In a photoelectric experiment, a monochromatic light is incident on the emitter plate $E$, as shown in the figure. When switch $S_1$ is closed and switch $S_2$ is open, the photoelectrons strike the collector plate $C$ with a maximum kinetic energy of $1 \mathrm{eV}$. If switch $S_1$…

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In a photoelectric experiment, a monochromatic light is incident on the emitter plate $E$, as shown in the figure. When switch $S_1$ is closed and switch $S_2$ is open, the photoelectrons strike the collector plate $C$ with a maximum kinetic energy of $1 \mathrm{eV}$. If switch $S_1$ is open and switch $S_2$ is closed and the frequency of the incident light is doubled the photoelectrons strike the collector plate with a maximum kinetic energy of $20 \mathrm{eV}$. The threshold wavelength of the emitter plate is

Aicharya · Accepted Answer

Let threshold frequency of emitter plate $=v_0$ Energy of photon in first case is $\mathrm{E}$. When switch $S_1$ is closed and switch $S_2$ is open, So, $E=h v_0+(5+1) \mathrm{eV}$ ...(i) For second case, when switch $S_1$ is open and switch $S_2$ is closed and frequency of incident light is doubled. then, $\quad 2 E=h v_0+(20-5) \mathrm{eV}$ ...(ii) From Eqs. (i) and (ii), we get $\begin{aligned} & \Rightarrow \quad 2\left(h v_0+6 \mathrm{eV}\right)=h v_0+15 \mathrm{eV} \ & \Rightarrow \quad 2 h v_0+12 \mathrm{eV}=h v_0+15 \mathrm{eV} \ & \Rightarrow \quad h v_0=3 \mathrm{eV} \ & \Rightarrow \quad v_0=\frac{3 \times 1.6 \times 10^{-19}}{6.62 \times 10^{-34}}=7.25 \times 10^{14} \mathrm{~Hz} \ & \lambda_0=\frac{c}{v_0}=\frac{3 \times 10^8}{7.25 \times 10^{14}} \ & \lambda_0=41333 Å \ \end{aligned}$ Hence, the threshold wavelength of the emitter plate is $4133.3 Å$.

In a photoelectric experiment, a monochromatic light is incident on the emitter plate \(E\), as shown in the…

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