Compute the energy released if an -particle is emitted by ²³⁸ U . 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 reaction for -decay is given by: ²³⁸ U ²³⁴ Th + ⁴ He + Q The mass defect m is calculated as: m = m(²³⁸ U ) - [m(²³⁴ Th ) + m(⁴ He )] Substituting the given values: m = 238.0508 u - (234.04363 u + 4.00260 u ) m = 238.0508 u - 238.04623 u = 0.00457 u The energy released Q is: Q = m 931 MeV/u Q = 0.00457 931 MeV 4.255 MeV