JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Modern Physics

An excited

He^{+}

ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm , in making a transition to ground state. The quantum number n , corresponding to its initial excited state is (for photon of wavelength

λ

, energy

E = \frac{1240 eV}{λ ( in nm )}

(A) $n = 4$
(B) $n = 6$
(C) $n = 5$ ✓ Correct
(D) $n = 7$

Answer: (C)

Solution

For a hydrogen-like ion the levels are

E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV

, and for

He X^{+}

we take

Z = 2

. The two emitted photons carry

E_{1} = \frac{1240}{108.5} = 11.43 eV

and

E_{2} = \frac{1240}{30.4} = 40.79 eV

, so the total drop from the initial level

n

to the ground state is

52.22 eV

. Hence

13.6 \times 4 (1 - \frac{1}{n ^{2}}) = 52.22

, i.e.

54.4 (1 - \frac{1}{n ^{2}}) = 52.22

, which gives

\frac{1}{n ^{2}} = 0.04

and

n^{2} = 25

, so

n = 5

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Modern Physics

Solution

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