Problem

Calculate the energy released in the $\beta$-decay reaction:

$${}_{90}\mathrm{Th}^{234} \rightarrow {}_{91}\mathrm{Pa}^{234} + {}_{-1}\beta + \bar{v} + Q$$

Data

• Mass of thorium isotope (${}_{90}\mathrm{Th}^{234}$): $234.0436 \, \mathrm{u}$
• Mass of protactinium isotope (${}_{91}\mathrm{Pa}^{234}$): $234.042762 \, \mathrm{u}$
• Mass of beta particle (${}_{-1}\beta$): $0.0005485 \, \mathrm{u}$
• Conversion factor:
  • $1 \, \mathrm{u} = 931.5 \, \mathrm{MeV}$

Prerequisite Concepts

1. Energy Released ($Q$):
   The energy released in the reaction is the difference in mass of the reactants and products, converted into energy:

$$Q = m_{{}_{90}\mathrm{Th}^{234}} - m_{{}_{91}\mathrm{Pa}^{234}} - m_{{}_{-1}\beta}$$

2. Mass-Energy Equivalence:
   The mass difference is converted into energy using the relation:

$$Q = \Delta m \cdot 931.5 \, \mathrm{MeV}$$

Solution

Step 1: Calculate the Mass Defect ($\Delta m$)

Using the formula:

$$\Delta m = m_{{}_{90}\mathrm{Th}^{234}} - m_{{}_{91}\mathrm{Pa}^{234}} - m_{{}_{-1}\beta}$$

Substitute the given values:

$$\Delta m = 234.0436 \, \mathrm{u} - 234.042762 \, \mathrm{u} - 0.0005485 \, \mathrm{u}$$

Perform the calculation:

$$\Delta m = 0.0002895 \, \mathrm{u}$$

---

Step 2: Convert Mass Defect to Energy ($Q$)

Using the formula:

$$Q = \Delta m \cdot 931.5 \, \mathrm{MeV}$$

Substitute $\Delta m = 0.0002895 \, \mathrm{u}$:

$$Q = 0.0002895 \cdot 931.5$$

Perform the calculation:

$$Q = 0.26967 \, \mathrm{MeV}$$

---

Answer

The energy released in the $\beta$-decay reaction is:

$$Q = 0.26967 \, \mathrm{MeV}$$

---