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Question:

If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be

1625λ

207λ

916λ

2013λ

Solution:

Wavelength of emitted photon:
1λ=R(1n22−1n12)
Transition from n1=3 to n2=2: ∴1λ=R(122−132)=5R36 (1)
Transition from n1=4 to n3=2: ∴1λ′=R(132−142)=7R144 (2)
Dividing (1) by (2) we get:
λ′λ=5R/367R/144=207
⇒λ′=207λ