A nucleus of mass M emits a photon of frequency \nu and the nucleus recoils. The recoil energy will be

  • Option 1)

    Mc^{2}-h \nu

  • Option 2)

    h^{2}\nu^{2}/2Mc^{2}

  • Option 3)

    0

  • Option 4)

    h\nu

 

Answers (1)
V Vakul

As we learnt in 

Energy emitted due to transition of electron -

Delta E= Rhcz^{2}left ( frac{1}{n_{f}, ^{2}}-frac{1}{n_{i}, ^{2}} 
ight )

frac{1}{lambda }= Rz^{2}left ( frac{-1}{n_{i}, ^{2}}+frac{1}{n_{f}, ^{2}} 
ight )

- wherein

R= R hydberg: constant

n_{i}= initial state \n_{f}= final : state

 

 From momentum conservation,

momentum of photon=momentum of recoil

\Rightarrow p=\frac{E}{c}=\frac{h\nu }{c}

\therefore \: Kinetic \: energy \: of \: recoil=\frac{p^{2}}{2M}=\frac{h^{2}\nu^{2}}{2Mc^{2}}


Option 1)

Mc^{2}-h \nu

Incorrect

Option 2)

h^{2}\nu^{2}/2Mc^{2}

Correct

Option 3)

0

Incorrect

Option 4)

h\nu

Incorrect

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