JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Modern Physics

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The energy required to ionize a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state?

(A)11.4 nm✓ Correct
(B)35.8 nm
(C)24.2 nm
(D)8.6 nm

Answer: (A)

Solution

The ground-state ionization energy of a hydrogen-like ion is

13.6 Z^{2} eV

; setting this equal to

9

Rydbergs,

13.6 Z^{2} = 9 \times 13.6

, gives

Z = 3

. The jump from the second excited state

(n = 3)

to the ground state

(n = 1)

releases

13.6 Z^{2} (1 - \frac{1}{9}) = 13.6 \times 9 \times \frac{8}{9} = 108.8 eV

. The emitted wavelength is

λ = \frac{1240}{108.8} \approx 11.4 nm

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option

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