The frequency of light emitted for the transition $n=4$ to $n=2$ of the $\mathrm{He}^{+}$ is equal to the transition in $\mathrm{H}$ atom corresponding to which of the following ?

Question

Aicharya · Accepted Answer

For He, $\bar{v}=\frac{1}{\lambda}=R_{H} Z^{2}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}ight)$
$=R_{H}(2)^{2}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}ight)=R_{H}\left(\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}ight)$
For H, $\bar{v}=\frac{1}{\lambda}=R_{H}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}ight)$
For same frequency, $R_{H}\left(\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}ight)=R_{H}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}ight)$
$\therefore \quad \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}=\frac{1}{1^{2}}-\frac{1}{2^{2}}$
$\therefore \quad n_{1}=1 \& n_{2}=2$ *

The frequency of light emitted for the transition $n=4$ to $n=2$ of the $\mathrm{He}^{+}$ is equal to the…

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