If the wavelength of the first member of the Lyman series of hydrogen is \( \lambda \). The wavelength of the second member will be:
- \( \frac{27}{32} \lambda \)
- \( \frac{32}{27} \lambda \)
- \( \frac{27}{5} \lambda \)
- \( \frac{5}{27} \lambda \)
The Correct Option is A
Approach Solution - 1
To determine the wavelength of the second member of the Lyman series, we need to understand the Lyman series and apply the Rydberg formula for hydrogen spectral lines. The Lyman series corresponds to electronic transitions where the electrons drop to the n=1 energy level from higher levels (n=2, n=3, etc.).
The Rydberg formula for calculating the wavelength \(\lambda\) of any transition in a hydrogen atom is given as:
\(\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\)
Where:
- \(R\) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\))
- \(n_1\) is the lower energy level (for Lyman series, \(n_1 = 1\))
- \(n_2\) is the higher energy level
The first member of the Lyman series corresponds to a transition from \(n_2 = 2\) to \(n_1 = 1\). Thus, the wavelength \(\lambda\) is given by:
\(\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4}\)
The second member of the Lyman series involves a transition from \(n_2 = 3\) to \(n_1 = 1\):
\(\frac{1}{\lambda_2} = R \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \cdot \frac{8}{9}\)
To find the relationship between \(\lambda\) and \(\lambda_2\), we divide the first equation by the second:
\(\frac{\frac{1}{\lambda}}{\frac{1}{\lambda_2}} = \frac{R \cdot \frac{3}{4}}{R \cdot \frac{8}{9}}\)
Simplifying, we have:
\(\frac{\lambda_2}{\lambda} = \frac{\frac{8}{9}}{\frac{3}{4}} = \frac{8}{9} \times \frac{4}{3} = \frac{32}{27}\)
Thus, the wavelength of the second member \(\lambda_2\) is:
\(\lambda_2 = \frac{27}{32} \lambda\)
Therefore, the correct answer is: \(\frac{27}{32} \lambda\).
Approach Solution -2
For the first member of the Lyman series:
\[ \frac{1}{\lambda} = \frac{13.6 \, z^2}{hc} \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] \quad \dots \text{(i)} \]
For the second member of the Lyman series:
\[ \frac{1}{\lambda'} = \frac{13.6 \, z^2}{hc} \left[ \frac{1}{1^2} - \frac{1}{3^2} \right] \quad \dots \text{(ii)} \]
Dividing equation (i) by (ii):
\[ \lambda' = \frac{27}{32} \lambda \]
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