A stationary thorium nucleus `(A=200 , Z=90)` emits an alpha particle with kinetic energy `E_(alpha)`. What is the kinetic energy of the recoiling nucleus

Options

A. `(E_(alpha))/(108)`
B. `(E_(alpha))/(110)`
C. `(E_(alpha))/(55)`
D. `(E_(alpha))/(54)`

Explanation

To find the kinetic energy of the recoiling nucleus after a stationary thorium nucleus emits an alpha particle, we can use the principle of conservation of momentum. Here’s the step-by-step solution:

### Step 1: Understand the Reaction
When a thorium nucleus (Th) emits an alpha particle, it transforms into a new nucleus (let's call it X). The thorium nucleus has a mass number \( A = 200 \) and atomic number \( Z = 90 \). The alpha particle has a mass number \( A = 4 \) and atomic number \( Z = 2 \).

### Step 2: Determine the Mass Numbers
After the emission of the alpha particle, the new nucleus X will have:
- Mass number: \( A_X = 200 - 4 = 196 \)
- Atomic number: \( Z_X = 90 - 2 = 88 \)

### Step 3: Apply Conservation of Momentum
Since the thorium nucleus is initially at rest, the total initial momentum is zero. According to the conservation of momentum:
\[ \text{Initial Momentum} = \text{Final Momentum} \]
This means:
\[ 0 = p_{\alpha} + p_X \]
Where \( p_{\alpha} \) is the momentum of the alpha particle and \( p_X \) is the momentum of the recoiling nucleus X.

### Step 4: Express Momentum in Terms of Kinetic Energy
The momentum of the alpha particle can be expressed as:
\[ p_{\alpha} = \sqrt{2 m_{\alpha} E_{\alpha}} \]
Where \( m_{\alpha} \) is the mass of the alpha particle and \( E_{\alpha} \) is its kinetic energy.

The momentum of the recoiling nucleus can be expressed similarly:
\[ p_X = \sqrt{2 m_X E_X} \]
Where \( m_X \) is the mass of the recoiling nucleus and \( E_X \) is its kinetic energy.

### Step 5: Set Up the Equation
From the conservation of momentum:
\[ \sqrt{2 m_{\alpha} E_{\alpha}} + \sqrt{2 m_X E_X} = 0 \]
This implies:
\[ \sqrt{2 m_{\alpha} E_{\alpha}} = \sqrt{2 m_X E_X} \]
Squaring both sides gives:
\[ 2 m_{\alpha} E_{\alpha} = 2 m_X E_X \]
This simplifies to:
\[ m_{\alpha} E_{\alpha} = m_X E_X \]

### Step 6: Solve for \( E_X \)
We can rearrange the equation to find \( E_X \):
\[ E_X = \frac{m_{\alpha}}{m_X} E_{\alpha} \]

### Step 7: Substitute Mass Values
The mass of the alpha particle \( m_{\alpha} \) is approximately 4 u (atomic mass units), and the mass of the recoiling nucleus \( m_X \) is approximately 196 u. Thus:
\[ E_X = \frac{4}{196} E_{\alpha} = \frac{E_{\alpha}}{49} \]

### Conclusion
The kinetic energy of the recoiling nucleus is:
\[ E_X = \frac{E_{\alpha}}{49} \]