Question

In a photoelectric effect experiment, irradiation of a metal with light of frequency $5.2\times {10}^{14}{s}^{-1}yiel{d}^{\prime }s$ electrons with maximum kinetic energy $1.3\times {10}^{-19}$. Calculate the threshold frequency $\left(v_e\right)$ for the metal?

Answer 1

Threshold frequency ${\nu }_0$,

${\nu }_0=\frac{(h\nu -E)}{h}$

$h=6.627\times 10^{-34}J{S}^{-1}$

${\nu }_0=\frac{(6.627\times 10^{-34}J{S}^{-1}\times 5.2\times 10^{14}{S}^{-1})-(1.3\times 10^{-19}J)}{6.627\times 10^{-34}J{S}^{-1}}$

$=3.238\times {10}^{14}{S}^{-1}$