Question

During the emission spectrum of electron in $H{e}^{+}$ ion, the electron on transition from same higher energy state drops to the ground state and first excited state emits two photon of ${\lambda }_1$=304 $A^{\circ }$ and ${\lambda }_2$=1085 $A^{\circ }$ respectively.The total number of photons emitted during the transition of electrons from highest excited to all lower energy level(Possible)

• A. 15
• B. 12
• C. 6
• D. 10

Answer 1

The correct option is D

10

$\frac{1}{{\lambda }_1}$=$R_n$$Z^2$($\frac{1}{n_1^2}$-$\frac{1}{n_2^2}$)

$n_1$=1 ${\lambda }_1$=304 $A^{\circ }$,z=2,$R_n$=1.1 $\times$ ${10}^7{m}^{-1}$

$n_2$=2 (lower excited state)

For first excited state

$\frac{1}{{\lambda }_2}$=$R_n$$Z^2$($\frac{1}{n_1^2}$-$\frac{1}{n_2^2})$

$n_1$=2 ${\lambda }_2$=1085 $A^{\circ }$

Therefore, total number of spectral lines=$\frac{(5-1)(5-1+1)}{2}$

=10