A photon and an electron have same de Broglie wavelength. Which one has greater total energy?

Explanation

The total energy of a photon of wavelength `lamda` is given by
`E_(p) = hf = (hc)/(lamda)`...(1)
de Broglie wavelength `lamda` of an electron (same as of photon) of mass m moving with velocity v is given by
`lamda = (h)/(mv) or, m = (h)/(lamda v)`
Total energy of an electron of mass m,
`E_(e) = mc^(2)` (according to the theory of relatively)
`= (hc^(2))/(lamda v)`...(2)
Dividing equation (2) by equation (1) we have,
`(E_(e))/(E_(p)) = ((hc^(2))/(lamda v))/((hc)/(lamda)) = (c)/(v)`
Since `c gt v, :. E_(e) gt E_(p)`
So the total energy of electron is greater than that of a photon.