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Hydrogen spectrum
Appears in
Concepts tested here
- Longest wavelength of spectral series 2
- Rydberg formula 2
- Number of spectral lines
- Recoil of an atom on photon emission
- Spectral series and energy
All Questions
2015 AIPMT-II 1 question
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In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:Lyman (2 → 1): $\frac{1}{\lambda_L} = R\left(\frac{1}{1^2} - \frac{1}{2^2}\right) \Rightarrow \lambda_L = \frac{4}{3R}$
Balmer (3 → 2): $\frac{1}{\lambda_B} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) \Rightarrow \lambda_B = \frac{36}{5R}$
$\frac{\lambda_L}{\lambda_B} = \frac{4}{3R}\times\frac{5R}{36} = \frac{5}{27}$
2014 AIPMT 1 question
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Hydrogen atom in ground state is excited by a monochromatic radiation of $\lambda$ = 975 Å. Number of spectral lines in the resulting spectrum emitted will beEnergy of the photon: $E = \frac{12400}{975}$ eV ≈ 12.75 eV
Energy of the excited state $= -13.6 + 12.75 = -0.85$ eV, which is the n = 4 level.
Number of spectral lines from n = 4: $\frac{n(n-1)}{2} = \frac{4\times3}{2} = 6$
2013 NEET 1 question
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Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is:$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$; the longest wavelength is for the smallest transition.
$\left(\frac{\lambda_{Lyman}}{\lambda_{Balmer}}\right)_{max} = \frac{\frac{1}{2^2} - \frac{1}{3^2}}{\frac{1}{1^2} - \frac{1}{2^2}} = \frac{5/36}{3/4} = \frac{5}{27}$
2012 AIPMT-MAINS 1 question
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The transition from the state n = 3 to n = 1 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition fromInfrared radiation has less energy than ultraviolet, so the transition must have a smaller energy gap than 3 → 1.
2 → 1 belongs to the Lyman series (ultraviolet); 3 → 2 and 4 → 2 belong to the Balmer series (visible).
4 → 3 belongs to the Paschen series, which lies in the infrared region.
2012 AIPMT-PRE 2 questions
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An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be (m is the mass of the electron, R, Rydberg constant and h, Planck's constant)Energy of the emitted photon: $E = hcR\left(\frac{1}{1^2} - \frac{1}{5^2}\right) = \frac{24}{25}hcR$
Momentum of the photon: $p = \frac{E}{c} = \frac{24hR}{25}$
By conservation of momentum the atom recoils with an equal momentum.
Velocity $= \frac{p}{m} = \frac{24hR}{25m}$ -
Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths $\lambda_1 : \lambda_2$ emitted in the two cases isThird excited state is n = 4, second excited state is n = 3 and first excited state is n = 2.
$\frac{1}{\lambda_1} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}$
$\frac{1}{\lambda_2} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$
$\frac{\lambda_1}{\lambda_2} = \frac{5R/36}{7R/144} = \frac{20}{7}$
2011 AIPMT-PRE 1 question
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The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion isRydberg formula: $\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
First line of Lyman series of hydrogen (Z = 1, $n_1 = 1$, $n_2 = 2$): $\frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$
Second line of Balmer series of the ion ($n_1 = 2$, $n_2 = 4$): $\frac{1}{\lambda} = RZ^2\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3RZ^2}{16}$
The wavelengths are equal: $\frac{3R}{4} = \frac{3RZ^2}{16}$
$Z^2 = 4$
Z = 2