Question

Find the quantum no 'n' corresponding to the excited state of He+ ion if on transition to the ground state that ion emits two photons in succession with wavelength 108.5 and 30.4nm

Answer 1

Hey,

Given :

λ2=30.4×10−7cm

λ1=108.5×10−7cm

Let excited state of He+ be n2.It comes from n2 to n1 and then n1 to 1 to emit two successive photon.

1λ2=RH.z2[112−1n21]

130.4×10−7=109678×4[112−1n21]

∴n1=2

Now for λ1:n1=2

1λ1=RH.z2[122−1n22]

1108.5×10−7=109678×4[122−1n22]

n2=5

∴ excited state of He is 5th orbit.

Hence (d) is the correct answer.