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 relate the energy of a photon to the kinetic energy of an electron, given that their wavelengths are equal. Here’s a step-by-step solution:

### Step 1: Write the energy of the photon
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 of the photon.
...