Gujarati
Hindi
12.Atoms
normal

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

$5$

Solution

Number of possible emission lines are $n(n-1) / 2$ when an electron jumps from $n^{t h}$ state to ground state. In this question, this value should be

$(n-1)(n-2) / 2$

Hence, $10=(n-1) \frac{n-2}{2}$

Solving this, we getn $=6$

Standard 12
Physics

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