The de-Broglie wavelength associated with proton changes by `0.25%` if its momentum is changed by `p_(0)`. The initial momentum was

Options

A. `100p_(0)`
B. `p_(0)//400`
C. `401 p_(0)`
D. `p_(0)//100`

Explanation

To solve the problem, we need to find the initial momentum \( p \) of the proton given that its de-Broglie wavelength changes by \( 0.25\% \) when its momentum is changed by \( p_0 \).

### Step-by-Step Solution:

1. **Understanding de-Broglie Wavelength**:
The de-Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Initial and Final Wavelengths**:
Let the initial momentum be \( p \). Then the initial wavelength is:
\[ \lambda = \frac{h}{p} \]
If the momentum changes by \( p_0 \), the new momentum becomes \( p - p_0 \). The new wavelength \( \lambda' \) can be expressed as:
\[ \lambda' = \frac{h}{p - p_0} \]

3. **Change in Wavelength**:
According to the problem, the change in wavelength is \( 0.25\% \) of the initial wavelength:
\[ \lambda' = \lambda + 0.0025\lambda = 1.0025\lambda \]
Therefore, we can write:
\[ \frac{h}{p - p_0} = 1.0025 \cdot \frac{h}{p} \]

4. **Canceling \( h \)**:
Since \( h \) is a constant and appears in both sides, we can cancel it out:
\[ \frac{1}{p - p_0} = 1.0025 \cdot \frac{1}{p} \]

5. **Cross Multiplying**:
Cross multiplying gives us:
\[ p = 1.0025(p - p_0) \]

6. **Expanding the Equation**:
Expanding the equation results in:
\[ p = 1.0025p - 1.0025p_0 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ p - 1.0025p = -1.0025p_0 \]
\[ -0.0025p = -1.0025p_0 \]

8. **Solving for \( p \)**:
Dividing both sides by \( -0.0025 \):
\[ p = \frac{1.0025}{0.0025} p_0 \]
\[ p = 401p_0 \]

### Final Answer:
The initial momentum \( p \) of the proton is:
\[ p = 401 p_0 \]