Compute the energy that needs to be supplied to ^ 238 U in order...

The process of emitting two protons and two neutrons is equivalent to breaking the ^{238}\text{U}$nucleus into a ^{234}\text{Th}$ nucleus, 2 protons, and 2 neutrons.

The overall reaction is:
^{238}\text{U} \rightarrow ^{234}\text{Th} + 2 ^{1}\text{H} + 2 ^{1}\text{n}$<br><br>Here, the mass of a hydrogen atom ($m_H$) is used instead of a bare proton to account for the electron mass, since the given masses of ^{238}\text{U}$ and ^{234}\text{Th}$are atomic masses.<br><br>Using the standard masses$m_H = 1.007825 \text{ u}$and$m_n = 1.008665 \text{ u}$, the total mass of the products is:<br>$M_{\text{products}} = M(^{234}\text{Th}) + 2m_H + 2m_n$<br>$M_{\text{products}} = 234.04363 + 2(1.007825) + 2(1.008665) = 238.07661 \text{ u}$<br><br>The mass defect$\Delta m$for the process is:<br>$\Delta m = M_{\text{products}} - M(^{238}\text{U})$<br>$\Delta m = 238.07661 - 238.0508 = 0.02581 \text{ u}$<br><br>The required energy is obtained by converting the mass defect into energy using$1 \text{ u} = 931 \text{ MeV}$:<br>$\Delta E = 0.02581 \times 931 \text{ MeV} = 24.029 \text{ MeV} \approx 24.03 \text{ MeV}$<br><br>Although the mass of ^{4}\text{He}$ is provided, it cancels out when combining the energy of alpha emission and the binding energy of the alpha particle, making it redundant for the overall calculation.

Answer: $24.03 \text{ MeV}$