A stationary nucleus of mass 24u emits a ¥ photon. The energy of the emited photon is 7MeV. Find the recoil energy of the nucleus?(ans:-1.1MeV).

in gama decay there is no change in mass number...

initially nucleous was at rest ,due to internal forces it will move after giving a gamma photon..

so we can apply conservation of momentam here

final momentam = initial momentam = 0

P(gamma photon) + P(nucleous) = 0  .......................1

  energy of photon = 7Mev

   p(photon) = E/c                        (c is speed of light , E =7Mev (given) )

 p(nucleous) = -E/c             ( from eq 1)

  kinetic energy = P²/2m = E²/2mc² 

now put E , C & m ,u will get the  1.1Mev ans