In a photoelectric experiment,a monochromatic | vedclass

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(B) Let the work function of the emitter plate be $\phi = h v_0$. The energy of the incident photon is $E = h v$.Case $1$: Switch $S_1$ is closed and

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$4133.3$

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(B) Let the work function of the emitter plate be $\phi = h v_0$. The energy of the incident photon is $E = h v$.Case $1$: Switch $S_1$ is closed and $S_2$ is open. The battery of $5 V$ is connected such that the collector $C$ is at a potential of $+5 V$ relative to $E$. The maximum kinetic energy of photoelectrons at the collector is $K_{max, C} = K_{max, E} + e V_{acc} = (E - \phi) + 5 eV = 1 eV$.Thus,$E - \phi = -4 eV$ (This implies the light frequency is below the threshold,but the problem states photoelectrons strike,so we interpret the potential as retarding). Looking at the circuit,$S_1$ closed means $C$ is at $-5 V$ relative to $E$. So,$K_{max, C} = (E - \phi) - 5 eV = 1 eV \Rightarrow E - \phi = 6 eV$.Case $2$: Switch $S_1$ is open and $S_2$ is closed. The battery of $5 V$ is connected such that $C$ is at $+5 V$ relative to $E$. The frequency is doubled,so the new energy is $2E$. The maximum kinetic energy at the collector is $K_{max, C} = (2E - \phi) + 5 eV = 20 eV \Rightarrow 2E - \phi = 15 eV$.We have the system of equations:$E - \phi = 6 eV$ ...$(i)$$2E - \phi = 15 eV$ ...(ii)Subtracting $(i)$ from (ii): $E = 9 eV$.Substituting $E = 9 eV$ into $(i)$: $9 eV - \phi = 6 eV \Rightarrow \phi = 3 eV$.The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \frac{hc}{\phi} = \frac{12400 eV Å}{3 eV} \approx 4133.3 Å$.

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