Question

A surface has light of wavelength ${{\lambda }_{1}}=550nm$ incident on it causing the ejection of photoelectrons for which the stopping potential is${{V}_{s1}}=0.19V$. If the radiation of wavelength ${{\lambda }_{2}}=190nm$ is now incident on the surface calculate the threshold frequency for the surface.
A. $500\times {{10}^{12}}Hz$
B. $480\times {{10}^{13}}Hz$
C. $520\times {{10}^{11}}Hz$
D. $460\times {{10}^{13}}Hz$

Answer 1

As a first step, you could note down the given values. Then, you could make use of Einstein’s law of photoelectric effect in order to find the work function of the surface. From the work function we could easily find the threshold frequency which is the minimum frequency required by the incident light to cause ejection of electrons from the surface.
Formula used:
Einstein’s law of photoelectric effect,
$E=h\nu =\dfrac{hc}{\lambda }=\phi +eV$

Complete answer:
In the question we are given a case of incidence of certain light of wavelength ${{\lambda }_{1}}=550nm$ on a surface resulting in the ejection of electrons with stopping potential${{V}_{s1}}=0.19V$. We are supposed to find the threshold frequency of the surface when a light wavelength ${{\lambda }_{2}}=190nm$ is incident.
We know that Einstein’s law of photoelectric effect is given by,
$E=h\nu =\dfrac{hc}{\lambda }=\phi +eV$ …………………………………… (1)
Where E is the energy of the incident light and $\phi$ is the work function
From (1),
$\phi =\dfrac{hc}{\lambda }-0.19eV$
$\Rightarrow \phi =\dfrac{1241}{550}eV-0.19eV=2.07eV$
Now we know that the work function is given by,
$\phi =h{{\nu }_{0}}$
$\Rightarrow {{\nu }_{0}}=\dfrac{\phi }{h}=\dfrac{2.07eV}{4.135\times {{10}^{-15}}eVH{{z}^{-1}}}=0.5006\times {{10}^{15}}Hz$
$\therefore {{\nu }_{0}}=500\times {{10}^{12}}Hz$
Therefore, we found the value of threshold frequency of the surface to be${{\nu }_{0}}=500\times {{10}^{12}}Hz$.

Hence, option A is the correct answer.

Note:
You may have noted that even though we are given a second case of another wavelength we have made the calculations from the first case itself. This is due to the fact that the work function and hence the threshold frequency is the property of the metal surface and hence is independent of the incident light’s wavelength. Also, note that we have substituted the standard values in electron volts.