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A student performs an experiment on photoelectric effect, using two meterials $\mathrm{A}$ and $\mathrm{B}$. A plot of $\mathrm{V}_{\text {stop }}$ versus $\mathrm{v}$ is given in figure.
(i) Which material A or B has a higher work function?
(ii) Given the electric charge of an electron $=1.6 \times 10^{-}$ ${ }^{19} \mathrm{C}$, find the value of $\mathrm{h}$ obtained from the experiment for both $\mathrm{A}$ and $\mathrm{B}$.
Comment on whether it is consistent with Einstein's theory.
(i) Which material A or B has a higher work function?
(ii) Given the electric charge of an electron $=1.6 \times 10^{-}$ ${ }^{19} \mathrm{C}$, find the value of $\mathrm{h}$ obtained from the experiment for both $\mathrm{A}$ and $\mathrm{B}$.
Comment on whether it is consistent with Einstein's theory.
Solution:
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Verified Answer
(i) As given that thresholed frequency of metal $\mathrm{A}$ is given by $v_{\mathrm{OA}}=5 \times 10^{14} \mathrm{~Hz}$ and
For metals $\mathrm{B}, v_{\mathrm{OB}}=10 \times 10^{14} \mathrm{~Hz}$
We know that
Work function, $\phi=\mathrm{hv} \mathrm{v}_{\mathrm{O}}$
$$
\begin{aligned}
\Rightarrow \quad \phi_{\mathrm{O}}=& \propto v_{\mathrm{O}} \\
\frac{\phi_{\mathrm{OA}}}{\phi_{\mathrm{OB}}}=\frac{\mathrm{h} v_{\mathrm{OA}}}{\mathrm{hv} v_{\mathrm{OB}}}
\end{aligned}
$$
So, $\quad \frac{\phi_{\mathrm{OA}}}{\phi_{\mathrm{OB}}}=\frac{5 \times 10^{14}}{10 \times 10^{14}}=\frac{1}{2}$
$$
\phi_{\mathrm{OB}}=2 \phi_{\mathrm{OA}}
$$
So, $\phi_{\mathrm{OA}} < \phi_{\mathrm{OB}}$
Hence, work function of B is higher than A.
(ii) For metal A, h= $\frac{2 \times \mathrm{e}}{(10-5) 10^{14}}$
or $\quad \mathrm{h}=\frac{2 \mathrm{e}}{5 \times 10^{14}}=\frac{2 \times 1.6 \times 10^{-19}}{5 \times 10^{14}}$ $\mathrm{h}=6.4 \times 10^{-34} \mathrm{JS}$
For metal B, h $=\frac{(2.5-0) \mathrm{e}}{(15-10) 10^{14}}$
or $\mathrm{h}=\frac{2.5 \times \mathrm{e}}{5 \times 10^{14}}=\frac{2.5 \times 1.6 \times 10^{-19}}{5 \times 10^{14}}=8 \times 10^{-34} \mathrm{JS}$
So, the value of $\mathrm{h}$ from experiment for metals A and B is different. Hence, experiment is not consistent with theory.
For metals $\mathrm{B}, v_{\mathrm{OB}}=10 \times 10^{14} \mathrm{~Hz}$
We know that
Work function, $\phi=\mathrm{hv} \mathrm{v}_{\mathrm{O}}$
$$
\begin{aligned}
\Rightarrow \quad \phi_{\mathrm{O}}=& \propto v_{\mathrm{O}} \\
\frac{\phi_{\mathrm{OA}}}{\phi_{\mathrm{OB}}}=\frac{\mathrm{h} v_{\mathrm{OA}}}{\mathrm{hv} v_{\mathrm{OB}}}
\end{aligned}
$$
So, $\quad \frac{\phi_{\mathrm{OA}}}{\phi_{\mathrm{OB}}}=\frac{5 \times 10^{14}}{10 \times 10^{14}}=\frac{1}{2}$
$$
\phi_{\mathrm{OB}}=2 \phi_{\mathrm{OA}}
$$
So, $\phi_{\mathrm{OA}} < \phi_{\mathrm{OB}}$
Hence, work function of B is higher than A.
(ii) For metal A, h= $\frac{2 \times \mathrm{e}}{(10-5) 10^{14}}$
or $\quad \mathrm{h}=\frac{2 \mathrm{e}}{5 \times 10^{14}}=\frac{2 \times 1.6 \times 10^{-19}}{5 \times 10^{14}}$ $\mathrm{h}=6.4 \times 10^{-34} \mathrm{JS}$
For metal B, h $=\frac{(2.5-0) \mathrm{e}}{(15-10) 10^{14}}$
or $\mathrm{h}=\frac{2.5 \times \mathrm{e}}{5 \times 10^{14}}=\frac{2.5 \times 1.6 \times 10^{-19}}{5 \times 10^{14}}=8 \times 10^{-34} \mathrm{JS}$
So, the value of $\mathrm{h}$ from experiment for metals A and B is different. Hence, experiment is not consistent with theory.
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