Compute the energy released if an α -particle is emitted by ^ 238 U...

The reaction for $\alpha$-decay is given by:

^{238}\text{U} \rightarrow ^{234}\text{Th} + ^{4}\text{He} + Q$<br><br>The mass defect$\Delta m$is calculated as:<br><br>$\Delta m = m(^{238}\text{U}) - [m(^{234}\text{Th}) + m(^{4}\text{He})]$<br><br>Substituting the given values:<br><br>$\Delta m = 238.0508 \text{ u} - (234.04363 \text{ u} + 4.00260 \text{ u})$<br><br>$\Delta m = 238.0508 \text{ u} - 238.04623 \text{ u} = 0.00457 \text{ u}$<br><br>The energy released$Q$is:<br><br>$Q = \Delta m \times 931 \text{ MeV/u}$<br><br>$Q = 0.00457 \times 931 \text{ MeV} \approx 4.255 \text{ MeV}