The wavelength `lambda` of a photon and the de-Broglie wavelength of an electron have the same value. Show that the energy of the photon is `(2 lambda mc)/h` times the kinetic energy of the electron, Where m,c and h have their usual meanings.

Explanation

To solve the problem, we need to show that the energy of a photon \( E_p \) is \( \frac{2 \lambda mc}{h} \) times the kinetic energy \( K \) of an electron, given that both the photon and the electron have the same wavelength \( \lambda \).

### Step-by-Step Solution:

1. **Energy of the Photon**:
The energy of a photon is given by the formula:
\[
E_p = \frac{hc}{\lambda}
...