The spectral series which corresponds to the electronic transition from the levels \({n}_{2}=5,6,\ldots \) to the level \({n}_{1}=4\mathrm~\) is:
1. Pfund series 2. Brackett series
3. Lyman series 4. Balmer series
Subtopic:  Spectral Series |
 81%
From NCERT
NEET - 2024
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Given that the value of the Rydberg constant is \(10^{7}~\text{m}^{-1},\) what will be the wave number of the last line of the Balmer series in the hydrogen spectrum?
1. \(0.5 \times 10^{7}~\text{m}^{-1}\)
2. \(0.25 \times 10^{7} ~\text{m}^{-1}\)
3. \(2.5 \times 10^{7}~\text{m}^{-1}\)
4. \(0.025 \times 10^{4} ~\text{m}^{-1}\)

Subtopic:  Spectral Series |
 87%
From NCERT
NEET - 2016
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When an electron transitions from \(n=4\) to \(n=2,\) then the emitted line in the spectrum will be:
1. the first line of the Lyman series.
2. the second line of the Balmer series.
3. the first line of the Paschen series.
4. the second line of the Paschen series.

Subtopic:  Spectral Series |
 87%
From NCERT
AIPMT - 2000
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In an atom, if the transition from \(n = 4\) to \(n=3\) gives ultraviolet radiation, then to obtain infrared radiation, the transition should be:
1. \(5\rightarrow 4\) 2. \(3\rightarrow 2\)
3. \(2\rightarrow 1\) 4. \(3\rightarrow 1\)
Subtopic:  Spectral Series |
 79%
From NCERT
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The transition from the state \(n=3\) to \(n=1\) in hydrogen-like atoms results in ultraviolet radiation. Infrared radiation will be obtained in the transition from:
1. \(3\rightarrow 2\)
2. \(4\rightarrow 2\)
3. \(4\rightarrow 3\)
4. \(2\rightarrow 1\)

Subtopic:  Spectral Series |
 81%
From NCERT
AIPMT - 2012
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Some energy levels of a molecule are shown in the figure with their wavelengths of transitions.
Then:
1. \(\lambda_{3}>\lambda_{2},\lambda_{1}=2\lambda_{2}\) 2. \(\lambda_{3}>\lambda_{2},\lambda_{1}=4\lambda_{2}\)
3. \(\lambda_{1}>\lambda_{2},\lambda_{2}=2\lambda_{3}\) 4. \(\lambda_{2}>\lambda_{1},\lambda_{2}=2\lambda_{3}\)
Subtopic:  Spectral Series |
 58%
From NCERT
NEET - 2024
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Match List I with List II.
List I
(Spectral Lines of Hydrogen for transitions from)
List II
(Wavelength (nm))
\(\mathrm{A.}\) \(n_2=3\)  to \(n_1=2\)  \(\mathrm{I.}\) \(410.2\)
\(\mathrm{B.}\) \(n_2=4\)  to \(n_1=2\)  \(\mathrm{II.}\) \(434.1\)
\(\mathrm{C.}\) \(n_2=5\)  to \(n_1=2\)  \(\mathrm{III.}\) \(656.3\)
\(\mathrm{D.}\) \(n_2=6\) to \(n_1=2\)  \(\mathrm{IV.}\) \(486.1\)
 
Choose the correct answer from the options given below:
1. \(\mathrm{A - III, B - IV, C - II, D - I}\)
2. \(\mathrm{A - IV, B - III, C - I, D - II}\)
3. \(\mathrm{A - I, B - II, C - III, D - IV}\)
4. \(\mathrm{A - II, B - I, C - IV, D - III}\)
Subtopic:  Spectral Series |
 64%
From NCERT
NEET - 2024
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In hydrogen spectrum, the shortest wavelength in the Balmer series is \(\lambda\). The shortest wavelength in the Bracket series is:
1. \(16\lambda\)
2. \(2\lambda\)
3. \(4\lambda\)
4. \(9\lambda\)
Subtopic:  Spectral Series |
 65%
From NCERT
NEET - 2023
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Let \(f_1\) be the maximum frequency of the Lyman series, \(f_2\) be the frequency of the first line of the Lyman series, and \(f_3\) be the frequency of the series limit of the Balmer series, then which of the following is correct?
1. \(f_1-f_2=f_3\)
2. \(f_2-f_1=f_3\)
3. \(f_1+f_2=f_3\)
4. \(2f_1 = f_2 + f_3\)

Subtopic:  Spectral Series |
 64%
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Electron in hydrogen atom first jumps from the third excited state to the second excited state and then from the second excited to the first excited state. The ratio of the wavelengths \(\lambda_1:\lambda_2\) emitted in the two cases is:
1. \(\frac{7}{5}\)
2. \(\frac{20}{7}\)
3. \(\frac{27}{5}\)
4. \(\frac{27}{20}\)

Subtopic:  Spectral Series |
 71%
From NCERT
AIPMT - 2012
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