Question

The work function of cesium is $2.14eV$. Find the threshold frequency for cesium. (Take $h=6.6\times {10}^{-34}Js$)

Answer 1

The work function is defined as $W=h{\nu }_0$ where ${\nu }_0$ is the threshold frequency.

Thus, ${\nu }_0=\frac{W}{h}=\frac{2.14\times (1.6\times {10}^{-19})J}{6.6\times {10}^{-34}Js}=5.19\times {10}^{14}{s}^{-1}orHz$ where $1eV=1.6\times {10}^{-19}J$