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))

n = 5

n = 4

n = 7

n = 6

Correct option is A. n = 5

1/λ = R (1/m² - 1/n²)z²

1/108.5 = R (1/m² - 1/n²)2²
1/30.4 = R (1/1² - 1/m²)2²

Therefore, m = 2
Therefore, n = 5