An electron and photon have same energy 100 eV. Which has greater associated wavelength ?

Explanation

de Broglie wavelength assciated with electron
`lambda_(e ) = ( h )/( sqrt( 2 m E_(c ))) rArr E_(e ) =( h^(2))/( 2mlambda_(e )^(2)) ` ....(1)
Also wavelength of photon of energy `E_(ph)` is
`E_(ph) = ( hc)/( lambda_(ph)) rArr E_(ph)^(2) = (h^(2) c^(2))/( lambda_(ph)^(2))` ...(2)
Given `E_(c ) = E_(ph) = E ` ( say ) = 100 eV ....(3)
Dividing ( 2) by (1) and using (3), we get
`E = (h^(2) c^(2)//lambda_(ph)^(2))/(h^(2) //2mlambda_(e )^(2))` or `E =( 2mc^(2) lambda_(e )^(2))/( lambda_(ph)^(2))`