Question

A student performs an experiment on photoelectric effect, using two materials $A$ and $B$. A plot of $V_{stop}$ vs $v$ is given in Figure.


A Which material $A$ or $B$ has a higher work function?

B Given the electric charge of an electron $=1.6\times {10}^{-19}C$, find the value of $h$ obtained from the experiment for both $A$ and $B$? Comment on whether it is consistent with Einstein’s theory:

Answer 1

(A) Given, threshold frequency of $A$,

$v_A=5\times {10}^{14}{H}_Z$

Threshold frequency of $B$,

$v_B=10\times {10}^{14}{H}_Z$

As we know work function,

$\varphi =h{v}_0$

$\Rightarrow {\varphi }_0\propto v_0$

Therefore,

$\frac{{\varphi }_A}{{\varphi }_B}=\frac{5\times {10}^{14}}{10\times {10}^{14}}<1$

$\Rightarrow {\varphi }_A<{\varphi }_B$ Thus, work function of $B$ is higher than$A$.

$\text{Final Answer}:$ Work function of $B$ is higher than$A$.

(B) Einstein’s photoelectric theory,

$e{V}_o=hv-{\varphi }_o$

$V_o=\frac{h}{e}v-\frac{{\varphi }_o}{e}$

Comparing with straight line equation,

$y=mx+c$

Slope, $m=\frac{h}{e}$

For metal $A$, slope $=\frac{h}{e}=\frac{2}{\left({10}^{–5}\right)\times {10}^{14}}$

$h=\frac{2e}{5\times {10}^{14}}=\frac{2\times 1.6\times {10}^{-19}}{5\times {10}^{14}}$

$=6.4\times {10}^{-34}Js$

For metal $B$, slope

$\frac{h}{e}=\frac{2.5}{(15-10)\times {10}^{14}}$

$h=\frac{2.5\times e}{5\times {10}^{14}}=\frac{2.5\times 1.6\times {10}^{-19}}{5\times {10}^{14}}$

$=8\times {10}^{-34}Js$

Since, the value of $h$ from experiment for metals $A$ and $B$ is different. Hence, experiment is not consistent with theory.

$\text{Final Answer}:$ Since $h$ works out differently, experiment is not consistent with the theory.