Suppose a R a 88 226 nucleus at rest and in ground state undergoes α -decay to a R n 86 222 nucleus in…

Suppose a Ra88226 nucleus at rest and in ground state undergoes α -decay to a Rn86222 nucleus in its excited state. The kinetic energy of the emitted α particle is found to be 4.44 MeV.Rn86222 nucleus then goes to its ground state by γ -decay. The energy of the emitted γ -photon is _______ keV,
[Given: atomic mass of  88226Ra=226.005 u, atomic mass of  86222Rn=222.000 u, atomic mass of α particle =4.000 u, 1u=931 MeV/c2,c is speed of the light]

Solution

$\Rightarrow$ Mass defect $\Delta m = 226.005 - 222.000 - 4.000$ for $_{88}^{226}\text{Ra} $ $\rightarrow$ $ \text{alpha-decay} \ _{86}^{222}\text{Rn} $+ $_{2}^{4}\text{He} $+ $\gamma$ $= 0.005 \, \text{amu}$ $\therefore Q \, \text{value} = 0.005 \times 931.5 = 4.655 \, \text{MeV}$ Also $\frac{K.E_{\alpha}}{K.E_{Rn}} = \frac{m_{Rn}}{m_{\alpha}}$ $\Rightarrow K.E_{Rn} = \frac{m_{\alpha}}{m_{Rn}} \cdot K.E_{\alpha} = \frac{4}{222} \times 4.44 = 0.08 \, \text{MeV}$ $\therefore$ Energy of $\gamma$-Photon $= 4.655 - (4.44 + 0.08)$ for $E_{\gamma} = Q - (K.E_{\alpha} + K.E_{Rn})$ $= 0.135 \, \text{MeV} = 135 \, \text{KeV}$

Asked in: JEE Advanced 2019 (Paper 2)

Practice more Nuclear Physics questions on Aicharya