The threshold frequency for a certain metal is `3.3 xx 10^14 Hz`. If light of frequency `8.2 xx 10^14 Hz` is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Explanation

Threshold frequency= `v_0 = 3.3 xx 10^(14)Hz`
Frequency of incident light = `v= 8.2 xx 10^14 Hz `
Cut off voltage `= V_s = ? `
`h = 7.625 xx 10^(-34)`
`e = 1.6 xx 10^(-19) C `
`eV_s = h(v - v_0)`
`V_s = (h(v - v_0))/(e)`
`V_s = (6.625 xx 10^(-34)(8.2 xx 10^(14) - 3.3 xx 10^(14)))/(1.6 xx 10^(-19))`
`V_s = 20.3 xx 10^(-1) = 2.03 V`