Energy of a photon = E , kinetic energy of a proton is also the same as energy of the photon. Corresponding wavelength of the photon and de Broglie wavelength of the proton are `lamda_(1) and lamda_(2)`, respectively. What is the relation between E and the ratio of `lamda_(1) and lamda_(2)`?

Explanation

In case of photon, `E = hf = (hc)/(lamda_(1))`
or, `lamda_(1) = (hc)/(E)`
In case of proton, `E = (p^(2))/(2m)` [p = momentum of proton]
Hence, `p = sqrt(2mE) and lamda_(2) = (h)/(p) = (h)/(sqrt(2mE))`
i.e., `(lamda_(2))/(lamda_(1)) = (h)/(sqrt(2mE)) .(E)/(hc) = (1)/(c) sqrt((E)/(2m)) or, (lamda_(2))/(lamda_(1)) prop sqrtE`