If the momentum of an electron is changed by `Delta p` , then the de - Broglie wavelength associated with it changes by `0.50 %`. The initial momentum of the electron will be

Options

A. `(Delta p)/(200)`
B. `(Delta p)/(199)`
C. `199 Delta p`
D. `400 Delta p`

Explanation

To solve the problem, we need to find the initial momentum of the electron given that the de Broglie wavelength changes by 0.5% when the momentum changes by Δp.

### Step-by-Step Solution:

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

2. **Express the change in wavelength**:
If the momentum changes by Δp, the new momentum \( p' \) can be expressed as:
\[ p' = p + \Delta p \]
The new wavelength \( \lambda' \) can then be expressed as:
\[ \lambda' = \frac{h}{p + \Delta p} \]

3. **Calculate the percentage change in wavelength**:
We know that the wavelength changes by 0.5%, which means:
\[ \lambda' = \lambda - 0.005\lambda = 0.995\lambda \]

4. **Set up the equation**:
Using the expressions for the wavelengths, we have:
\[ \frac{h}{p + \Delta p} = 0.995 \cdot \frac{h}{p} \]

5. **Simplify the equation**:
Cancel \( h \) from both sides:
\[ \frac{1}{p + \Delta p} = 0.995 \cdot \frac{1}{p} \]
Cross-multiplying gives:
\[ p = 0.995(p + \Delta p) \]

6. **Expand and rearrange**:
Expanding the right side:
\[ p = 0.995p + 0.995\Delta p \]
Rearranging terms gives:
\[ p - 0.995p = 0.995\Delta p \]
\[ 0.005p = 0.995\Delta p \]

7. **Solve for initial momentum \( p \)**:
Dividing both sides by 0.005:
\[ p = \frac{0.995\Delta p}{0.005} \]
\[ p = 199\Delta p \]

### Final Answer:
The initial momentum of the electron is:
\[ p = 199\Delta p \]