An excited H e + ion emits two photons in succession, with wavelengths 108.5 n m and 30.4 n m in making a transition to the ground state. The quantum number n , corresponding to its initial excited state is (for a photon of wavelength λ , energy E = 1240 eV λ ( in nm ) )

Question

Aicharya · Accepted Answer

E 1 = 1240 λ 1 = 1240 108.5 ≈ 11.43 e V E 2 = 1240 λ 2 = 1240 30.4 ≈ 40.79 e V E T o t a l = E 1 + E 1 = 52.22 e V 52.22 = 13.6 2 2 1 - 1 n 2 52.22 = 54.4 1 - 1 n 2 0.96 = 1 - 1 n 2 1 n 2 = 0.04 n 2 = 1 0.04 = 100 4 = 25 n = 5

An excited H e + ion emits two photons in succession, with wavelengths 108.5 n m and 30.4 n m in making a…

Solution