An electron and a photon have same de Broglie wavelength (say 1 Å). Which one possesses more kinetic energy?

Kinetic energy of electron is  $E_e=\frac{1}{2}m{v}^2=\frac{1}{2}\frac{(mv{)}^2}{m}=\frac{p^2}{2m}=\frac{h^2}{2m{\lambda }^2} \left\{\because p=\frac{h}{\lambda }\right\}$

Energy of a photon is given by  $E_p=h\nu =\frac{hc}{\lambda }$

$\Rightarrow \frac{E_p}{E_e}=\frac{hc}{\lambda }·\frac{2m{\lambda }^2}{h^2}=\frac{2mc\lambda }{h}=\frac{2\times 9.11\times {10}^{-31}\times 3\times {10}^8\times 1\times {10}^{-10}}{6.63\times {10}^{-34}}=82.44$

Thus, the KE of a photon is greater than that of an electron of same wavelength.