Consider the following nuclear decay: (initially `.^(236)U_(92)` is at rest)
`._(92)^(236)rarr._(90)^(232)ThrarrX`
Following atomic masses and conversion factor are provided
`._(92)^(236)U=236.045 562 u`,
`._(90)^(232)Th=232.038054 u`,
`._(0)^(1)n=1.008665 ,. _(1)^(1)p=1.007277 u`,
`._(2)^(4)He=4.002603 u` and
`1 u=1.5xx10^(-10)J`
The amount of energy released in this decay is equal to:

Options

A. `3.5xx10^(-8)J`
B. `4.6 xx10^(-12)J`
C. `6.0xx10^(-10)J`
D. `7.4xx10^(-13)J`

Explanation

To find the energy released in the nuclear decay of Uranium-236 to Thorium-232 and Helium-4, we will follow these steps:

### Step 1: Write down the mass-energy equivalence formula
The energy released (E) during a nuclear decay can be calculated using the mass defect (Δm) and the equation:
\[ E = \Delta m \cdot c^2 \]
However, since we are given a conversion factor for atomic mass units (u) to joules, we can directly use:
\[ E = \Delta m \cdot (1.5 \times 10^{-10} \, \text{J}) \]

### Step 2: Calculate the mass defect (Δm)
The mass defect is the difference between the initial mass of the Uranium and the total mass of the decay products (Thorium and Helium).

The masses are given as:
- Mass of Uranium-236: \( m(U) = 236.045562 \, u \)
- Mass of Thorium-232: \( m(Th) = 232.038054 \, u \)
- Mass of Helium-4: \( m(He) = 4.002603 \, u \)

Now, we calculate Δm:
\[ \Delta m = m(U) - (m(Th) + m(He)) \]
\[ \Delta m = 236.045562 \, u - (232.038054 \, u + 4.002603 \, u) \]

### Step 3: Perform the calculation for Δm
Calculating the total mass of the products:
\[ m(Th) + m(He) = 232.038054 \, u + 4.002603 \, u = 236.040657 \, u \]

Now, substituting back into the equation for Δm:
\[ \Delta m = 236.045562 \, u - 236.040657 \, u = 0.004905 \, u \]

### Step 4: Convert Δm to energy (E)
Now, we can calculate the energy released:
\[ E = \Delta m \cdot (1.5 \times 10^{-10} \, \text{J}) \]
\[ E = 0.004905 \, u \cdot (1.5 \times 10^{-10} \, \text{J}) \]

### Step 5: Calculate the energy
Calculating the energy:
\[ E = 0.004905 \times 1.5 \times 10^{-10} \]
\[ E = 7.3575 \times 10^{-13} \, \text{J} \]

### Step 6: Round the result
Rounding to two significant figures, we get:
\[ E \approx 7.4 \times 10^{-13} \, \text{J} \]

Thus, the amount of energy released in this decay is approximately \( 7.4 \times 10^{-13} \, \text{J} \).

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