Question

Question #6a4bb

Answer 1

Energy of a photon of wavelength `lambda_p` is given by the expression

`E_p=(hc)/lambda_p` ......(1)

de Brogile wavelength of an electron is expressed as

`lambda_e=h/p_e`
`=>p_e=h/lambda_e` ........(2)

Kinetic energy equation for a particle can also be written as

`KE=p^2/(2m)`

Using (2), for electron

`KE_e=(h/lambda_e)^2/(2m)`

Given that wavelength of a photon and the de Brogile wavelength of an electron has same value. Let it be equal to `lambda`. Re-writing above with the help of (1) we get

`KE_e=(h/lambdaxxE_p/c)/(2m)`
`=>KE_e=(hE_p)/(2lambdamc)`

`=>E_p=(2lambdamc)/hxxKE_e`