Question

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

Answer 1

The correct answer is: 2.0292V.
Threshold frequency of the metal, $v_0=3.3×10^{14}Hz.$
Frequency of light incident on the metal, $v=8.2×10^{14}Hz$
Charge on an electron, $e=1.6×10^{−19}C$
Planck’s constant, $h=6.626×10^{−34}Js$
Cut-off voltage for the photoelectric emission from the metal$=V_0$
The equation for the cut-off energy is given as:
$eV_0=h(v-v_0)$
$v_0=\frac{h(v-v_0)}{e}$
$=\frac{6.62×10^{-34}×(8.2×10^{14}-3.3×10^{14})}{1.6×10^{-19}}=2.0292V$
Therefore,the cut-off voltage for the photoelectric emission is 2.0292V.