Electronic transition in He^+ ion takes from n_2 to n_1 shell suc… - Vidyadip

MCQ

Electronic transition in $He^+$ ion takes from $n_2$ to $n_1$ shell such that :
$2n_2 + 3n_1 = 18,\,\,\,$ $\,\,\,2n_2 -3n_1 = 6$
What will be the total number of photons emitted when electrons transit to $n_1$ shell ?

A
$21$
B
$15$
C
$20$
✓
$10$

✓

Answer

Correct option: D.

$10$

d
$2 n_{2}+3 n_{1}=18$

$2 n_{2}-3 n_{1}=6$

Add equations (i) and (ii) $4 n_{2}=24$

$n_{2}=6_{-\ldots}(i i i)$

Substitute value of (iii) in (ii)

$2 n_{2}-3 n_{1}=6$

$2(6)-3 n_{1}=6$

$3 n_{1}=6$

$n_{1}=2$

Hence, the transition is from $6 \rightarrow 2$

The total number of photons emitted

$=\frac{\left(n_{2}-n_{1}\right)\left(n_{2}-n_{1}+1\right)}{2}=\frac{(6-2)(6-2+1)}{2}=10$

Therefore, the correct option is $D$

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