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What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n=4 to n=2 of $He^+$ spectrum ?


1 Answer

For $He^+,\large\frac{1}{\lambda}$$=R_Hz^2\bigg[\large\frac{1}{2^2}-\frac{1}{4^2}\bigg]$
For $H,\large\frac{1}{\lambda}$$=R_H\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
Since $\lambda$ is same
$\therefore Z^2\bigg[\large\frac{1}{2^2}-\frac{1}{4^2}\bigg]=\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
$n_1=1$ and $n_2=2$
Hence (d) is the correct answer.
answered Jan 15, 2014 by sreemathi.v

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