Question

The threshold frequency for a certain metal is $3.3\times {10}^{14}cycle/sec$. If incident light on the metal has a cut-off frequency $8.2\times {10}^{14}cycle/sec$, calculate the cut-off potential for the photoelectron.

• A. $2$ volt.
• B. $0.5$ volt.
• C. $1$ volt.
• D. $4$ volt.

Answer 1

The correct option is A $2$ volt.

$KE=h(v-v_0)$

Where $v=$ incident frequency,

$v_0=$ threshold frequency.

$\therefore KE=6.626\times {10}^{-34}(8.2-3.3)\times {10}^{14}=3.246\times {10}^{-19}J=\frac{3.246\times {10}^{-19}J}{1.6\times {10}^{-19}}eV=2eV$

$\therefore$ cut off potential $=2v$.