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Light of wavelength $488 \mathrm{~nm}$ is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
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Given: $\lambda=488 \mathrm{~nm}=488 \times 10^{-9} \mathrm{~m}, \mathrm{~V}_0=0.38 \mathrm{~V}$
By Einsteins equation $\mathrm{hv}=\phi_0+\frac{1}{2} \mathrm{mv}_{\max }^2$
$$
\begin{aligned}
\Rightarrow \frac{\mathrm{hc}}{\lambda}=\phi_0+\mathrm{eV}_0 \quad\left(\because \mathrm{k}_{\max }=\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}_0\right) \\
\Rightarrow \phi_0=\frac{\mathrm{hc}}{\lambda}-\mathrm{eV}_0 &=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}} \\
&-1.6 \times 10^{-19} \times 0.38
\end{aligned}
$$
$=4.07 \times 10^{-19}-0.61 \times 10^{-19}=3.46 \times 10^{-19}$
$=\frac{3.46 \times 10^{-19}}{1.6 \times 10^{-19}}=2.16 \mathrm{eV}$
By Einsteins equation $\mathrm{hv}=\phi_0+\frac{1}{2} \mathrm{mv}_{\max }^2$
$$
\begin{aligned}
\Rightarrow \frac{\mathrm{hc}}{\lambda}=\phi_0+\mathrm{eV}_0 \quad\left(\because \mathrm{k}_{\max }=\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}_0\right) \\
\Rightarrow \phi_0=\frac{\mathrm{hc}}{\lambda}-\mathrm{eV}_0 &=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}} \\
&-1.6 \times 10^{-19} \times 0.38
\end{aligned}
$$
$=4.07 \times 10^{-19}-0.61 \times 10^{-19}=3.46 \times 10^{-19}$
$=\frac{3.46 \times 10^{-19}}{1.6 \times 10^{-19}}=2.16 \mathrm{eV}$
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