For the alpha-decay reaction ²³⁸U → ²³⁴Th + ⁴He, the respective atomic masses are 238.050 u, 234.043 u, and 4.003 u. Calculate the energy released Q in keV. (1 u = 931.5 MeV/c².)
Options

Question

For the alpha-decay reaction ²³⁸U → ²³⁴Th + ⁴He, the respective atomic masses are 238.050 u, 234.043 u, and 4.003 u. Calculate the energy released Q in keV. (1 u = 931.5 MeV/c².)

G1 Physics · Accepted Answer

Step 1: Initial mass = 238.050 u. Final mass = 234.043 + 4.003 = 238.046 u. Step 2: Mass defect Δm = 238.050 − 238.046 = 0.004 u. Step 3: Q = Δmc² = 0.004 × 931.5 MeV = 3.726 MeV. Step 4: Convert MeV to keV: 3.726 × 1000 = 3726 keV. Final answer: 3726 keV (Option A).