The distance of closest approach of an `alpha`-particle fired at nucleus with momentum p is d. The distance of closest approach when the `alpha`-particle is fired at same nucleus with momentum 3p will be

Options

A. 3d
B. `(d)/(3)`
C. 9d
D. `(d)/(9)`

Explanation

To solve the problem of finding the distance of closest approach of an alpha particle fired at a nucleus with momentum \(3p\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Distance of Closest Approach**:
The distance of closest approach \(d\) for an alpha particle is determined by the balance between its kinetic energy and the potential energy due to the electrostatic force from the nucleus.

2. **Kinetic Energy in Terms of Momentum**:
The kinetic energy \(KE\) of the alpha particle can be expressed in terms of its momentum \(p\):
\[ KE = \frac{p^2}{2m} \]
where \(m\) is the mass of the alpha particle.

3. **Potential Energy at Closest Approach**:
At the distance of closest approach \(d\), the kinetic energy is equal to the potential energy \(U\) due to the nucleus:
\[ KE = U \]
The potential energy can be expressed as:
\[ U = \frac{K}{d} \]
where \(K\) is a constant related to the charge of the nucleus.

4. **Setting Up the Equation**:
From the above, we can equate the two expressions:
\[ \frac{p^2}{2m} = \frac{K}{d} \]
Rearranging gives us:
\[ K = \frac{p^2 d}{2m} \quad \text{(Equation 1)} \]

5. **Considering the New Momentum**:
Now, when the alpha particle is fired with momentum \(3p\), we can express the new kinetic energy \(KE'\):
\[ KE' = \frac{(3p)^2}{2m} = \frac{9p^2}{2m} \]
Setting this equal to the new potential energy \(U'\) at the new distance of closest approach \(d'\):
\[ \frac{9p^2}{2m} = \frac{K}{d'} \]
Rearranging gives us:
\[ K = \frac{9p^2 d'}{2m} \quad \text{(Equation 2)} \]

6. **Equating the Two Expressions for K**:
From Equation 1 and Equation 2, we can set them equal to each other:
\[ \frac{p^2 d}{2m} = \frac{9p^2 d'}{2m} \]
The \(p^2\) and \(2m\) terms cancel out:
\[ d = 9d' \]

7. **Solving for \(d'\)**:
Rearranging gives:
\[ d' = \frac{d}{9} \]

### Final Answer:
The distance of closest approach when the alpha particle is fired with momentum \(3p\) is:
\[ d' = \frac{d}{9} \]

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