The threshold frequency of a certain metal is `3.3xx10^(14)Hz`. If light of frequency `8.2xx10^(14)Hz` is incident on the metal, predict the cut off voltage for photoelectric emission. Given Planck's constant, `h=6.62xx10^(-34)Js`.

Explanation

Threshold frequency of the metal, `v_0=3.3xx10^14` Hz
Frequency of light incident on the metal, `v=8.2xx10^14` Hz
Charge on an electron , e=`1.6xx10^(-19)C`
Planck's constant, h=`6.626xx10^(-34)Js`
Cut-off voltage for the photoelectric emission from the metal =`V_0` the equation for the cut-of energy is given as : `eV_0=h(v-v_0)`
`V_0=(h(v-v_0))/e`
`=(6.626xx10^(-34)xx(8.2xx10^14 -3.3xx10^14))/(1.6xx10^(-19))=2.0292 V`
Therefore, the cut-off voltage for the photoelectric emission is 2.0292 V