TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Given that, work-function, $\phi_{0}=1.5\ \mathrm{eV}$ Let $\lambda_{0}=$ threshold wavelength. Then, $\;\;\phi_{0}=\;\frac{h c}{\lambda_{0}}$ $\Rightarrow 1.5\ \mathrm{eV} = \;\frac{1242}{\lambda_{0}}(\mathrm{eV} - \mathrm{nm})$ $\lambda_{0} = 828\ \mathrm{nm}$ Now, photocurrent is directly proportional to intensity of incident light.$V_{1}, V_{2}, V_{3}$ and $V_{4}$ are cut off voltage, which is required to stop the electron having maximum kinetic energy. Given that, $\lambda_{A}=200\ \mathrm{nm}$, its corresponding cut off voltage is $V_{2}$. $\lambda_{B}=400\ \mathrm{nm}$, its corresponding cut off voltage is $V_{3}$. $\lambda_{C}=600\ \mathrm{nm}$, its corresponding cut off voltage is $V_{4}$. and corresponding photocurrent, $I_{A} = 1.8\ \mathrm{W/m}^{2} \rightarrow \mathrm{III}$ $I_{B} = 1\ \mathrm{W/m}^{2} \rightarrow \mathrm{II}$ $I_{C} = 0.5\ \mathrm{W/m}^{2} \rightarrow \mathrm{IV}$ Hence, consequently option (d) is correct.