devarshi-dt-logo

Question:

Suppose a ^226_88 Ra nucleus at rest and in ground state undergoes α-decay to a ^222_86 Rn nucleus in its excited state. The kinetic energy of the emitted α particle is found to be 4.44 MeV. ^222_86 Rn 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_88Rn = 222.000 u, atomic mass of α particle = 4.000 u, 1 u = 931 MeV/c^2, c is speed of light]

135

Solution:

Correct option is A. 135
Mass defect Δm = 226.005 - 222.000 - 4.000= 0.005 amu
∴ Q value = 0.005 × 931.5 = 4.655 MeV
Also K.E_α + K.E_Rn = m_Rn/m_α × K.E_α = 4/222 × 4.44 = 0.08 MeV
∴ Energy of γ-Photon = 4.655 - (4.44 + 0.08)= 0.135 MeV = 135 KeV