JEE Physics Atoms MCQs Set 07

Answer: (a) λmin changes

Explanation:
1. λmin = hc/eV changes with the electron energy
2. Kα and Kβ are fixed by the energy levels of molybdenum, so they remain unchanged.

Teacher's Note:
1. Characteristic lines do not shift with voltage.
2. Only the continuous spectrum shifts.

Answer: (d) \(\lambda_{min} \propto \frac{1}{V}\)

Explanation:
1. λmin = hc/eV
2. So λmin is inversely proportional to V.

Teacher's Note:
1. Double the voltage and the cut-off wavelength halves.
2. It is not a square-root relation.

Answer: (a) 1.45 × 10¹⁹ Hz

Explanation:
1. ν_max = eV/h = (1.6 × 10⁻¹⁹ × 6 × 10⁴)/(6.63 × 10⁻³⁴) = 1.45 × 10¹⁹ Hz.

Teacher's Note:
1. Frequency in Hz equals voltage × 2.42 × 10¹⁴.
2. X-ray frequencies are of the order of 10¹⁸ to 10¹⁹ Hz.

Answer: (b) 122 nm

Explanation:
1. The least energetic Lyman photon comes from n = 2 → n = 1
2. 1/λ = R(1 − 1/4) = 3R/4, so λ = 4/(3 × 1.097 × 10⁷) = 121.5 nm ≈ 122 nm.

Teacher's Note:
1. Least energy means the longest wavelength of the series.
2. Lyman series lies in the ultraviolet region.

Answer: (a) 4 : 1

Explanation:
1. Shortest wavelength is for n = ∞: Balmer 1/λ = R/4 so λ = 4/R; Lyman 1/λ = R so λ = 1/R
2. Ratio = 4 : 1.

Teacher's Note:
1. The series limit is the transition from n = ∞.
2. Balmer limit is 365 nm and Lyman limit is 91 nm.

Answer: (b) 13.6 eV

Explanation:
1. In a Bohr orbit KE = −E (total energy)
2. Total energy of n = 1 is −13.6 eV, so KE = +13.6 eV.

Teacher's Note:
1. KE is always positive.
2. PE = −27.2 eV and total = KE + PE = −13.6 eV.

Answer: (c) \(-\frac{Z^2}{n^2}\)

Explanation:
1. E_n = −13.6 Z²/n² eV, which is proportional to −Z²/n².

Teacher's Note:
1. Energy grows as Z² and falls as n².
2. The negative sign shows that the electron is bound.

Answer: (a) \(\frac{n^2}{Z}\)

Explanation:
1. r_n = 0.529 n²/Z Å, so r ∝ n²/Z.

Teacher's Note:
1. The radius grows as n² and shrinks as Z rises.
2. For hydrogen r₁ = 0.529 Å.

Answer: (c) 54.4 V

Explanation:
1. Ionisation energy ∝ Z²
2. For He⁺, Z = 2, so the value is 13.6 × 4 = 54.4 V.

Teacher's Note:
1. He⁺ is hydrogen-like with one electron.
2. The ionisation potential of neutral helium is 24.6 V.

Answer: (c) The Paschen series lies in the infrared region

Explanation:
1. Lyman lines are in the ultraviolet and Balmer lines are in the visible region
2. The Paschen (n → 3), Brackett (n → 4) and Pfund (n → 5) series all lie in the infrared region.

Teacher's Note:
1. Each series consists of discrete lines that converge to a limit.
2. Only Balmer lines are visible.

Answer: (a) \(\frac{1}{n}, \frac{1}{n^2}, n^2\)

Explanation:
1. v = 2.18 × 10⁶ Z/n m/s, so v ∝ 1/n
2. E_n = −13.6 Z²/n² eV, so E ∝ 1/n² (in magnitude)
3. r ∝ n².

Teacher's Note:
1. Speed falls and radius rises as n increases.
2. Energy magnitude falls as 1/n².

Answer: (b) −1.51 eV

Explanation:
1. E₃ = −13.6/3² = −1.51 eV.

Teacher's Note:
1. Divide 13.6 by n².
2. n = 2 gives −3.4 eV.

Answer: (c) e⁵

Explanation:
1. Current I = e × frequency = e v/(2πr)
2. v ∝ e² and r ∝ 1/e², so v/r ∝ e⁴
3. I ∝ e × e⁴ = e⁵.

Teacher's Note:
1. Count the powers of e in the speed and the radius.
2. The factor e from the charge itself is easy to forget.

Answer: (c) A is true, but R is false.

Explanation:
1. Photoelectric emission of electrons by X-rays can only be explained if X-rays are photons (particles of energy): A is true
2. X-ray photons are electrically neutral, not positively charged, so R is false.

Teacher's Note:
1. Photons carry energy and momentum but no charge.
2. X-rays are not deflected by electric or magnetic fields.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. X-rays need transitions between inner shells with energy gaps of keV, but hydrogen's largest gap is only 13.6 eV, so A is true
2. The hydrogen atom does have a single electron, so R is true
3. The single electron is not the reason; the small energy gaps are, so R is not the explanation.

Teacher's Note:
1. Characteristic X-rays arise from inner-shell vacancies in heavy atoms.
2. Hydrogen has no inner shells.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Electrons slow down by different amounts in the field of target nuclei and radiate photons of different energies (bremsstrahlung): A is true
2. A decelerating charge radiates electromagnetic waves: R is true and explains A.

Teacher's Note:
1. This radiation is called braking radiation.
2. It gives a continuous spectrum with a sharp cut-off wavelength.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. X-ray photons have energies from about 100 eV to 100 keV, which is high compared to visible light: A is true
2. α-particles are helium nuclei: R is true
3. The two statements are unrelated, so R is not the explanation.

Teacher's Note:
1. Photon energy E = hc/λ is large for the short wavelengths of X-rays.
2. α-particles are massive charged particles, not photons.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Excited atoms (with inner-shell vacancies) emit X-rays: A is true
2. Excited nuclei emit γ-rays: R is true
3. R describes a different process (nuclear de-excitation), so it does not explain A.

Teacher's Note:
1. X-rays come from the electron shells, γ-rays from the nucleus.
2. Both are electromagnetic radiation.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Crystal diffraction needs wavelengths comparable to atomic spacings (about 1 Å), as X-rays have: A is true
2. R states exactly this, so R is the correct explanation of A.

Teacher's Note:
1. Bragg's law 2d sin θ = nλ needs λ ≤ 2d.
2. Visible light is too long for this purpose.

Answer: (c) A is true, but R is false.

Explanation:
1. Soft and hard X-rays have different frequencies (and wavelengths) but are the same kind of radiation, and A is accepted as true
2. All X-rays travel at the speed of light in vacuum, so hard X-rays are not faster: R is false.

Teacher's Note:
1. Speed of electromagnetic waves in vacuum is the same for all frequencies.
2. Hardness refers to penetrating power.

Answer: (d) A is false, but R is true.

Explanation:
1. Ejecting the electron needs at least 13.6 eV, so 10.6 eV cannot ionise the atom (it can only excite it to n = 2 using 10.2 eV): A is false
2. The energy difference between n = 1 and n = 2 is indeed 10.2 eV: R is true.

Teacher's Note:
1. Ionisation needs 13.6 eV from the ground state.
2. An incident electron can give up part of its energy for excitation.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Chemical energy is the energy stored in the arrangement of the electrons of atoms and molecules: A is true
2. It arises from the potential and kinetic energy of the electrons: R is true and explains A.

Teacher's Note:
1. Chemical bonds store energy through electrons.
2. Energy is released when electrons rearrange to lower energy states.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Bohr assumed electrons with exact radius and momentum in orbits, which violates ΔxΔp ≥ h/4π: A is true
2. The model could not predict relative line intensities: R is true
3. The failure to explain intensities does not explain the inconsistency with uncertainty principle, so R is not the explanation.

Teacher's Note:
1. Quantum mechanics replaced fixed orbits by probability orbitals.
2. Other failures of the Bohr model include multi-electron atoms.