The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $6630\ \text{Å}$ is $0.42\ \text{V}$. If the threshold frequency is $x\times10^{13}\ \text{s}^{-1}$, where $x$ is (nearest integer): (Given, speed of light $= 3\times10^8\ \text{m s}^{-1}$, Planck's constant $= 6.63\times10^{-34}\ \text{J s}$)
Numerical value type. Enter your answer.
Answer: 35
Photon energy: $E = \dfrac{hc}{\lambda} = \dfrac{6.63\times10^{-34}\times3\times10^8}{6630\times10^{-10}} = 3\times10^{-19}\ \text{J}$.
Work function: $\phi = E - eV_0 = 3\times10^{-19} - 0.42\times1.6\times10^{-19} = 2.328\times10^{-19}\ \text{J}$.
$$\nu_0 = \frac{\phi}{h} = \frac{2.328\times10^{-19}}{6.63\times10^{-34}}\approx3.5\times10^{14} = 35\times10^{13}\ \text{s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics