Question

In a hydrogen atom, the electron is in $n^{th}$ excited state. It may come down to second excited state by emitting ten different wavelengths. What is the value of n?

• A. 6
• B. 7
• C. 8
• D. 5

Answer 1

The correct option is A 6

As we know that when electron comes down from $n^{th}$ excited state to any lower state.
$\frac{\left(\right(n+1)-3)\left(\right(n+1)-3+1)}{2}=10$
(n - 2) (n - 1) = 20
$n^2-3n-18=0$
n = 6.