The threshold wavelength of a metal is `230 nm` calculate the `KE` of the electrons from that metal surface by using UV radiation of wavelength `180 nm`

Explanation

From Einstein's law ,we have
`(hc)/(lambda) = (hc)/(lambda_(0)) + (1)/(2) mv^(2)`
`or hc((1)/(lambda) - (1)/(lambda_(0))) = KE`
`or 6.63 xx 10^(-34) xx 3 xx 10^(8) ((1)/(180 xx 10^(-9)) - (1)/(230 xx 10^(-9))) = KE`
`(therefore" 1 nm = 10^(-9) m)`
`or 6.63 xx 3 xx 10^(-26) ((230 xx 10^(-9) - 180 xx 10^(-9))/( 41400 xx 10^(-19))) = KE`
`or 6.63 xx 3 xx 10^(-26)(( 230 xx 10^(-8) - 180 xx 10^(-8))/(41400 xx 10^(-19))) = KE`
`or KE = 6.63 xx 3 xx 10^(-26)[(230 xx 18)/(41400)]`
`= 19.89 xx 10^(-20) xx (50)/(41400)`
`= (99.45)/(4140) xx 10^(-26) J = 0.0244 xx 10^(-26) J`
`= 24.4 xx 10^(-20)J`