Question

A metal has a work function of $2.5\times {10}^{-19}J.$The threshold frequency beyond which photo electrons will come out is.

• A. $2.3\times {10}^{14}Hz$
• B. $3.3\times {10}^{14}Hz$
• C. $2.8\times {10}^{14}Hz$
• D. $3.8\times {10}^{14}Hz$

Answer 1

The correct option is D $3.8\times {10}^{14}Hz$

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.

Energy of a photon is $h\nu$.

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

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

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

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

So, the threshold frequency is 3.8$\times {10}^{14}$ Hz.