Question:
The ratio of wavenumber of first line of Balmer series and wavenumber of second line of Lyman series of hydrogen atom is:
The ratio of wavenumber of first line of Balmer series and wavenumber of second line of Lyman series of hydrogen atom is:
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First line means transition from the next higher orbit. Second line means transition from the second higher orbit.
Updated On: Jun 18, 2026
- \(5:32\)
- \(32:5\)
- \(22:7\)
- \(7:22\)
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The Correct Option is A
Solution and Explanation
Concept:
The wavenumber of a spectral line in hydrogen spectrum is given by the Rydberg formula:
\[
\bar{\nu}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)
\]
where \(n_2>n_1\).
Step 1: Find the wavenumber of the first Balmer line.
For Balmer series, \[ n_1=2 \] The first line corresponds to \[ n_2=3. \] Therefore, \[ \bar{\nu}_B = R\left(\frac14-\frac19\right) \] \[ = R\left(\frac{5}{36}\right). \]
Step 2: Find the wavenumber of the second Lyman line.
For Lyman series, \[ n_1=1. \] The second line corresponds to \[ n_2=3. \] Hence, \[ \bar{\nu}_L = R\left(1-\frac19\right) \] \[ = R\left(\frac89\right). \]
Step 3: Calculate the ratio.
\[ \bar{\nu}_B:\bar{\nu}_L = \frac{5}{36}:\frac89 \] \[ = \frac{5}{36}\times\frac98 \] \[ = \frac5{32}. \] Therefore, \[ \boxed{5:32} \]
Step 1: Find the wavenumber of the first Balmer line.
For Balmer series, \[ n_1=2 \] The first line corresponds to \[ n_2=3. \] Therefore, \[ \bar{\nu}_B = R\left(\frac14-\frac19\right) \] \[ = R\left(\frac{5}{36}\right). \]
Step 2: Find the wavenumber of the second Lyman line.
For Lyman series, \[ n_1=1. \] The second line corresponds to \[ n_2=3. \] Hence, \[ \bar{\nu}_L = R\left(1-\frac19\right) \] \[ = R\left(\frac89\right). \]
Step 3: Calculate the ratio.
\[ \bar{\nu}_B:\bar{\nu}_L = \frac{5}{36}:\frac89 \] \[ = \frac{5}{36}\times\frac98 \] \[ = \frac5{32}. \] Therefore, \[ \boxed{5:32} \]
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