λ1=R[n121−n221]
⇒λ=R(221−321)1
⇒λ′=R(321−421)1
⇒λλ′=(321−421)(221−321)⇒λ′=720λ.
If an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength λ. When it jumps from the fourth orbit to the third orbit, the corresponding wavelength of the photon will be
Held on 30 Apr 2016 · Verified 9 Jul 2026.
2516λ
169λ
720λ
1320λ
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De-Broglie wavelength of an electron orbiting in the $n=2$ state of hydrogen atom is close to (Given Bohr radius $=0.052 \mathrm{~nm}$ )
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