Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths $\lambda_1: \lambda_2$ emitted in the two cases is

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Aicharya · Accepted Answer

Here, for wavelength $\lambda_1$
$n_1=4$ and $n_2=3$
and for $\lambda_2, n_1=3$ and $n_2=2$
We have $\frac{h c}{\lambda}=-13.6\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]$
So, for $\lambda_1$
$\begin{aligned}
\Rightarrow \frac{h c}{\lambda_1} & =-13.6\left[\frac{1}{(4)^2}-\frac{1}{(3)^2}\right] \\
\frac{h c}{\lambda_1} & =13.6\left[\frac{7}{144}\right] \quad \ldots (i)
\end{aligned}$
Similarly, for $\lambda_2$
$\begin{aligned}
\Rightarrow \frac{h c}{\lambda_2} & =-13.6\left[\frac{1}{(3)^2}-\frac{1}{(2)^2}\right] \\
\frac{h c}{\lambda_2} & =13.6\left[\frac{5}{36}\right] \quad \ldots (ii)
\end{aligned}$
Hence, from Eqs. (i) and (ii), we get
$\frac{\lambda_1}{\lambda_2}=\frac{20}{7}$

Electron in hydrogen atom first jumps from third excited state to second excited state and then from second…

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