Question

What electronic transition in the $H{e}^{+}$ ion would emit the radiation of the same wavelength as that of the first Lyman transition of hydrogen atom ?

• A. $2\to 1$
• B. $4\to 2$
• C. $3\to 1$
• D. $2\to 4$

Answer 1

The correct option is B $4\to 2$

For the first Lyman transition for hydrogen,
$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})=R{1}^2(\frac{1}{1^2}-\frac{1}{2^2})=\frac{3R}{4}$
Since, wavelength of the radiation emitted in electronic transition in $H{e}^{+}$ is the same as for hydrogen,
$\frac{1}{{\lambda }_{H{e}^{+}}}=R{.2}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})=\frac{3R}{4}$

Hence, $n_1$ and $n_2$ have to be 2 and 4 respectively as,
$\frac{1}{\lambda }=4R(\frac{1}{2^2}-\frac{1}{4^2})=\frac{3R}{4}$