The de-Broglie wavelength `lambda` of a particle is related to ites kinetic energy E as:

Explanation

Wavelength of photon= Wavelength of electron `=lamda= 0.75Å`
`rArr lamda= 0.75 xx 10^(-10)m`
We know that de-Broglie wavelength of an electron `=lamda= (h)/(sqrt(2mE))`
where E= kinetic energy of electron
`rArr lamda^(2)= (h)/(2mE)`
`E= (h)/(2m lamda^(2))`
Also the K.E of the photon is given by E `=(hc)/(lamda)`
Now `(E.)/(E )= (hc)/(lamda) xx (2m lamda^(2))/(h^(2))= (2mc lamda)/(h)`
`m= 9.1 xx 10^(-31)kg, h= 6.62 xx 10^(-34)Js, c= 3 xx 10^(8) ms^(-1)`
`rArr (E.)/(E )= (2 xx 9.1 xx 10^(-31) xx 3 xx 10^(8) xx 0.75 xx 10^(-10))/(6.62 xx 10^(-34))`
`=6.186 xx 10^(1)= 61.86`