Question

the work function of silver is 4.73eV EM radiation with a frequency of 1.2x10^15 Hz strikes a piece of pure silver,what is the speed of the electrons that are emitted

Answer 1

The photoelectric effect equation says

$$hf=\phi+\frac{mv^2}{2}$$

Hence, the speed of the electrons that are emitted

$$v=\sqrt{\frac{2(hf-\phi)}{m}}$$

$$=\sqrt{\frac{2(6.67\times 10^{-34}\times 1.2\times 10^{15}-4.73\times 1.6\times 10^{-19})}{9.1\times 10^{-31}}}$$

$$=3.1\times 10^{5}\:\rm m/s$$