Question

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $6630\mathring{A}$ is $0.42V$. If the threshold frequency is $x\times$ $10^{13}/s$, where $x$ is _____ (nearest integer).
(Given, speed light $=3\times 10^{8}m/s$, Planck's constant $=6.63\times 10^{-34}Js$ )

Answer 1

Answer: 35

Stopping potential $V_0=0.42V$
$\lambda =6630\mathring{A}$
$E=\varphi +e{V}_0$
$E$ : energy of incident photon
$V_0$ : Stopping potential
$\varphi =E-e{V}_0$
$E=\frac{12400}{6630}eV=1.87eV$
$\varphi =(1.87-0.42)=1.45eV$
$\varphi =h{v}_0;v_0:$threshold frequency
$1.45\times 1.6\times 10^{-19}=6.63\times 10^{-34}\times v_0$
$v_0=0.35\times 10^{15}$
$=35\times 10^{13}{\sec }^{-1}$
$=35$