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Q.

A photon of energy 10.2 eV corresponds to light of wavelength λ0.  Due to an electron transition from n = 2 to n = 1 in a Hydrogen atom, light of wavelength  λ is emitted. If we take into account the recoil of the atom when the photon is emitted, then

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Detailed Solution

Energy of photon + K.E. of atom =  13.6  1−122eV  = 10.2eV∴    hcλ  =  hcλ0−  KE of atom  ⇒   λ  >  λ0
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