chemistry : class11 Atomic Structure of Hydrogen : Properties of hydrogen : Hydrogen MCQ Questions and Answers
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Atomic Structure of Hydrogen
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MCQ Questions
1. For a hydrogen atom, the energy of an electron in the nth Bohr orbit is given by E_n = -13.6/n^2 eV. What is the energy required to excite the electron from n = 2 to n = 4?
A.
2.55 eV
B.
10.20 eV
C.
12.75 eV
D.
13.60 eV
Correct Answer:
2.55 eV
Explanation:
E_2 = -13.6/4 = -3.40 eV and E_4 = -13.6/16 = -0.85 eV. Therefore, the excitation energy is E_4 - E_2 = (-0.85) - (-3.40) = 2.55 eV.
2. Which transition in the hydrogen atom produces radiation having the shortest wavelength?
A.
n = 3 to n = 2
B.
n = 4 to n = 2
C.
n = 5 to n = 3
D.
n = 6 to n = 1
Correct Answer:
n = 6 to n = 1
Explanation:
The wavelength is shortest when the energy difference between the initial and final levels is greatest. Among the given transitions, n = 6 to n = 1 has the largest energy difference.
3. An electron in a hydrogen atom absorbs a photon of energy 12.09 eV from the ground state. To which principal energy level is it excited?
A.
n = 2
B.
n = 3
C.
n = 4
D.
n = 5
Correct Answer:
n = 3
Explanation:
The energy required to excite hydrogen from n = 1 to n = 3 is E_3 - E_1 = -1.51 - (-13.60) = 12.09 eV. Hence, the electron reaches n = 3.
4. The ratio of the radii of the third and second Bohr orbits of a hydrogen atom is
A.
2:3
B.
3:2
C.
9:4
D.
4:9
Correct Answer:
9:4
Explanation:
For hydrogen, the radius of the nth orbit is proportional to n^2. Therefore, r_3:r_2 = 3^2:2^2 = 9:4.
5. If the electron in a hydrogen atom falls from n = 5 to n = 2, the emitted radiation belongs to which spectral series and lies in which region?
A.
Lyman series, ultraviolet region
B.
Balmer series, visible region
C.
Paschen series, infrared region
D.
Brackett series, infrared region
Correct Answer:
Balmer series, visible region
Explanation:
All transitions terminating at n = 2 form the Balmer series. Hydrogen Balmer lines generally lie in the visible region of the electromagnetic spectrum.
6. According to the Bohr model, the angular momentum of the electron in the fourth orbit of hydrogen is
A.
h/4π
B.
h/2π
C.
2h/π
D.
4h/2π
Correct Answer:
4h/2π
Explanation:
Bohr's quantization condition is mvr = nh/2π. For n = 4, the angular momentum is 4h/2π.
7. What is the wave number of the radiation emitted when an electron in a hydrogen atom transitions from n = 3 to n = 2? Use R = 1.097 × 10^7 m^-1.
8. The ionization energy of a hydrogen atom in its second excited state is
A.
13.6 eV
B.
6.8 eV
C.
3.4 eV
D.
1.51 eV
Correct Answer:
1.51 eV
Explanation:
The second excited state corresponds to n = 3. The ionization energy from this level is 13.6/n^2 = 13.6/9 ≈ 1.51 eV.
9. Which statement correctly compares the speed of the electron in the first and third Bohr orbits of hydrogen?
A.
The speed in the third orbit is three times that in the first orbit
B.
The speed in the first orbit is three times that in the third orbit
C.
The speeds in both orbits are equal
D.
The speed in the first orbit is nine times that in the third orbit
Correct Answer:
The speed in the first orbit is three times that in the third orbit
Explanation:
In the Bohr model, electron speed is inversely proportional to the principal quantum number, v_n proportional to 1/n. Thus, v_1/v_3 = 3.
10. A hydrogen atom absorbs radiation of wavelength 102.6 nm. Between which two energy levels does the corresponding transition occur?
A.
n = 1 and n = 2
B.
n = 1 and n = 3
C.
n = 2 and n = 3
D.
n = 2 and n = 4
Correct Answer:
n = 1 and n = 3
Explanation:
A wavelength of approximately 102.6 nm corresponds to the energy difference between n = 1 and n = 3 in hydrogen. This is the reverse of the Lyman-alpha-to-higher transition associated with the 1 to 3 energy gap.
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