ground state only, because almost all atoms are in n = 1.
2. Emission after excitation to n = 4 gives n(n − 1)/2 = 6 lines.

Answer: (a) 1 : 9

Explanation:
1. For a hydrogen-like atom 1/λ = RZ²(1 − 1/4), so λ ∝ 1/Z²
2. Li²⁺ has Z = 3 and deuterium has Z = 1 (the small change due to nuclear mass is ignored)
3. λ(Li²⁺) : λ(D) = 1/9 : 1 = 1 : 9.

Teacher's Note:
1. Deuterium is chemically hydrogen with Z = 1.
2. For the same transition the wavelength falls as Z² rises.

Answer: (c) 16 : 1

Explanation:
1. Orbit radius r ∝ n², so area ∝ r² ∝ n⁴
2. First excited state is n = 2 and ground state is n = 1
3. Ratio = 2⁴ : 1 = 16 : 1.

Teacher's Note:
1. The first excited state is n = 2.
2. Area depends on the fourth power of n.

Answer: (d) 12.09 V

Explanation:
1. Energy of the n = 3 level is −13.6/9 = −1.51 eV
2. Excitation energy = E₃ − E₁ = −1.51 − (−13.6) = 12.09 eV
3. So the excitation potential is 12.09 V.

Teacher's Note:
1. Excitation potential is measured from the ground state, not from infinity.
2. Ionisation potential is 13.6 V, excitation to n = 2 is 10.2 V.

Answer: (b) 122 nm

Explanation:
1. 1/λ = R(1/1² − 1/2²) = 3R/4
2. λ = 4/(3 × 1.097 × 10⁷) = 1.215 × 10⁻⁷ m ≈ 122 nm.

Teacher's Note:
1. This is the first line of the Lyman series, in the ultraviolet region.
2. Energy of the photon is 10.2 eV.

Answer: (a) 49.6 pm and 99.2 pm

Explanation:
1. Initial kinetic energy = 50 keV
2. First collision: loses 25 keV, so λ₁ = 12400/25000 Å = 0.496 Å = 49.6 pm
3. Second collision: remaining 25 keV, loses half = 12.5 keV, so λ₂ = 12400/12500 = 0.992 Å = 99.2 pm.

Teacher's Note:
1. Use λ(Å) = 12400/E(eV) for each photon separately.
2. The energy lost in the second collision is half of what is left, not half of the original.

Answer: (c) 18.6 kV and 0.667 Å

Explanation:
1. λmin = 12400/V, so when V becomes 2V/5 the new wavelength is 2.5 λ₁
2. Shift = 2.5λ₁ − λ₁ = 1.5λ₁ = 1 Å, so λ₁ = 0.667 Å
3. V = 12400/0.667 ≈ 18600 V = 18.6 kV.

Teacher's Note:
1. λmin is inversely proportional to the applied voltage.
2. Write the shift as a multiple of the original wavelength.

Answer: (b) 7.25 × 10¹⁸ Hz and 0.413 Å

Explanation:
1. eV = hν_max, so ν_max = (1.6 × 10⁻¹⁹ × 3 × 10⁴)/(6.63 × 10⁻³⁴) = 7.25 × 10¹⁸ Hz
2. λmin = c/ν_max = 3 × 10⁸/7.25 × 10¹⁸ = 4.13 × 10⁻¹¹ m = 0.413 Å.

Teacher's Note:
1. The photon takes all the kinetic energy of the electron at the cut-off.
2. Check with λmin = 12400/30000 Å = 0.413 Å.

Answer: (d) 27.55 keV and about 27.5 kV

Explanation:
1. E_max = hc/λmin = 12400/0.45 = 27550 eV = 27.55 keV
2. An electron gains this energy through V = E/e, so V ≈ 27.5 kV.

Teacher's Note:
1. Maximum photon energy corresponds to the minimum wavelength.
2. Energy in eV equals the accelerating voltage in volts.

Answer: (c) 0.4 mm⁻¹

Explanation:
1. I = I₀e^−μx and I/I₀ = 1/4, so μx = ln 4 = 1.386
2. μ = 1.386/3.5 mm = 0.396 mm⁻¹ ≈ 0.4 mm⁻¹.

Teacher's Note:
1. Take the natural logarithm of the intensity ratio.
2. Keep the thickness in the same unit as 1/μ.

Answer: (b) 1.52 Å

Explanation:
1. Moseley's law gives λ ∝ 1/(Z − 1)²
2. λ(Cu) = 0.71 × (41/28)² = 0.71 × 2.144 = 1.52 Å.

Teacher's Note:
1. For K_α lines use the screening constant 1, so the factor is (Z − 1).
2. A lighter target gives a longer wavelength.

Answer: (a) 1.2 × 10⁵ eV

Explanation:
1. Maximum photon energy equals the kinetic energy gained by an electron, eV
2. Energy = 120 kV × e = 1.2 × 10⁵ eV.

Teacher's Note:
1. An electron accelerated through V volts gains V electron volts.
2. The photon cannot have more energy than the electron.

Answer: (a) 0.49 Å

Explanation:
1. λmin = 12400/V (Å, with V in volts)
2. λmin = 12400/25000 = 0.496 Å, the nearest option is 0.49 Å.

Teacher's Note:
1. The constant hc/e is 12400 V Å (about 12.4 kV Å).
2. Higher voltage gives a shorter cut-off wavelength.

Answer: (b) 20 Å

Explanation:
1. Bragg's law: 2d sin θ = nλ
2. λ is largest when sin θ = 1 and n = 1, so λmax = 2d = 20 Å.

Teacher's Note:
1. Diffraction is impossible if λ > 2d.
2. The same crystal can diffract shorter wavelengths at smaller angles.

Answer: (a) X-rays

Explanation:
1. X-ray wavelengths (about 1 Å) are comparable to the spacing of atoms in crystals
2. So crystals act as natural diffraction gratings for X-rays.

Teacher's Note:
1. Diffraction needs a wavelength comparable to the spacing.
2. This is the basis of X-ray crystallography.

Answer: (b) more than 50%

Explanation:
1. A higher potential difference makes the X-rays harder, with shorter wavelength
2. Harder X-rays are absorbed less, so a larger fraction passes through.

Teacher's Note:
1. Absorption coefficient falls as the wavelength falls.
2. Penetrating power increases with voltage.

Answer: (b) 0.5 Å

Explanation:
1. K_α is a characteristic line, determined by the energy levels of the target atom
2. It does not depend on the accelerating voltage, so it stays at 0.5 Å.

Teacher's Note:
1. Only the continuous spectrum and the cut-off wavelength depend on the voltage.
2. Characteristic lines act like a fingerprint of the target element.

Answer: (d) the intensity of X-rays increases

Explanation:
1. More filament current heats the filament more and releases more electrons per second
2. More electrons strike the target, so more photons are produced and the intensity increases
3. The energy of each photon (set by the voltage) does not change.

Teacher's Note:
1. Filament current controls intensity; the tube voltage controls hardness.
2. Wavelength and penetration depend only on voltage.

Answer: (c) X-rays

Explanation:
1. α-rays, β-rays and cathode rays are streams of charged particles and are deflected
2. X-rays are electromagnetic waves (photons) with no charge, so they are not deflected.

Teacher's Note:
1. Only charged particles are deflected in electric and magnetic fields.
2. γ-rays are also undeflected.

Answer: (b) 6.2 kV

Explanation:
1. V = 12400/λmin = 12400/2 = 6200 V = 6.2 kV.

Teacher's Note:
1. Use eV = hc/λmin.
2. Shorter wavelength needs higher voltage.

Answer: (c) of all wavelengths above a minimum wavelength

Explanation:
1. The continuous spectrum starts sharply at the cut-off wavelength λmin = hc/eV and extends to longer wavelengths
2. Sharp characteristic lines are superimposed on it, so the X-rays are not monochromatic.

Teacher's Note:
1. There is a minimum wavelength but no sharp maximum.
2. Characteristic lines appear above a critical voltage.

Answer: (d) neither electric charge nor magnetic moment

Explanation:
1. A photon is electrically neutral
2. Photons have no magnetic moment either, so they are not deflected by fields.

Teacher's Note:
1. Electromagnetic waves carry energy and momentum but no charge.
2. This is why X-rays pass undeviated through fields.

Answer: (c) the energy difference between the energy levels of hydrogen is much less

Explanation:
1. Characteristic X-rays come from transitions to inner shells of heavy atoms, with energy differences of keV
2. In hydrogen the largest energy gap is only 13.6 eV, so the photons are in the UV and visible region.

Teacher's Note:
1. X-ray photon energy is thousands of times larger than the energy of hydrogen spectral lines.
2. Inner shell vacancies need heavy atoms.