An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths
associated with them is  (c being  velocity of light)

Since, it is given that electron has mass m. de-Broglie wavelength for an electron will be given as

${\lambda }_e=\frac{h}{p}$            …(i)

where, h = Planck's constant
and p = linear momentum of electron.
As, kinetic energy of electron,  $E=\frac{p^2}{2m}$

$\Rightarrow p=\sqrt{2mE}$    …(ii)

From Eqs. (i) and (ii), we get

${\lambda }_e=\frac{h}{\sqrt{2mE}}$         …(iii)

Energy of a photon can be given as E = hv

$\Rightarrow E=\frac{hc}{{\lambda }_p}$

$\Rightarrow {\lambda }_p=\frac{hc}{E}$           …(iv)

where, ${\lambda }_{\rho }$ = de-Broglie wavelength of photon.
Now, divide Eq. (iii) by Eq. (iv), we get

$\frac{{\lambda }_e}{{\lambda }_p}=\frac{h}{\sqrt{2mE}}·\frac{E}{hc}$

$\Rightarrow \frac{{\lambda }_e}{{\lambda }_p}=\frac{1}{c}·\sqrt{\frac{E}{2m}}$