A photon of frequency `5 xx 10^(14)Hz` is incident on the surface of a metal. If the photoelectric thresh-old frequency for the metal is `3 xx 10^(14)Hz`, Calculate the following
Energy of the photon

Explanation

Energy of the photon, E=hv
`h=6.625 xx 10^(-34)J.s, v=5 xx 10^(14)Hz`
`therefore E= (6.625 xx 10^(-34) xx 5 xx 10^(14))J`
`1.6 xx 10^(-19) J=1` electron volt
`therefore` E in electron volts
`=(6.625 xx 10^(-34) xx 5 xx 10^(14))/(1.6 xx 10^(-19))eV`
`E= (6.625 xx 5)/(1.6)xx 10^(-34) xx 10^(+33)`
= 2.07eV