Compute the energy that needs to be supplied to ²³⁸ U in order for two protons and two neutrons to be emitted one by one. The atomic masses of ²³⁸ U , ²³⁴ Th , and ⁴ He are 238.0508 u , 234.04363 u , and 4.00260 u respectively.

Step-by-step solution

The process of emitting two protons and two neutrons is equivalent to breaking the ²³⁸ U nucleus into a ²³⁴ Th nucleus, 2 protons, and 2 neutrons. The overall reaction is: ²³⁸ U ²³⁴ Th + 2 ¹ H + 2 ¹ n 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 U and ²³⁴ Th are atomic masses. Using the standard masses m_H = 1.007825 u and m_n = 1.008665 u , the total mass of the products is: M_ products = M(²³⁴ Th ) + 2m_H + 2m_n M_ produ