Question

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 = 1240 eV/λ(in nm) :)
(A) n = 4 (B) n = 6
(C) n = 5 (D) n = 7

Answer 1

Answer:

sorry don't know the answer

Answer 2

Thus the value of n = 5

Option (C) is correct.

Explanation:

Given data:

• Wavelength λ1 = 108.5 nm

• Wavelength λ2 = 30.4 nm

• Energy E = 1240 eV/λ(in nm)

Solution:

1/λ = R (1 / m^2 - 1 / n^2) z^2

1/1085 = R (1/m^2 - 1 / n^2)

1 / 304 = R ( 1/ 1^2 - 1/ m^2) 2^2

m = 2

n = 5

Thus the value of n = 5