Question

Suppose a $$^{226}_{88} Ra$$ nucleus at rest and in ground state undergoes $$\alpha $$-decay to a $$^{222}_{86} Rn$$ nucleus in its excited state. The kinetic energy of the emitted $$\alpha$$ particle is found to be $$4.44 MeV$$. $$^{222}_{86} Rn$$ nucleus then goes to its ground state by $$\gamma$$-decay. The energy of the emitted $$\gamma$$ photon is ____ keV..
[Given : atomic mass of $$^{226}_{88}Ra = 226.005 u$$, atomic mass of $$^{222}_{88}Rn = 222.000 u$$, atomic mass of $$\alpha$$ particle = $$4.000 u, \, 1 u = 931 MeV/c^2, c$$ is speed of light]

Answer 1

Correct option is A. 135
Mass defect $$\Delta m = 226.005 - 222.000 - 4.000$$
$$= 0.005 \, amu$$
$$\therefore Q $$ value = $$0.005 \times 931.5 = 4.655 MeV$$
Also $$\dfrac{K.E_{\alpha}}{K.E_{Rn}} = \dfrac{m_{Rn}}{m_{\alpha}}$$
$$\Rightarrow K.E_{Rn} = \dfrac{m_{\alpha}}{m_{Rn}} . K.E_{\alpha} = \dfrac{4}{222} \times 4.44 = 0.08 MeV$$
$$\therefore$$ Energy of $$\gamma$$-Photon = $$4.655 - (4.44 + 0.08)$$
$$= 0.135 MeV = 135 KeV$$