Questions & explanations
1. Which one of the following statements correctly defines the ionization energy and excitation energy for a hydrogen atom?
- Ionization energy is the minimum energy required to remove the electron from the ground state, while excitation energy is the energy required to raise the electron from a lower energy level to a higher energy level.
- Ionization energy is the energy released when an electron is captured by a proton to form a hydrogen atom, while excitation energy is the energy emitted when an electron jumps from a higher to a lower energy level.
- Ionization energy is the energy required to excite the electron from the first orbit (n=1) to the second orbit (n=2), while excitation energy is the energy required to completely remove the electron from the atom.
- Ionization energy and excitation energy are equal for hydrogen because it has only one electron, both being 13.6 eV.
Answer: Ionization energy is the minimum energy required to remove the electron from the ground state, while excitation energy is the energy required to raise the electron from a lower energy level to a higher energy level.
Ionization energy is the energy needed to remove an electron from an atom in its ground state (for hydrogen, 13.6 eV). Excitation energy is the energy needed to promote an electron from a lower orbit to a higher orbit (e.g., 10.2 eV from n=1 to n=2). Option A correctly states both definitions. Option B describes electron affinity (ionization energy is not released) and emission (excitation energy is not emitted). Option C swaps the two definitions. Option D is false because ionization energy (ground state to infinity) and excitation energy (ground state to n=2) are not equal; only the ground state energy is 13.6 eV. Hence, A is correct.
2. Which of the following is a major limitation of Rutherford’s nuclear model of the atom?
- It could not explain the scattering of α-particles by a thin gold foil.
- It predicted that electrons would rapidly lose energy and spiral into the nucleus.
- It assumed that the atom has a positively charged nucleus at its centre.
- It failed to account for the existence of protons in the nucleus.
Answer: It predicted that electrons would rapidly lose energy and spiral into the nucleus.
Rutherford’s model, based on classical electrodynamics, predicted that an accelerating electron (moving in a curved path) would continuously radiate energy, causing its orbit to shrink and eventually fall into the nucleus. This instability contradicted the stability of atoms, which is a key limitation. The other options are either correct features (nucleus) or not a limitation (α-scattering was explained, and protons were not yet known).
3. Which of the following is a correct postulate of Bohr’s model of the hydrogen atom?
- Electrons can revolve in any orbit around the nucleus without radiating energy.
- The angular momentum of the electron in an allowed orbit is an integral multiple of h/2π.
- The energy of the electron increases when it jumps from a higher to a lower orbit.
- The radius of the electron’s orbit is directly proportional to the principal quantum number n.
Answer: The angular momentum of the electron in an allowed orbit is an integral multiple of h/2π.
Bohr’s second postulate states that only those orbits are allowed in which the angular momentum of the electron is an integer multiple of h/(2π). Option A is wrong because only stationary orbits (not any orbit) are non-radiating. Option C is wrong because energy decreases when going from higher to lower orbit. Option D is wrong because radius is proportional to n², not n.
4. The Lyman series of hydrogen corresponds to transitions where the final state is n = 1. Which of the following transitions produces the longest wavelength in the Lyman series?
- From n = 2 to n = 1
- From n = 3 to n = 1
- From n = 4 to n = 1
- From n = ∞ to n = 1
Answer: From n = 2 to n = 1
The wavelength of a spectral line is given by 1/λ = R(1/nf² – 1/ni²). For Lyman series (nf = 1), the longest wavelength corresponds to the smallest energy difference, which occurs for the smallest ni, i.e., ni = 2. The transition from n=2 to n=1 gives the longest wavelength (121.6 nm). Other transitions have shorter wavelengths as ni increases.
5. In a hydrogen atom, the ground state energy is –13.6 eV. An electron is in an excited state with energy –1.51 eV. What is this excited state?
- First excited state (n = 2)
- Second excited state (n = 3)
- Third excited state (n = 4)
- Fourth excited state (n = 5)
Answer: Second excited state (n = 3)
Energy levels in hydrogen: n = 1: –13.6 eV; n = 2: –3.4 eV; n = 3: –1.51 eV; n = 4: –0.85 eV. The state with energy –1.51 eV corresponds to n = 3, which is the second excited state (first excited is n=2). Options A, C, D give energies –3.4 eV, –0.85 eV, and –0.54 eV respectively, so only n=3 matches.
6. A hydrogen atom makes a transition from the n = 4 state to the n = 2 state. If the energy of the n = 4 state is –0.85 eV and that of n = 2 state is –3.40 eV, what is the energy and wavelength range of the emitted photon? (Given: 1 eV = 1.6 × 10⁻¹⁹ J, h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, and 1 nm = 10⁻⁹ m.)
- Energy = 2.55 eV, wavelength = 487 nm (visible)
- Energy = 4.25 eV, wavelength = 292 nm (ultraviolet)
- Energy = 2.55 eV, wavelength = 487 nm (infrared)
- Energy = 0.85 eV, wavelength = 1460 nm (infrared)
Answer: Energy = 2.55 eV, wavelength = 487 nm (visible)
ΔE = E₄ – E₂ = (–0.85) – (–3.40) = 2.55 eV. λ = hc/ΔE = (1240 eV·nm)/(2.55 eV) ≈ 486 nm, which lies in the visible region (Balmer series). Option B gives wrong energy; Option C is correct energy but wrong region; Option D uses wrong energy and wavelength. Only option A matches.
7. In Bohr’s model of hydrogen, the radius of the electron in the nth orbit (rn) is related to the radius of the first orbit (r₁) as:
- rn = n r₁
- rn = n² r₁
- rn = r₁ / n
- rn = r₁ / n²
Answer: rn = n² r₁
The radius of the nth orbit in hydrogen is given by rn = (n²h²ε₀)/(πme²) = n² r₁, where r₁ ≈ 0.529 Å. This follows from the quantisation of angular momentum and Coulomb’s law. The other options do not match the n² dependence.
8. According to Bohr’s model, the angular momentum of an electron in the third allowed orbit of a hydrogen atom is:
- h/(2π)
- h/π
- 3h/(2π)
- 2h/π
Answer: 3h/(2π)
Bohr’s quantisation condition states that angular momentum L = n(h/2π). For n = 3, L = 3h/(2π). The other options correspond to n = 1, n = 2, and n = 4 respectively.
9. The energy of an electron in the nth orbit of a hydrogen atom is given by En = –13.6/n² eV. What is the energy of the electron in the n = 4 orbit?
- –3.40 eV
- –1.51 eV
- –0.85 eV
- –0.54 eV
Answer: –0.85 eV
Using En = –13.6/n² eV, for n = 4: E₄ = –13.6/16 = –0.85 eV. Option A is for n = 2, option B is for n = 3, and option D is for n = 5. Only –0.85 eV is correct.