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

We know, `eV_(0) = hf - hf_(0)`
or, `V_(0) = (h)/(e) (f - f_(0))`
`= (6.626 xx 10^(-34) (8.2 xx 10^(14) - 3.3 xx 10^(14)))/(1.6 xx 10^(-19)) = 2.03 V`