Hydrogen Spectrum — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Hydrogen Spectrum MCQs with step-by-step solutions (21 questions). Part of Structure of Atoms and Nuclei. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Hydrogen Spectrum · easy · theory
The Lyman series of the hydrogen spectrum lies in the:
A. Visible region
B. Microwave region
C. Infrared region
D. Ultraviolet region ✓ Correct
Solution: All Lyman transitions end at $n = 1$ and so involve the largest energy differences.
Q2 — Hydrogen Spectrum · easy · theory
The Balmer series of the hydrogen spectrum lies mainly in the:
A. Visible region ✓ Correct
B. Infrared region
C. X-ray region
D. Ultraviolet region
Solution: Transitions ending at $n = 2$ give the familiar red, blue-green and violet hydrogen lines.
Q3 — Hydrogen Spectrum · easy · theory
The Paschen series of the hydrogen spectrum lies in the:
A. Ultraviolet region
B. Visible region
C. Gamma ray region
D. Infrared region ✓ Correct
Solution: These transitions terminate at $n = 3$ and involve comparatively small energy differences.
Q4 — Hydrogen Spectrum · medium · theory
The Rydberg formula for the hydrogen spectrum is:
A. $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_i^2} - \dfrac{1}{n_f^2}\right)$
B. $\dfrac{1}{\lambda} = R(n_i^2 - n_f^2)$
C. $\lambda = R\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$
D. $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ ✓ Correct
Solution: Here $n_i > n_f$ for emission, and $R$ is the Rydberg constant.
Q5 — Hydrogen Spectrum · medium · theory
The series limit of a spectral series corresponds to a transition from:
A. $n_i = n_f + 1$
B. $n_i = \infty$ ✓ Correct
C. $n_i = 1$
D. $n_i = 2$
Solution: It marks the shortest wavelength of the series, where the electron comes from the ionisation limit.
Q6 — Hydrogen Spectrum · medium · theory
The value of the Rydberg constant is approximately:
A. $3 \times 10^8\text{ m}^{-1}$
B. $1.097 \times 10^7\text{ m}^{-1}$ ✓ Correct
C. $6.63 \times 10^{-34}\text{ m}^{-1}$
D. $1.097 \times 10^{-7}\text{ m}^{-1}$
Solution: Its reciprocal, about $912\text{ \AA}$, is the series limit of the Lyman series.
Q7 — Hydrogen Spectrum · medium · theory
The longest wavelength in any spectral series corresponds to the transition with the:
A. Highest initial quantum number
B. Smallest energy difference between the levels ✓ Correct
C. Largest energy difference between the levels
D. Lowest final quantum number
Solution: Wavelength and energy are inversely related.
Q8 — Hydrogen Spectrum · medium · theory
The Balmer series was the first to be discovered because:
A. It has the shortest wavelengths
B. It lies in the visible region and could be observed easily ✓ Correct
C. It requires the least energy to produce
D. It contains only one spectral line
Solution: The other series needed ultraviolet or infrared detectors developed later.
Q9 — Hydrogen Spectrum · hard · numerical
The ratio of the longest to the shortest wavelength in the Lyman series is:
A. $9 : 5$
B. $4 : 3$ ✓ Correct
C. $16 : 9$
D. $3 : 4$
Solution: Longest ($n = 2 \to 1$): $\dfrac{1}{\lambda} = \dfrac{3R}{4}$. Shortest ($\infty \to 1$): $\dfrac{1}{\lambda} = R$. The ratio is $\dfrac{4}{3}$.
Q10 — Hydrogen Spectrum · hard · numerical
The longest wavelength of the Balmer series is approximately ($R = 1.097 \times 10^7\text{ m}^{-1}$):
A. $4861\text{ \AA}$
B. $6563\text{ \AA}$ ✓ Correct
C. $3646\text{ \AA}$
D. $1215\text{ \AA}$
Solution: For $n = 3 \to 2$: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = \dfrac{5R}{36}$, giving $\lambda \approx 6563\text{ \AA}$ — the red H-alpha line.
Q11 — Hydrogen Spectrum · hard · numerical
The shortest wavelength of the Lyman series is approximately:
A. $3646\text{ \AA}$
B. $6563\text{ \AA}$
C. $1215\text{ \AA}$
D. $912\text{ \AA}$ ✓ Correct
Solution: At the series limit $\dfrac{1}{\lambda} = R$, so $\lambda = \dfrac{1}{1.097 \times 10^7} \approx 912\text{ \AA}$.
Q12 — Hydrogen Spectrum · hard · numerical
The series limit of the Balmer series is approximately:
A. $8204\text{ \AA}$
B. $3646\text{ \AA}$ ✓ Correct
C. $6563\text{ \AA}$
D. $912\text{ \AA}$
Solution: $\dfrac{1}{\lambda} = \dfrac{R}{4}$, so $\lambda = \dfrac{4}{R} \approx 3646\text{ \AA}$.
Q13 — Hydrogen Spectrum · medium · numerical
The energy of the photon emitted in the transition $n = 3 \to n = 2$ in hydrogen is approximately:
A. $1.89\text{ eV}$ ✓ Correct
B. $12.09\text{ eV}$
C. $10.2\text{ eV}$
D. $2.55\text{ eV}$
Solution: $\Delta E = 13.6\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 13.6 \times \dfrac{5}{36} \approx 1.89\text{ eV}$.
Q14 — Hydrogen Spectrum · medium · numerical
The energy of the photon emitted in the transition $n = 2 \to n = 1$ in hydrogen is:
A. $10.2\text{ eV}$ ✓ Correct
B. $13.6\text{ eV}$
C. $1.89\text{ eV}$
D. $3.4\text{ eV}$
Solution: $\Delta E = 13.6 - 3.4 = 10.2\text{ eV}$.
Q15 — Hydrogen Spectrum · hard · numerical
The wavelength of the photon emitted in the $n = 2 \to n = 1$ transition is approximately ($hc = 1240\text{ eV}\cdot\text{nm}$):
A. $91.2\text{ nm}$
B. $121.6\text{ nm}$ ✓ Correct
C. $656\text{ nm}$
D. $364.6\text{ nm}$
Solution: $\lambda = \dfrac{1240}{10.2} \approx 121.6\text{ nm}$, the Lyman-alpha line.
Q16 — Hydrogen Spectrum · hard · numerical
The series limit of the Paschen series is approximately:
A. $8204\text{ \AA}$ ✓ Correct
B. $3646\text{ \AA}$
C. $6563\text{ \AA}$
D. $912\text{ \AA}$
Solution: $\dfrac{1}{\lambda} = \dfrac{R}{9}$, so $\lambda = \dfrac{9}{R} \approx 8204\text{ \AA}$.
Q17 — Hydrogen Spectrum · medium · numerical
The energy of the photon emitted in the transition $n = 4 \to n = 2$ in hydrogen is:
A. $10.2\text{ eV}$
B. $1.89\text{ eV}$
C. $0.85\text{ eV}$
D. $2.55\text{ eV}$ ✓ Correct
Solution: $\Delta E = 13.6\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = 13.6 \times \dfrac{3}{16} = 2.55\text{ eV}$.
Q18 — Hydrogen Spectrum · medium · numerical
The energy required to ionise a hydrogen atom already in the first excited state is:
A. $13.6\text{ eV}$
B. $10.2\text{ eV}$
C. $3.4\text{ eV}$ ✓ Correct
D. $1.51\text{ eV}$
Solution: The electron must be raised from $-3.4\text{ eV}$ to zero.
Q19 — Hydrogen Spectrum · hard · numerical
The number of spectral lines emitted when hydrogen atoms de-excite from the $n = 4$ level is:
A. $6$ ✓ Correct
B. $4$
C. $3$
D. $10$
Solution: The number of possible transitions is $\dfrac{n(n-1)}{2} = \dfrac{4 \times 3}{2} = 6$.
Q20 — Hydrogen Spectrum · medium · numerical
The number of spectral lines emitted when hydrogen atoms de-excite from the $n = 3$ level is:
A. $3$ ✓ Correct
B. $2$
C. $6$
D. $1$
Solution: $\dfrac{3 \times 2}{2} = 3$ lines.
Q21 — Hydrogen Spectrum · hard · numerical
The longest wavelength of the Lyman series is approximately:
A. $6563\text{ \AA}$
B. $3646\text{ \AA}$
C. $912\text{ \AA}$
D. $1215\text{ \AA}$ ✓ Correct
Solution: For $n = 2 \to 1$: $\dfrac{1}{\lambda} = \dfrac{3R}{4}$, giving $\lambda = \dfrac{4}{3R} \approx 1215\text{ \AA}$.