₉₂²³⁸A → ₉₀²³⁴B + ₂⁴D + Q
In the given nuclear reaction, the approximate amount of energy released will be:
[Given, mass of ₉₂²³⁸ A=238.05079 × 931.5 MeV / c²,
mass of ₉₀²³⁴ B=234 · 04363 × 931 · 5 MeV / c²,
mass of ₂⁴ D=4 · 00260 × 931 · 5 MeV / c²]

Solution

Physics sir · 4 steps

Subtract the product masses from the parent mass; that missing mass times 931.5 is your Q.

1. Where Q comes from. In this alpha decay the products weigh a little less than the parent nucleus. That lost mass, the mass defect, comes out as the energy Q.

   Q = (m_A - m_B - m_D) × 931.5 MeV

2. Find the mass defect. The products together have 234.04363 + 4.00260 = 238.04623 u. So the mass defect is 238.05079 - 238.04623 = 0.00456 u.
3. Convert to energy. Q = 0.00456 × 931.5 = 4.248 MeV.
4. So. About 4.25 MeV is released, which is option (b).

The idea to keep. For any decay, Q is the parent mass minus the total product mass, times 931.5 MeV per u. Subtract first and multiply once at the end, so you keep every significant digit.

Answer from the NTA final answer key. Solution by PrepWiser.