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Photoelectric effect experiments are performed using three different metal plates $\mathrm{p}, \mathrm{q}$ and $\mathrm{r}$ having work functions $\phi_p=2.0 \mathrm{eV}, \phi_q=2.5 \mathrm{eV}$ and $\phi_r=3.0 \mathrm{eV}$, respectively. A light beam containing wavelengths of $550 \mathrm{~nm}, 450 \mathrm{~nm}$ and $350 \mathrm{~nm}$ with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is [Take $h c=1240 \mathrm{eV} \mathrm{nm}$ ]
Solution
Given $\phi_p=2.0 \ eV , \phi_{ q }=2.5 \ eV$ and $\phi_{ r }=3.0 \ eV$
From the above we can find the cutoff wavelength for each plate $\lambda_p=621 \ nm$, $\lambda_q=496 \ nm$ and $\lambda_r=414 \ nm$
It's mentioned that each plate is illuminated by a light that consists of $3$ different wavelengths i.e. $550 \ nm , 450 \ nm , 350 \ nm$ of equal intensities.
Now since the cut off wavelength of plate $p$ is higher than all $3$ incident wavelengths so all $3$ types of photons will contribute to photoelectrons. Whereas in the case of plate q incident photons of wavelength $450 \ nm$ won't be able to contribute in photocurrent as it is higher than it's cut off wavelength of $496 \ nm$. So naturally its photo current will be lesser than the plate $p$ since its being excited by only $2$ wavelegths. In case of plate $r, 2$ incident wavelengths i.e. $550 \ nm$ and $450 \ nm$ won't be able to excite the photo electrons. Hnece its photocurrent will be only due to photon of wavelength $350 \ nm$ and hence will be least.
$I_p<I_q<I_r$
