Question

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

Short answer

$300\%$

Step-by-step solution

Identification of the given data

The given data is listed below as-

There is a 75 % loss of photon energy

Significance of the momentum of a photon

The momentum of a photon is related to its energy and is given by-

$\mathbf{p}\mathbf{=}\frac{E}{c}$

Here, E is the rest energy of a photon which is given by$E=m_e{c}^2$.

$m_e$ Is the mass of electron and is the speed of light.

To find the percentage increase in wavelength

Let E is the energy of photon initially and$E\text{'}$is the energy of photon after collision. The fractional energy loss is given by-

$\frac{\Delta E}{E}=\frac{E-E\text{'}}{E}=\frac{\Delta \lambda }{\lambda +\Delta \lambda }$

Now,

$\begin{array}{c}\frac{\Delta \lambda }{\lambda }=\frac{\Delta E/E}{1-\Delta E/E}\\ =\frac{0.75}{1-0.75}\\ =3\\ =300\%\end{array}$

Thus, the percentage increase in wavelength is $300\%$.