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

A
2013λ
B
1625λ
C
916λ
D
207λ

Solution

## The correct option is C 207λWavelength of emitted photon,        1λ=R(1n22−1n21) Transition from  n1=3  to  n2=2: ∴1λ=R(122−132)=5R36_____(1) Transition from  n1=4  to  n2=3 ∴1λ′=R(132−142)=7R144 ____(2) Dividing (1) by (2), we get      λ′λ=5R/367R/144=207 ⟹λ′=207λ Co-Curriculars

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