Question

A transition for $H$-atom from $II$ to $I$ orbit has same wavelength as from $n^{th}$ orbit to $2^{nd}$ orbit for $H{e}^{+}$ ion. The value of ${}^{\prime }{n}^{\prime }$ is :

• A. $7$
• B. $2$
• C. $5$
• D. $4$

Answer 1

The correct option is D $4$

$\left[Z^2\right(\frac{1}{n_1^2}-\frac{1}{n_2^2}){]}_{H_2}=[Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2}){]}_{He}$
$\left[1^2\right(\frac{1}{1^2}-\frac{1}{2^2}\left)\right]=\left[2^2\right(\frac{1}{2^2}-\frac{1}{n_2^2}\left)\right]$
$1-0.25=4(0.25-\frac{1}{n_2^2})$
$0.1875=4(0.25-\frac{1}{n_2^2})$
$\frac{1}{n_2^2}=00625$
$n_2^2=4$