Find the frequency of light which ejects electrons from a metal surface, fully stopped by a retarding potential of 3 V. The photoelectric effect brings in this metal at a frequency of `6 xx 10^14 Hz`. Find the work function for this metal `(Given h=6.63 xx 10^-34 Js)`/

Explanation

`v = ?`
Stopping potential = `V_s = 3V`
Threshold frequency = `v_0 = 6 xx 10^14 s^(-1)`
`h = 6.6 xx 10^(-34) Js`
`c = 3 xx 10^8 ms^(-1)`
`e = 1.6 xx 10^(-19)C`
We know `eV_s = h(v - v_0)`
`= 1.6 xx 10^(-19) xx 3`
`= 6.6 xx 10^(-34) (v - 6 xx 10^(14))`
`(v - 6 xx 10^14) = (1.6 xx 3 xx 10^(-19))/(6.6 xx 10^(-34))`
`v - 6 xx 10^(14) = 0.73 xx 10^(-19 + 34)`
`v = 0.73 xx 10^(15) + 6 xx 10^(14)`
`= 7.3 xx 10^(14) + 6 xx 10^14`
`v = (7.3 + 6) 10^14`
`= 13.3 xx 10^14 s^(-1)`
Work function of the metal `= phi_0 = hv_0`
`phi_0 = 6.6 xx 10^(-34) xx 6 xx 10^(14)`
`phi_0 = 39.6 xx 10^(-20)`
`= 3.96 xx 10^(-20)`
`= 3.96 xx 10^(-19) J`
`phi_0 = (3.96 xx 10^(-19))/(1.6 xx 10^(-19)) eV`
`phi_0 = 2.48 eV`