If the energy of the photon is ` (2 lambda_(p)mc)/(h) ` times the kinetic energy of the electron then show then the wavelength ` lambda ` of photon and the de Broglie wavelength of an electron have the same value
(Where m, c and h have their usual meanings .)

Explanation

To show that the wavelength \( \lambda \) of the photon and the de Broglie wavelength of the electron have the same value given that the energy of the photon is \( \frac{2 \lambda_p mc}{h} \) times the kinetic energy of the electron, we can follow these steps:

### Step 1: Write the expression for the energy of the photon
The energy \( E_p \) of a photon can be expressed in terms of its wavelength \( \lambda_p \):
\[ E_p = \frac{hc}{\lambda_p} \]

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