Question

Suppose a ²²⁶ ₈₈Ra nucleus at rest and in ground state undergoes α-decay to a ²²² ₈₆Rn nucleus in its excited state. The kinetic energy of the emitted α particle is found to be 4.44 MeV. ²²² ₈₆Rn nucleus then goes to its ground state by γ-decay. The energy of the emitted γ photon is _______ keV. 

[Given: atomic mass of ²²⁶ ₈₈Ra = 226.005 u, atomic mass of ²²² ₈₆Rn = 222.000 u, atomic mass of α particle 

= 4.000 u, 1 u = 931 MeV/c², c is speed of the light]

Answer 1

Δm = [226.005 – 222 – 4]

= 0.005 amu

Q = Δmc²

= 931.5 × 0.005 = 4.655 MeV

Since momentum is conserved, kinetic energy is in inverse ratio of masses.