Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength ` 6800` Å . Calculate threshold frequency `( v_0)` and work function `(W_0)` of the metal.

Answer

`4.41 xx 10^(14) s^(-1), 2.92 xx 10^(-19) J`

Explanation

K.E. = `hv - hv_0`
If the velocity is zero, K.E. = 0
`:. " " 0 = hv - hv_0`
or `" " hv = hv_0 "or" v = v_0 `
`:. v = c/(lambda) = (3.0 xx 10^8 ms^(-1))/(6800 xx 10^(-10)m)`
`= 4.41 xx 10^(14) s^(-1)`
Threshold frequency = `4.41 xx 10^(14) s^(-1)`
Work function = `hv_0 = 6.626 xx 10^(-34) xx 4.41 xx 10^(14)`
`= 2.92 xx 10^(-19) J`.