Questions & explanations
1. In Bohr's model, the Rydberg formula 1/λ = R(1/n_f² - 1/n_i²) is derived. Which step is correct?
- From angular momentum quantization and Coulomb force, we get E_n = -13.6/n² eV. Then using hν = E_i - E_f gives the formula.
- From angular momentum quantization and Coulomb force, we get E_n = +13.6/n² eV. Then using hν = E_i - E_f gives the formula.
- From angular momentum quantization and Coulomb force, we get E_n = -13.6/n eV. Then using hν = E_i - E_f gives the formula.
- From angular momentum quantization and Coulomb force, we get E_n = -13.6 n² eV. Then using hν = E_i - E_f gives the formula.
Answer: From angular momentum quantization and Coulomb force, we get E_n = -13.6/n² eV. Then using hν = E_i - E_f gives the formula.
Bohr's second postulate quantizes angular momentum: mvr = nħ. Coulomb force provides centripetal force: ke²/r² = mv²/r. Solving gives radius r_n = n²a₀ and total energy E_n = -13.6/n² eV. The third postulate says hν = E_i - E_f. Substituting E_n = -R h c / n² yields the Rydberg formula. Option a correctly states the energy expression.
2. The Rydberg constant R is given by m e⁴ / (8 ε₀² h³ c). Using m = 9.1×10⁻³¹ kg, e = 1.6×10⁻¹⁹ C, ε₀ = 8.85×10⁻¹² C²/N·m², h = 6.63×10⁻³⁴ J·s, c = 3×10⁸ m/s, the value of R (in m⁻¹) is closest to:
- -1.097 × 10⁷
- 1.097 × 10⁷
- 1.097 × 10⁵
- 1.097 × 10⁹
Answer: 1.097 × 10⁷
Using R = m e⁴ / (8 ε₀² h³ c), substitute values: e⁴ = (1.6×10⁻¹⁹)⁴ = 6.5536×10⁻⁷⁶, ε₀² = (8.85×10⁻¹²)² = 7.832×10⁻²³, h³ = (6.63×10⁻³⁴)³ = 2.915×10⁻¹⁰⁰. Denominator = 8 × 7.832×10⁻²³ × 2.915×10⁻¹⁰⁰ × 3×10⁸ = 5.48×10⁻¹¹⁴. Numerator = 9.1×10⁻³¹ × 6.5536×10⁻⁷⁶ = 5.96×10⁻¹⁰⁶. R = 5.96×10⁻¹⁰⁶ / 5.48×10⁻¹¹⁴ = 1.088×10⁷ ≈ 1.097×10⁷ m⁻¹.
3. According to Bohr's correspondence principle, for large n, the frequency of radiation emitted in a transition from n to n-1 approaches:
- the orbital frequency of the electron in the nth orbit
- the orbital frequency of the electron in the (n-1)th orbit
- the average of orbital frequencies in the nth and (n-1)th orbits
- the frequency of the electron's revolution in the ground state
Answer: the orbital frequency of the electron in the nth orbit
Bohr's correspondence principle states that for large quantum numbers, quantum predictions match classical physics. For a transition from n to n-1 with large n, the emitted radiation frequency equals the classical orbital frequency of the electron in the nth orbit, as both scale as 1/n³ and become equal in the limit.
4. The emission spectrum of hydrogen consists of bright lines on a dark background. This indicates that hydrogen atoms emit light only at
- all wavelengths
- wavelengths in the ultraviolet region only
- wavelengths in the visible region only
- specific wavelengths
Answer: specific wavelengths
The emission spectrum of hydrogen is a line spectrum, meaning it contains only certain discrete wavelengths. This is because electrons in hydrogen atoms can only occupy specific energy levels, and transitions between these levels produce photons of fixed energies, hence specific wavelengths.
5. A beam of electrons with kinetic energy 12.5 eV strikes ground-state hydrogen atoms. What is the maximum principal quantum number to which the atoms can be excited, and how many distinct spectral lines can subsequently appear?
- n_max = 3; 3 lines (Lyman: 2, Balmer: 1)
- n_max = 2; 1 line (Lyman: 1)
- n_max = 4; 6 lines (Lyman: 3, Balmer: 2, Paschen: 1)
- n_max = 3; 6 lines (Lyman: 3, Balmer: 2, Paschen: 1)
Answer: n_max = 3; 3 lines (Lyman: 2, Balmer: 1)
Using Bohr's model, excitation energies from n=1: to n=2 is 10.2 eV, to n=3 is 12.09 eV, to n=4 is 12.75 eV. Since 12.5 eV > 12.09 eV but < 12.75 eV, maximum excitation is to n=3. From n=3, possible downward transitions: 3→2, 3→1, 2→1, giving 3 distinct spectral lines (Lyman: 2, Balmer: 1).
6. According to Bohr's first postulate, an electron in a hydrogen atom revolves in certain stationary orbits without radiating energy. This postulate was introduced to explain
- the stability of the atom
- the line spectrum of hydrogen
- the quantisation of angular momentum
- the energy of the ground state
Answer: the stability of the atom
Bohr's first postulate states that electrons can revolve in certain allowed orbits without radiating energy. This directly addresses the classical collapse problem, where an accelerating electron would radiate and spiral into the nucleus, thus explaining the stability of the atom.
7. The ratio of the wavelength of the first Balmer line to the first Lyman line in hydrogen is:
- 5/27
- 27/5
- 4/3
- 3/4
Answer: 27/5
Using the Rydberg formula 1/λ = R(1/n_f² - 1/n_i²). For first Lyman line: n_i=2, n_f=1 → 1/λ_L = R(1 - 1/4)=3R/4. For first Balmer line: n_i=3, n_f=2 → 1/λ_B = R(1/4 - 1/9)=5R/36. Taking ratio λ_B/λ_L = (3R/4)/(5R/36) = (3/4)*(36/5)=27/5. So λ_B/λ_L = 27/5.
8. In the Lyman series of hydrogen, the shortest wavelength line corresponds to which transition?
- n=2 to n=1
- n=3 to n=1
- n=∞ to n=1
- n=4 to n=1
Answer: n=∞ to n=1
The Lyman series involves transitions to n=1. The shortest wavelength (series limit) occurs when the electron falls from infinity (n=∞) to n=1, giving the maximum energy difference. Using the Rydberg formula, 1/λ = R(1/1² - 1/∞²) = R, so λ = 1/R ≈ 912 Å.
9. The Rydberg constant for deuterium is slightly larger than that for hydrogen. What is the ratio R_D / R_H? (Given: m_p = 1836 m_e)
- 1.00109
- 1.00054
- 0.99973
- 1.00027
Answer: 1.00027
Rydberg constant R ∝ reduced mass μ. For H: μ_H = m_e m_p/(m_e+m_p) = (1836/1837)m_e. For D: μ_D = m_e·2m_p/(m_e+2m_p) = (3672/3673)m_e. Ratio R_D/R_H = μ_D/μ_H = (3672/3673)/(1836/1837) = (3672×1837)/(3673×1836) = 2×1837/3673 = 3674/3673 ≈ 1.00027.
10. At room temperature, which spectral series is predominantly observed in the absorption spectrum of atomic hydrogen?
- All series equally
- Balmer series
- Paschen series
- Lyman series
Answer: Lyman series
At room temperature, most hydrogen atoms are in the ground state (n=1). Absorption requires upward transitions from n=1, which belong to the Lyman series. Other series require atoms to be in excited states, which are negligible at room temperature.
11. In the Franck–Hertz experiment, the collector current drops sharply at an accelerating voltage of 4.9 V. What does this indicate?
- Mercury atoms are ionized at 4.9 eV.
- The work function of the collector is 4.9 eV.
- Electrons are absorbed by mercury atoms at 4.9 eV.
- The first excitation energy of mercury is 4.9 eV.
Answer: The first excitation energy of mercury is 4.9 eV.
The drop in current occurs because electrons lose 4.9 eV of kinetic energy in inelastic collisions with mercury atoms, exciting the atoms to their first excited state. This confirms discrete energy levels with a first excitation energy of 4.9 eV.
12. Which of the following transitions in hydrogen gives the H-α line of the Balmer series?
- n=4 to n=2
- n=6 to n=2
- n=5 to n=2
- n=3 to n=2
Answer: n=3 to n=2
The Balmer series corresponds to transitions ending at n=2. H-α is the first line (longest wavelength) of the series, given by the transition from n=3 to n=2. Using Rydberg formula, 1/λ = R(1/2² - 1/3²) yields λ ≈ 656 nm, which is the H-α line.