When the momentum of a photon is changed by an amount p'. The corresponding change in the di-Broglie wavelength is found to be 0.2% . Then, the original momentum of the photon was

Options

A. 300 p'
B. 500 p'
C. 400 p'
D. 100 p'

Answer

To solve the problem, we need to determine the original momentum of the photon given that the change in the de Broglie wavelength is 0.2%.

### Step-by-Step Solution:

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

2. **Differentiate the equation**:
To find the change in wavelength with respect to momentum, we differentiate both sides of the equation:
\[ d\lambda = -\frac{h}{p^2} dp \]
This shows that a change in momentum (\( dp \)) will result in a change in wavelength (\( d\lambda \)).

3. **Relate the changes**:
We can express the relative change in wavelength as:
\[ \frac{d\lambda}{\lambda} = -\frac{h}{p^2} \cdot \frac{dp}{p} \]

4. **Substitute the given values**:
We know that the percentage change in wavelength is given as 0.2%, which can be expressed as:
\[ \frac{d\lambda}{\lambda} = \frac{0.2}{100} = 0.002 \]
Therefore, we can write:
\[ 0.002 = -\frac{h}{p^2} \cdot \frac{dp}{p} \]

5. **Substituting \( dp \)**:
Let \( dp = p' \) (the change in momentum). Rearranging gives:
\[ 0.002 = -\frac{h}{p^2} \cdot \frac{p'}{p} \]

6. **Rearranging the equation**:
Rearranging the equation to solve for \( p \):
\[ p = -\frac{h}{0.002 p'} \cdot p^2 \]
This can be simplified to:
\[ p = \frac{h}{0.002 p'} \]

7. **Substituting Planck's constant**:
The value of Planck's constant \( h \) is approximately \( 6.626 \times 10^{-34} \, \text{Js} \). Substituting this value:
\[ p = \frac{6.626 \times 10^{-34}}{0.002 p'} \]

8. **Final calculation**:
Simplifying further gives:
\[ p = \frac{6.626 \times 10^{-34}}{0.002} \cdot \frac{1}{p'} = 3.313 \times 10^{-32} \cdot \frac{1}{p'} \]
Therefore, the original momentum \( p \) can be expressed as:
\[ p = 500 p' \]

### Conclusion:
The original momentum of the photon was \( 500 p' \).