Question

In a photocell circuit the stopping potential $V_{0}$ is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency $v$ of incident light.

The values of Planck's constant and the work function as determined from the graph are (taking, the magnitude of electronic charge to be $e=1.6 \times 10^{-19}\, C$ )

• A. $6.4 \times 10^{-34} \,Js , 2.0\, eV$
• B. $6.0 \times 10^{-34}\, Js , 2.0\, eV$
• C. $6.4 \times 10^{-34} \,Js , 3.2\, eV$
• D. $6.0 \times 10^{-34}\, Js , 3.2 \,eV$

Answer 1

Answer: (B)

For a photoelectron,
$e V_{0}=h v-\phi_{0}$
$ \Rightarrow V_{0}=\frac{h}{e} \cdot v-\frac{\phi_{0}}{e}$
So, in a $V_{0}$ versus $v$ graph,
slope $=\frac{h}{e}$ and intercept $=\frac{\phi_{0}}{e} .$
Now from given graph,

Slope $=\frac{h}{e}=\frac{V_{0_{2}}-V_{0_{1}}}{v_{2}-v_{1}}=\frac{4-1}{(1.6-0.8) \times 10^{15}} $
$\Rightarrow h=\left(\frac{3 \times 1.6 \times 10^{-19}}{0.8 \times 10^{15}}\right)=6 \times 10^{-34}\, Js$
Also, from graph intercept is - $2\, V$.
Hence, work function is $\phi=2\, eV$.