Question

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

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

Answer 1

The correct option is A $n=2$ to $n=1$

The frequency of light emitted for the transition $n=4$ to $n=2$ of helium ion is equal to that of $H$ atom will be $n=2$ to $n=1$

For ${He}^{+}$ ion,

$\frac{1}{\lambda }=Z^2{R}_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$={\left(2\right)}^2{R}_H(\frac{1}{2^2}-\frac{1}{4^2})=R_H\left(\frac{3}{4}\right)⟶\left(I\right)$

For $H$ atom,

$\frac{1}{\lambda }=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})⟶\left(II\right)$