What percentage increase in wavelength leads to `75%` loss of photon energy in a photon-electron collision?

Options

A. a.200
B. b.100
C. c.67
D. d.300

Explanation

To solve the problem of finding the percentage increase in wavelength that leads to a 75% loss of photon energy in a photon-electron collision, we can follow these steps:

### Step 1: Understand the relationship between energy and wavelength
The energy \( E \) of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. As the wavelength increases, the energy decreases.

### Step 2: Define the initial and final energies
Let the initial energy of the photon be \( E \). A 75% loss of energy means that the final energy \( E' \) is:
\[ E' = E - 0.75E = 0.25E \]

### Step 3: Relate the final energy to the final wavelength
Using the energy-wavelength relationship for the final energy:
\[ E' = \frac{hc}{\lambda'} \]
Substituting \( E' = 0.25E \) into the equation gives:
\[ 0.25E = \frac{hc}{\lambda'} \]

### Step 4: Substitute the initial energy expression
Now, substitute \( E = \frac{hc}{\lambda} \) into the equation:
\[ 0.25 \left(\frac{hc}{\lambda}\right) = \frac{hc}{\lambda'} \]

### Step 5: Simplify the equation
Cancel \( hc \) from both sides:
\[ 0.25 \cdot \frac{1}{\lambda} = \frac{1}{\lambda'} \]
This leads to:
\[ \lambda' = 4\lambda \]

### Step 6: Calculate the percentage increase in wavelength
The increase in wavelength is:
\[ \Delta \lambda = \lambda' - \lambda = 4\lambda - \lambda = 3\lambda \]
The percentage increase in wavelength is given by:
\[ \text{Percentage Increase} = \left(\frac{\Delta \lambda}{\lambda}\right) \times 100 = \left(\frac{3\lambda}{\lambda}\right) \times 100 = 300\% \]

### Conclusion
The percentage increase in wavelength that leads to a 75% loss of photon energy is:
\[ \boxed{300\%} \]