Question

Electron transition in ${He}^{+}$ ion takes from $n_2$ to $n_1$ shell such that:

$2n_2+3n_1=18$ ....(i)
$2n_2-3n_1=6$ ....(ii)

What will be the total number of photons emitted when electrons transit to $n_1$ shell?

• A. $21$
• B. $15$
• C. $20$
• D. $10$

Answer 1

The correct option is D $10$

$2n_2+3n_1=18$ ......(i)
$2n_2-3n_1=6$......(ii)
Add equations (i) and (ii)
$4n_2=24$
$n_2=6$......(iii)

Substitute value of (iii) in (ii)
$2n_2-3n_1=6$
$2\left(6\right)-3n_1=6$
$3n_1=6$
$n_1=2$

Hence, the transition is from $6\to 2$
The total number of photons emitted

$=\frac{(n_2-n_1)(n_2-n_1+1)}{2}=\frac{(6-2)(6-2+1)}{2}=10$

Therefore, the correct option is $D$