If the wavelength of the first member of Lyman series of hydrogen is $\lambda$. The wavelength of the second member will be
Solution
<p>$$\begin{aligned}
& \frac{1}{\lambda}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{2^2}\right] ...... \text{(i)}\\
& \frac{1}{\lambda^{\prime}}=\frac{13.6 \mathrm{z}^2}{\mathrm{hc}}\left[\frac{1}{1^2}-\frac{1}{3^2}\right] ...... \text{(ii)}
\end{aligned}$$</p>
<p>On dividing (i) & (ii)</p>
<p>$\lambda^{\prime}=\frac{27}{32} \lambda$</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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