In the atomic spectrum of hydrogen, the wavel | vedclass

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(C) Using the Rydberg formula: $\frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. For hydrogen, $Z=1$. For transition $(i

Vedclass · Accepted Answer

$\frac{1}{R_H} \left[ \frac{101}{24} \right]$

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(C) Using the Rydberg formula: $\frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. For hydrogen, $Z=1$. For transition $(i)$ $n=4$ to $n=2$: $\frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \left( \frac{3}{16} \right)$. Thus, $\lambda_1 = \frac{16}{3R_H}$. For transition $(ii)$ $n=3$ to $n=1$: $\frac{1}{\lambda_2} = R_H \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R_H \left( 1 - \frac{1}{9} \right) = R_H \left( \frac{8}{9} \right)$. Thus, $\lambda_2 = \frac{9}{8R_H}$. Calculating $(\lambda_1 - \lambda_2)$: $\frac{16}{3R_H} - \frac{9}{8R_H} = \frac{1}{R_H} \left( \frac{128 - 27}{24} \right) = \frac{1}{R_H} \left( \frac{101}{24} \right)$.

In the atomic spectrum of hydrogen, the wavelengths of the spectral lines corresponding to electronic transitions $(i)$ $n=4$ to $n=2$ and $(ii)$ $n=3$ to $n=1$ are $\lambda_1$ and $\lambda_2$ $\mathring{A}$ respectively. The value of $(\lambda_1-\lambda_2)$ (in cm) is ($R_H$ = Rydberg constant)

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