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(C) For the Lyman series, the Rydberg formula is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$, where $n_1 = 1$.

Vedclass · Accepted Answer

$4$

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(C) For the Lyman series, the Rydberg formula is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$, where $n_1 = 1$. Given $\lambda = \frac{16}{15 R}$, so $\frac{1}{\lambda} = \frac{15 R}{16}$. Substituting the values: $\frac{15 R}{16} = R \left[ \frac{1}{1^2} - \frac{1}{n_2^2} \right]$. Dividing both sides by $R$: $\frac{15}{16} = 1 - \frac{1}{n_2^2}$. Rearranging the terms: $\frac{1}{n_2^2} = 1 - \frac{15}{16} = \frac{1}{16}$. Therefore, $n_2^2 = 16$, which gives $n_2 = 4$.

The wavelength of a spectral line emitted by a hydrogen atom in the Lyman series is $\frac{16}{15 R} \ cm$. What is the value of $n_2$? $(R = \text{Rydberg constant})$

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