An `alpha`-particle colliding with one of the electrons in a gold atom looses

Options

A. Most of its momentum
B. About `1/3`rd of its momentum
C. Little of its energy
D. Most of its energy

Answer

To solve the problem of an alpha particle colliding with an electron in a gold atom, we can follow these steps:

### Step 1: Understand the scenario
An alpha particle (which consists of 2 protons and 2 neutrons) collides with an electron in a gold atom. We need to analyze the collision in terms of momentum and energy.

### Step 2: Define the variables
Let:
- \( m_1 \) = mass of the alpha particle
- \( m_2 \) = mass of the electron
- \( V \) = initial velocity of the alpha particle
- \( V_1 \) = final velocity of the alpha particle after the collision
- \( V_2 \) = final velocity of the electron after the collision

### Step 3: Apply conservation of linear momentum
The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Thus, we can write:

\[ m_1 V = m_1 V_1 + m_2 V_2 \tag{1} \]

### Step 4: Consider the nature of the collision
Since the collision is elastic, we can also use the coefficient of restitution, which for elastic collisions is equal to 1. This gives us the equation:

\[ V = V_2 - V_1 \tag{2} \]

### Step 5: Rearrange the equations
From equation (2), we can express \( V_2 \) in terms of \( V \) and \( V_1 \):

\[ V_2 = V + V_1 \tag{3} \]

### Step 6: Substitute equation (3) into equation (1)
Now, substitute equation (3) into equation (1):

\[ m_1 V = m_1 V_1 + m_2 (V + V_1) \]

This simplifies to:

\[ m_1 V = m_1 V_1 + m_2 V + m_2 V_1 \]

### Step 7: Combine like terms
Rearranging gives us:

\[ m_1 V = (m_1 + m_2) V_1 + m_2 V \]

### Step 8: Isolate \( V_1 \)
Now, isolate \( V_1 \):

\[ (m_1 + m_2) V_1 = m_1 V - m_2 V \]

\[ V_1 = \frac{m_1 V - m_2 V}{m_1 + m_2} \]

\[ V_1 = \frac{(m_1 - m_2)V}{m_1 + m_2} \tag{4} \]

### Step 9: Analyze the result
Since the mass of the electron \( m_2 \) is much smaller than the mass of the alpha particle \( m_1 \), we can conclude that:

\[ V_1 \approx \frac{m_1 V}{m_1} = V \text{ (approximately)} \]

This indicates that the speed of the alpha particle after the collision is slightly less than its initial speed, meaning it loses very little energy.

### Conclusion
Thus, the alpha particle loses little of its energy in the collision with the electron.

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