A nucleus X initially at rest, undergoes alpha decay according to the equation
`_Z^232Xrarr_90^AY+alpha`
What fraction of the total energy released in the decay will be the kinetic energy of the alpha particle?

Options

A. (a) `90/92`
B. (b) `228/232`
C. (c) `sqrt(228/232)`
D. (d) `1/2`

Explanation

To solve the problem of finding the fraction of the total energy released in the alpha decay that will be the kinetic energy of the alpha particle, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the decay process**: The alpha decay of nucleus X can be represented as:
\[ _Z^{232}X \rightarrow _{90}^{A}Y + \alpha \]
Here, nucleus X decays into nucleus Y and an alpha particle.

2. **Conservation of momentum**: Since the initial nucleus X is at rest, the total initial momentum is zero. Therefore, the momentum of nucleus Y and the alpha particle must be equal and opposite:
\[ 0 = P_Y + P_{\alpha} \]
This implies:
\[ P_Y = -P_{\alpha} \]

3. **Relate kinetic energies and masses**: The momentum \( P \) of an object is given by \( P = mv \), where \( m \) is mass and \( v \) is velocity. Thus, we can write:
\[ m_Y v_Y = m_{\alpha} v_{\alpha} \]
Squaring both sides gives:
\[ m_Y^2 v_Y^2 = m_{\alpha}^2 v_{\alpha}^2 \]
Using kinetic energy \( K = \frac{1}{2}mv^2 \), we have:
\[ K_Y m_Y = K_{\alpha} m_{\alpha} \]

4. **Express kinetic energies**: Rearranging gives:
\[ \frac{K_{\alpha}}{K_Y} = \frac{m_Y}{m_{\alpha}} \]

5. **Determine the masses**: From the decay equation, the mass of nucleus Y can be found:
\[ 232 = A + 4 \implies A = 232 - 4 = 228 \]
Thus, the mass of nucleus Y is 228 and the mass of the alpha particle is 4.

6. **Substitute the masses**: Now we can substitute the masses into the kinetic energy ratio:
\[ \frac{K_{\alpha}}{K_Y} = \frac{228}{4} = 57 \]

7. **Total energy released**: The total energy released in the decay is the sum of the kinetic energies:
\[ K_{total} = K_{\alpha} + K_Y \]
We can express \( K_Y \) in terms of \( K_{\alpha} \):
\[ K_Y = \frac{4}{228} K_{\alpha} \]
Therefore:
\[ K_{total} = K_{\alpha} + \frac{4}{228} K_{\alpha} = K_{\alpha} \left(1 + \frac{4}{228}\right) = K_{\alpha} \left(\frac{232}{228}\right) \]

8. **Find the fraction**: Now we can find the fraction of the total energy that is the kinetic energy of the alpha particle:
\[ \frac{K_{\alpha}}{K_{total}} = \frac{K_{\alpha}}{K_{\alpha} \left(\frac{232}{228}\right)} = \frac{228}{232} \]

### Final Answer:
The fraction of the total energy released in the decay that is the kinetic energy of the alpha particle is:
\[ \frac{228}{232} \]