Question

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

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

Answer 1

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

$hv=DE={13.62}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$v_{H{e}^{+}}=v_4\times 2^2[\frac{1}{{\left(\frac{n_1}{2}\right)}^2}-\frac{1}{{\left(\frac{n_2}{2}\right)}^2}]$
$=v_H[\frac{1}{{\left(\frac{2}{2}\right)}^2}-\frac{1}{{\left(\frac{4}{2}\right)}^2}]$
$n_1=1,n_2=2$