Atoms – Bohr's Model and Hydrogen Spectra: NEET MCQs with Answers

15 free questions on Atoms – Bohr's Model and Hydrogen Spectra for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Bohr's Model and Hydrogen Spectra: Bohr model, energy levels, hydrogen spectrum.

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Question 1

Previous-year question
For which one of the following is the Bohr model not valid? [NEET 2020]
  1. Singly ionised helium atom (He⁺)
  2. Deuteron atom
  3. Singly ionised neon atom (Ne⁺)
  4. Hydrogen atom
Show answer
Answer: (C) Singly ionised neon atom (Ne⁺)
Explanation: Bohr’s model applies only to one-electron (hydrogen-like) systems. He⁺, deuterium and hydrogen each have a single electron, but Ne⁺ still has nine electrons, so electron–electron repulsion makes the model invalid.

Question 2

Previous-year question
The total energy of an electron in an orbit of an atom is –3.4 eV. Its kinetic and potential energies are respectively [NEET 2019]
  1. –3.4 eV, –3.4 eV
  2. –3.4 eV, –6.8 eV
  3. 3.4 eV, –6.8 eV
  4. 3.4 eV, 3.4 eV
Show answer
Answer: (C) 3.4 eV, –6.8 eV
Explanation: In a Bohr orbit K = –E and U = 2E. With E = –3.4 eV we get K = +3.4 eV and U = –6.8 eV. (Check: K + U = 3.4 – 6.8 = –3.4 eV = E.)

Question 3

Previous-year question
The ratio of the kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is [NEET 2018]
  1. 1 : 1
  2. 1 : –1
  3. 1 : –2
  4. 2 : –1
Show answer
Answer: (B) 1 : –1
Explanation: In a Bohr orbit K = –E, so K : E = 1 : –1. (Equivalently, the total energy is the negative of the kinetic energy.)

Question 4

Previous-year question
The ratio of the wavelengths of the last line of the Balmer series and the last line of the Lyman series is [NEET 2017]
  1. 1
  2. 4
  3. 0.5
  4. 2
Show answer
Answer: (B) 4
Explanation: Balmer limit: λ = 4/R. Lyman limit: λ = 1/R. Their ratio is therefore 4 : 1.

Question 5

Previous-year question
Given Rydberg’s constant as 10⁷ m⁻¹, the wave number of the last line of the Balmer series in the hydrogen spectrum is [NEET 2016]
  1. 0.025 × 10⁴ m⁻¹
  2. 0.5 × 10⁷ m⁻¹
  3. 0.25 × 10⁷ m⁻¹
  4. 2.5 × 10⁷ m⁻¹
Show answer
Answer: (C) 0.25 × 10⁷ m⁻¹
Explanation: The last line of the Balmer series is the series limit (∞ → 2): ṽ = R(1/2²) = 10⁷/4 = 0.25 × 10⁷ m⁻¹.

Question 6

Previous-year question
In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is [AIPMT 2015]
  1. 9/4
  2. 27/5
  3. 5/27
  4. 4/9
Show answer
Answer: (C) 5/27
Explanation: Longest Lyman line: λ = 4/(3R). Longest Balmer line: λ = 36/(5R). Ratio = (4/3)/(36/5) = 20/108 = 5/27.

Question 7

Previous-year question
A hydrogen atom in the ground state is excited by monochromatic radiation of λ = 975 Å. The number of spectral lines in the resulting emission spectrum will be [AIPMT 2014]
  1. 3
  2. 2
  3. 6
  4. 10
Show answer
Answer: (C) 6
Explanation: Photon energy = 12400/975 ≈ 12.72 eV, so Eₙ = –13.6 + 12.72 = –0.88 eV ⇒ n = 4. Number of lines = 4 × 3/2 = 6.
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Question 8

An electron is in the nth orbit of a hydrogen-like ion of atomic number Z. According to Bohr's theory, how do its orbital frequency f, angular momentum L, orbital magnetic moment μ and equivalent current I depend on Z and n?
  1. f ∝ Z/n², L ∝ n, μ ∝ n/Z, I ∝ Z/n²
  2. f ∝ Z²/n³, L ∝ n, μ ∝ n, I ∝ Z²/n³
  3. f ∝ n³/Z², L ∝ n, μ ∝ Z²/n³, I ∝ n³/Z²
  4. f ∝ Z²/n³, L ∝ Zn, μ ∝ Zn, I ∝ Z/n²
Show answer
Answer: (B) f ∝ Z²/n³, L ∝ n, μ ∝ n, I ∝ Z²/n³
Explanation: Since v ∝ Z/n and r ∝ n²/Z, the frequency f = v/2πr varies as Z²/n³. The angular momentum L = nh/2π does not involve Z. The magnetic moment μ = (e/2m)L therefore also varies as n alone. The equivalent current is I = ef, so it varies as Z²/n³.

Question 9

For the electron in the nth Bohr orbit of hydrogen, let E, p and r be the total energy, the magnitude of the linear momentum and the orbital radius. Match each quantity in List I with its dependence on n in List II. List I: (A) E × r (B) E × p × r (C) p × r (D) E × p. List II: (P) Proportional to n. (Q) Independent of n. (R) Proportional to 1/n. (S) Proportional to 1/n³.
  1. A–Q, B–R, C–P, D–S
  2. A–Q, B–S, C–P, D–R
  3. A–Q, B–P, C–R, D–S
  4. A–R, B–Q, C–P, D–S
Show answer
Answer: (A) A–Q, B–R, C–P, D–S
Explanation: In the Bohr model E ∝ 1/n², p ∝ 1/n and r ∝ n². So E × r does not depend on n, E × p × r varies as 1/n, and E × p varies as 1/n³. The product p × r varies as n, as expected, since it is the angular momentum.

Question 10

In a hydrogen atom, transition D takes the electron from the level n = 2 up to the level n = 4, and transition E takes the electron from the level n = 1 to n = ∞. Transitions D and E respectively represent
  1. An emission line of the Paschen series, and absorption of a wavelength longer than the Lyman series limit
  2. An absorption line of the Lyman series, and the series limit of the Paschen series
  3. An absorption line of the Balmer series, and ionisation of the atom from its ground state
  4. An emission line of the Balmer series, and the longest-wavelength line of the Lyman series
Show answer
Answer: (C) An absorption line of the Balmer series, and ionisation of the atom from its ground state
Explanation: An upward jump needs a photon to be absorbed, and a line whose lower level is n = 2 belongs to the Balmer series. Taking the electron from n = 1 to n = ∞ removes it completely from a ground-state atom. This needs 13.6 eV, the ionisation energy of hydrogen.

Question 11

For the hydrogen spectrum, what is the ratio of the longest wavelength to the shortest wavelength within the Paschen series?
  1. 9/5
  2. 25/9
  3. 16/7
  4. 16/9
Show answer
Answer: (C) 16/7
Explanation: Paschen lines end on n = 3. The longest wavelength comes from n = 4 to n = 3, with 1/λ = R(1/9 − 1/16) = 7R/144. The shortest comes from n = ∞ to n = 3, with 1/λ = R/9 = 16R/144. The ratio of the wavelengths is therefore 16/7.

Question 12

Statement I: Bohr's model gives the wavelengths of the hydrogen spectral lines well, but it cannot say why some lines are more intense than others. Statement II: Applied without change, the same model gives the energy levels of the neutral helium atom correctly.
  1. Statement I is true and Statement II is false
  2. Statement I is true and Statement II is true
  3. Statement I is false and Statement II is false
  4. Statement I is false and Statement II is true
Show answer
Answer: (A) Statement I is true and Statement II is false
Explanation: Bohr's model predicts the frequencies of hydrogen lines but says nothing about their relative intensities, so the first statement is true. The model works only for hydrogen-like systems with a single electron. In neutral helium each electron also feels the other electron, which the model does not handle, so the second statement is false.

Question 13

In the de Broglie picture of the Bohr atom, the n-th orbit holds exactly n electron wavelengths around its circumference. What is the ratio of the electron's de Broglie wavelength in the n = 3 orbit of hydrogen to that in the n = 1 orbit of He⁺?
  1. 3 : 1
  2. 6 : 1
  3. 3 : 2
  4. 9 : 2
Show answer
Answer: (B) 6 : 1
Explanation: The standing-wave condition gives 2πr = nλ, so λ = 2πr/n. With r = a₀n²/Z this becomes λ = 2πa₀n/Z, so λ varies as n/Z. For hydrogen with n = 3 this factor is 3, and for He⁺ with n = 1 it is 1/2. The ratio is therefore 6 : 1.

Question 14

The energy needed to remove the first two electrons from a neutral lithium atom (Z = 3) is 81.0 eV in total. The ionisation energy of a hydrogen atom in its ground state is 13.6 eV. What is the total energy needed to remove all three electrons?
  1. 94.6 eV
  2. 122.4 eV
  3. 121.8 eV
  4. 203.4 eV
Show answer
Answer: (D) 203.4 eV
Explanation: Once two electrons are gone, Li²⁺ is a one-electron ion, so the Bohr result applies. Its ionisation energy is 13.6 Z² = 13.6 × 9 = 122.4 eV. The total is 81.0 + 122.4 = 203.4 eV.

Question 15

Hydrogen gas in the ground state is exposed to radiation containing photons of all energies from 10 eV to 12.8 eV. The levels n = 1, 2, 3, 4, 5 lie at −13.6, −3.40, −1.51, −0.85 and −0.54 eV. How many absorption lines appear in the radiation, and how many different lines does the excited gas then emit?
  1. 3 absorption lines and 3 emission lines
  2. 3 absorption lines and 6 emission lines
  3. 6 absorption lines and 3 emission lines
  4. 6 absorption lines and 6 emission lines
Show answer
Answer: (B) 3 absorption lines and 6 emission lines
Explanation: Absorption starts from n = 1 only. The photons of 10.2, 12.09 and 12.75 eV raise atoms to n = 2, 3 and 4, while n = 5 needs 13.06 eV, which is not available. That gives 3 absorption lines. Atoms in n = 4 can return by any route, so the number of emission lines is 4 × 3/2 = 6.

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