A nucleus of mass ( 20 u ) emits a ( y ) photon of energy ( 6 MeV ). If the emission assume to occur when nucleus is free and rest, then the nucleus will have kinetic energy nearest to (take ( 1 u =1.6 10⁻²⁷ ~kg ) )

Step-by-step solution

Given, nucleus of mass ( 20 u ) emits a ( y ) photon of energy ( 6 MeV ) Now, Momentum of photon ( p_ p = E C = 6 1.6 10⁻¹⁹ 10⁶ 3 10⁸ =3.2 10⁻²¹ kgms ⁻¹ ) Kinetic energy of nucleus ( = Momentum of photon 2 m = 3.2 10⁻²¹ 2 20 1.6 10⁻²⁷ ) ( =1.6 10⁻¹⁶ ~J ) ( = 1.6 10⁻¹⁶ 1.6 10⁻¹⁹ eV =10³ eV =1 keV ) Therefore, nucleus will have kinetic energy nearest to ( 1 keV )