Q87 / 90 · JEE Main 2024 (01 Feb Shift 1) · Physics

The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly :

Solution

For Balmer series, transition is from $n = 3, 4, 5, 6, e t c$ to $n = 2$ .

For transition to $n = 2$ , the minimum state to transition from is $n = 3$ .

Therefore, energy required $Δ E = E_{3} - E_{1} = - \frac{13.6}{9} - (- 13.6) = 12 .1 eV$

Hence, the correct answer is option: 'D'.