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

Answer 1

Answer: (B)

For a photoelectron,
$e{V}_0=hv-{\varphi }_0$
$\Rightarrow V_0=\frac{h}{e}\cdot v-\frac{{\varphi }_0}{e}$
So, in a $V_0$ versus $v$ graph,
slope $=\frac{h}{e}$ and intercept $=\frac{{\varphi }_0}{e}.$
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 - $2V$.
Hence, work function is $\varphi =2eV$.