The threshold frequency for 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 the photoelectric emission.

Explanation

Given , threshold frequency , `v_0=3.3xx10^(14)Hz`
Incident frequency , `v=8.2xx10^(14)Hz`
Let `V_0` be the cut - off voltage . The
`eV_0=hv-hv_0`
`impliesV_0=h/e(v-v_0)`
`=(6.62xx10^(-34))/(1.6xx10^(-19))xx(8.2xx10^(14)-3.3xx10^(14))`
`=4.14xx10^(-15)xx4.9xx10^(14)`
= 2.02V
= 2V