Question: Q. 8. In the study of a photoelectric effect the graph between the stopping potential $V$ and frequency of the incident radiation on two different metals $P$ and $Q$ is shown below.
(i) Which one of the two metals has higher threshold frequency?
(ii) Determine the work function of the metal which has greater value.
(iii) Find the maximum kinetic energy of electron emitted by light of frequency $8 \times 10^{14} \mathrm{~Hz}$ for this metal.
U] [Delhi II 2017]
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Solution:
Ans. (i) $Q$ has higher threshold frequency
(ii) Work function,
$$ \begin{align*} \phi_{0}= & h v_{0} \ h v_{0}= & \left(6.6 \times 10^{-34}\right) \ & \times \frac{6 \times 10^{14}}{1.6 \times 10^{-19}} \mathrm{eV} 1 / 2 \ = & 2.5 \mathrm{eV} \ K_{\max }= & h\left(v-v_{0}\right) \ = & \frac{6.6 \times 10^{-34} \times 2 \times 10^{14}}{1.6 \times 10^{-19}} \mathrm{eV} \ K_{\max }= & 0.82 \mathrm{eV} \end{align*} $$
[CBSE Marking Scheme, 2017]
Detailed Answer :
(ii) Work function of $Q \quad \phi_{0}=h v_{0}$
$$ \begin{align*} & =\frac{6.6 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}} \mathrm{eV} \ & =2.47 \mathrm{eV} \ h v_{0} & =\frac{6.6 \times 10^{-34} \times 8 \times 10^{14}}{1.6 \times 10^{-19}} \mathrm{eV} \tag{iii}\ & =3.3 \mathrm{eV} \ K E_{\max } & =h v-h v_{0} \end{align*} $$
$$ \begin{align*} & =(3.3-2.47) \mathrm{eV} \ & =0.83 \mathrm{eV} \tag{1} \end{align*} $$
