Question

The value of $n_2+n_1$ and $n_2^2-n_1^2$ for $H{e}^{+}$ ion in atomic spectrum are 4 and 8 respectively. The wavelenght of emitted photon when electron jump from $n_2$ to $n_1$ is:

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$n_2+n_1$ = $4$

$n_2^2-n_1^2$ = $8$

$(n_2-n_1)(n_2+n_1)$ = $8$

$(n_2-n_1)$ = $2$

Therefore $n_2=3$ and $n_1=1$

$\frac{1}{\lambda }$=$R_h$$4$($\frac{1}{1}-\frac{1}{9}$)

$\lambda$ = $\frac{9}{32}$$R_h$