the threashold frequency `v_(0)` for a metal is `7xx10^(14)s^(-1)` . Calculate the kinetic energy of an electron emitted when radiation of fequency `v = 1.0 xx 10^(15)s^(-1)` hits the metal .

Answer

`1.99 xx 10^(-19) J`

Explanation

K.E. = `hv - hv_0 = h(v - v_0)`
`h = 6.63 xx 10^(-34) Js`,
`v = 1.0 xx 10^(15) s^(-1) , v_0 = 7.0 xx 10^(14) s^(-1)`
`:. K.E. = 6.63 xx 10^(-34) (10.0 xx 10^(14) - 7.0 xx 10^(14))`
`= 6.63 xx 10^(-34) xx 3.0 xx 10^(14) = 1.99 xx 10^(-19) J`
`= 3.44 xx 10^(-22) J`.