If the momentum of an electron is changed by `p`, then the de - Broglie wavelength associated with it changes by `0.5 %`. The initial momentum of electron will be

Options

A. 200p
B. 400p
C. `(p)/(200)`
D. 100p

Explanation

To solve the problem step by step, we will use the de Broglie wavelength formula and the relationship between changes in momentum and wavelength.

### Step-by-Step Solution:

1. **Understanding the de Broglie Wavelength**:
The de Broglie wavelength (\( \lambda \)) of 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. **Change in Wavelength**:
According to the problem, the change in momentum (\( \Delta p \)) causes a change in the de Broglie wavelength (\( \Delta \lambda \)). The relationship between the change in wavelength and the change in momentum can be expressed as:
\[ \frac{\Delta \lambda}{\lambda} = -\frac{\Delta p}{p} \]
The negative sign indicates that an increase in wavelength corresponds to a decrease in momentum and vice versa.

3. **Given Information**:
We know that the change in wavelength is 0.5%, which can be expressed as:
\[ \frac{\Delta \lambda}{\lambda} = \frac{0.5}{100} = 0.005 \]

4. **Substituting Values**:
We can substitute this value into the equation:
\[ 0.005 = -\frac{\Delta p}{p} \]
This implies:
\[ \Delta p = -0.005p \]

5. **Finding Initial Momentum**:
Rearranging the equation gives us:
\[ \frac{\Delta p}{\Delta \lambda/\lambda} = p \]
Substituting \( \Delta p = p \) and \( \Delta \lambda/\lambda = 0.005 \):
\[ p = \frac{p}{0.005} \]
Thus, we can express the initial momentum \( p \) in terms of the change in momentum \( p \):
\[ p = 200 \Delta p \]

6. **Final Result**:
Therefore, the initial momentum of the electron is:
\[ p = 200p \]

### Conclusion:
The initial momentum of the electron is \( 200p \).