A student performs an experiment on photoelectric effect, using t… - Vidyadip

Question

A student performs an experiment on photoelectric effect, using two materials $A$ and $B. A$ plot of $V$ stop vs $ν$ is given in Fig.

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

From the given figure, we have

Threshold frequency of $A, v_{0A} = 5 \times 10^{14}Hz$ and
Threshold frequency of $B, v_{0B} = 10 \times 10^{14}Hz$
Now,Work function, $\phi=\text{hv}_0$
$\Rightarrow\ \phi_0\propto\text{v}_0 \ [\because h \ ($Planck's constant$)$ is constant$]$
Now, $\frac{\phi_{0\text{A}}}{\phi_{0\text{B}}}=\frac{5\times10^{14}}{10\times10^{14}}$
$\Rightarrow\ \frac{\phi_{0\text{A}}}{\phi_{0\text{B}}}=\frac{1}{2}$
$\therefore\ {\phi_{0\text{A}}}<{\phi_{0\text{B}}}$
Hence, the work function of $B$ is higher than $A$.

For metal $A,$

The slope is given by $\frac{\text{h}}{\text{e}}=\frac{2}{(10-5)10^{14}}$
$\Rightarrow\ \text{h}=\frac{2\text{e}}{5\times10^{14}}$
$=\frac{2\times1.6\times10^{-19}}{5\times10^{14}}\ \ \big[\because\ \text{e}=1.6\times10^{-19}\big]$
$=6.4\times10^{-34}\text{Js}$
Similarly, for metal $B,$
The slopw is given by slope $\frac{\text{h}}{\text{e}}=\frac{2.5}{(15-10)10^{14}}$
$\Rightarrow\ \text{h}=\frac{2.5\times\text{e}}{5\times10^{14}}$
$=\frac{2.5\times1.6\times10^{-19}}{5\times10^{14}}\ \ \big[\because\ 1.6\times10^{-19}\big]$
$=8\times10^{-34}\text{js}$
Since, the experimental value of $h$ for metals $A$ and $B$ is different.
Hence, the given experiment is not consistent with theory.

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