Question

A metal has a work function of $2.5\times {10}^{-19}$ J. What should be the threshold frequency beyond which if I shine light on the metal, photo electrons will come out?

• A. $3.8\times {10}^{14}$Hz.
• B. $2.3\times {10}^{14}$Hz.
• C. $3\times {10}^{17}$Hz.
• D. none of these

Answer 1

The correct option is A

$3.8\times {10}^{14}$Hz.

So, the work function of the metal is $2.5\times {10}^{-19}$J.

This means that we need to give an electron a minimum of $2.5\times {10}^{-19}$J of energy for it to come out of the metal. This energy needs to be supplied by light, in this case a photon.

So if the photon has a minimum energy of more than $2.5\times {10}^{-19}$J, then I am done.

Energy of a photon is hv

So hv$\ge 2.5\times {10}^{-19}$J.

v$\ge \frac{2.5\times {10}^{-19}}{h}$

v$\ge \frac{2.5\times {10}^{-19}}{6.626\times {10}^{-34}}$

v$\ge 3.8\times {10}^{14}$Hz

So cut off frequency is 3.8$\times {10}^{14}$Hz.