In an experiment on photoelectric emission from a metallic surface, wavelength of incident light is $2 imes 10^{-7} \mathrm{~m}$ and stopping potential is $2.5 \mathrm{~V}$. The threshold frequency of the metal (in $\mathrm{Hz}$ ) approximately (charge of electron $e=1.6 imes 10^{-19}…

Question

In an experiment on photoelectric emission from a metallic surface, wavelength of incident light is $2 	imes 10^{-7} \mathrm{~m}$ and stopping potential is $2.5 \mathrm{~V}$. The threshold frequency of the metal (in $\mathrm{Hz}$ ) approximately (charge of electron $e=1.6 	imes 10^{-19} \mathrm{C}$, Planck's constant $\left.h=6.6 	imes 10^{-34} \mathrm{~J}-\mathrm{s}ight)$

Aicharya · Accepted Answer

$e V_0=h v-h v_0$ $	herefore$ Threshold frequency, $\begin{aligned} v_0 & =v-\frac{e V_0}{h} \ & =\frac{c}{\lambda}-\frac{e V_0}{h} \ v_0 & =\frac{3 	imes 10^8}{2 	imes 10^{-7}}-\frac{1.6 	imes 10^{-19} 	imes 2.5}{6.6 	imes 10^{-34}} \ & =9.0 	imes 10^{14} \mathrm{~Hz}\end{aligned}$

In an experiment on photoelectric emission from a metallic surface, wavelength of incident light is $2…

Solution