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 a photon and the de Broglie wavelength of an electron have the same value under the given condition, we can follow these steps:

### 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.
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