One of naturally occurring radioactive decay series begins with `_(90)^(232)Th` and ends with a stable `_(82)^(208)Pb`.
How much energy in MeV is released in the complete chain?

Answer

To find the total energy released in the radioactive decay series starting from \( _{90}^{232}Th \) and ending with \( _{82}^{208}Pb \), we can follow these steps:

### Step 1: Identify the decay process
The decay series involves the transformation of thorium-232 into lead-208. During this process, thorium emits particles, including alpha particles (helium nuclei) and beta particles (electrons).

### Step 2: Determine the number of emitted particles
From the video transcript, it is mentioned that the decay series emits:
- 6 alpha particles (each \( _{2}^{4}He \))
- 4 beta particles (each \( e^- \))

### Step 3: Write the mass-energy equivalence formula
The energy released in a nuclear reaction can be calculated using Einstein's mass-energy equivalence formula:
\[ E = \Delta m \cdot c^2 \]
Where:
- \( E \) is the energy released,
- \( \Delta m \) is the change in mass,
- \( c \) is the speed of light.

### Step 4: Calculate the change in mass
The change in mass (\( \Delta m \)) can be calculated as follows:
\[ \Delta m = \text{mass of initial nucleus} - \text{mass of final nucleus} - \text{mass of emitted particles} \]

1. The mass of thorium-232 is approximately 232.038 amu.
2. The mass of lead-208 is approximately 207.976 amu.
3. The mass of 6 alpha particles:
- Each alpha particle has a mass of approximately 4.0026 amu.
- Total mass of 6 alpha particles = \( 6 \times 4.0026 \approx 24.0156 \) amu.
4. The mass of 4 beta particles (negligible, approximately 0 amu).

Now, substituting these values:
\[ \Delta m = 232.038 - 207.976 - 24.0156 \]
\[ \Delta m \approx 232.038 - 231.9916 \]
\[ \Delta m \approx 0.0464 \text{ amu} \]

### Step 5: Convert mass change to energy
To convert the mass change to energy in MeV, we use the conversion factor:
1 amu = 931.5 MeV.

Thus,
\[ E = \Delta m \cdot 931.5 \text{ MeV/amu} \]
\[ E \approx 0.0464 \text{ amu} \cdot 931.5 \text{ MeV/amu} \]
\[ E \approx 43.28 \text{ MeV} \]

### Step 6: Conclusion
The total energy released in the complete decay chain from \( _{90}^{232}Th \) to \( _{82}^{208}Pb \) is approximately **43.28 MeV**.

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