Find the ratio of the respective kinetic energies of a photon to that of an electron if it is known that the de Broglie wavelength of the electron moving with a velocity of `2 xx 10^(8 )m s^(-1)` is equal to that of the photon.

Answer

`03.00`

Explanation

Speedof photon (c) `= 3 xx 10 ^(8)ms ^(-1)`
`therefore lamda = ( h)/(mc) and lamda _(e) = ( h )/( m _(c) v)`
`implies lamda = lamda_(e)`
Now,` ("K.E. of photon")/("K.E. of electron") = (hv)/((1//2) m_(e) v ^(2))= ( 2 hc)/( m _(e) v ^(2) lamda ) = ( 2 c)/(v) .....(because lamda _(c) = (h)/(m _(e) v )) = ( 2 xx 3 xx 10 ^(8))/(2 x 10 ^(3)) = 3`