A metal has a threshold wavelength of 6000 Å. Calculate : Threshold frequency
given : `h= 6.62 xx 10^(-34) Js, e = 1.6 xx 10^(-19) C`.

Explanation

Threshold wavelength = `lambda_0 = 6000 Å `
`lambda_0 = 6000 xx 10^(-10)m`
(a) `c = lambda_0 v_0`
`v_0 = c/(lambda_0) = (3 xx 10^8)/(6 xx 10^(-7))`
`v_0 = 0.5 xx 10^(15)`
Threshold frequency = `v_0 = 5 xx 10^14 Hz`
(b) Work function of metal = `phi_0 = hv_0`
`phi_0 = 6.62 xx 10^(-34) xx 5 xx 10^14`
`= 33.10 xx 10^(-20)`
`phi_0 = 3.31 xx 10^(-19) J`
`phi_0 = (3.31 xx 10^(-19))/(1.6 xx 10^(-19)) eV`
`phi_0 = 2.07 eV`