An electron and a photon have same de Broglie wavelength `lamda = 10^(-1)m`. Compare the kinetic energy of the electron with the total energy of the photon.

Explanation

the kintic energy of an electron having mass m and velocity v, `K = (1)/(2) mv^(2)`
Wavelength, `lamda = (h)/(mv) or, v = (h)/(m lamda)`
`:. K = (1)/(2) m. (h^(2))/(m^(2) lamda^(2)) = (h^(2))/(2m lamda^(2))`
The energy of the photon of wavelength `lamda, E = (hc)/(lamda)`
`:. (K)/(E) = (h^(2))/(2m lamda^(2)). (lamda)/(hc) = (h)/(2m lamdac)`
`= (6.6 xx 10^(-34))/(2 xx 9.1 xx 10^(-31) xx 10^(-10) xx 3 xx 10^(8)) = 0.012 lt 1`
Hence, the kinetic energy of the electron is lower than the total energy of the photon.