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 find the original momentum of the photon given the change in its de Broglie wavelength is 0.2%. Let's break down the solution step by step.

### Step 1: Understand the relationship between de Broglie wavelength and momentum
The de Broglie wavelength (λ) is related to the momentum (p) of a particle by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant.

### Step 2: Express the change in wavelength in terms of momentum
The fractional change in wavelength can be expressed as:
\[ \frac{\Delta \lambda}{\lambda} = -\frac{\Delta p}{p} \]
where:
- \( \Delta \lambda \) is the change in wavelength,
- \( \lambda \) is the original wavelength,
- \( \Delta p \) is the change in momentum,
- \( p \) is the original momentum.

### Step 3: Convert the percentage change in wavelength to a fraction
Given that the change in wavelength is 0.2%, we convert this percentage to a fraction:
\[ \frac{\Delta \lambda}{\lambda} = \frac{0.2}{100} = 0.002 \]

### Step 4: Substitute the fractional change into the equation
Substituting the fractional change into the equation from Step 2:
\[ 0.002 = -\frac{\Delta p}{p} \]

### Step 5: Rearranging the equation
From the equation, we can express the change in momentum in terms of the original momentum:
\[ \Delta p = -0.002 p \]

### Step 6: Relate the change in momentum to the given change \( p' \)
We know that the change in momentum is given as \( p' \):
\[ p' = -0.002 p \]

### Step 7: Solve for the original momentum \( p \)
Rearranging the equation gives:
\[ p = -\frac{p'}{0.002} \]
Calculating this gives:
\[ p = -500 p' \]

### Conclusion
Thus, the original momentum of the photon was:
\[ p = 500 p' \]