The work function of caesium (Cs) metal is 2 eV. When monochromatic...

Given data: - Work function of Caesium ((\phi)=2 \mathrm{eV}=) (2 \times 1.6 \times 10^{-19} \mathrm{ J}=3.2 \times 10^{-19} \mathrm{ J}) - Frequency of radiation ((f)=6 \times 10^{14}) Hz - Area of the plate = (6.6, \mathrm{ cm}^2=6.6 \times 10^{-4}, \mathrm{ m}^2) - Intensity of radiation = (60, \mathrm{ kW} / \mathrm{m}^2=60 \times 10^3, \mathrm{ W} / \mathrm{m}^2) Step 1: Energy of incident photons The energy of each photon is given by: $E=h f$ (E=6.626 \times 10^{-34} \times 6 \times 10^{14}=3.97 €) Step 2: Maximum kinetic energy of ejected electrons The maximum kinetic energy of the ejected electrons is given by the photoelectric equation: (K_{\max }=E_{\text{photon }}-\phi) (K_\max = 3.9756 \times 10^{-19}) - 3.2 (\times 10^{-19}=7.756 \times 10^{-20}, \text{J}) This value is very close to (8 \times 10^{-20}, \mathrm{ J}), so Statement (a) is approximately correct. Step 3: Stopping potential The stopping potential $V_0$ is given by: (e V_0=K_{\max }) The stopping potential $V_0$ is given by: ($$\begin{aligned} & e V_0=K_{\max } \\ & V_{-0}= \frac{K_{-}\max}{\mathrm{e}}= \frac{7.756 \times 10^{-20}}{1.6 \times 10^{-19}} \text{l } \approx 0.48475, \text{V} \end{aligned}$$) This is very close to 0.48 V, so Statement (b) is correct. Step 4: Saturation current To find the saturation current, we first calculate the total power incident on the plate: ($$\begin{aligned} & P=\text{ Intensity } \times \text{ Area }=60 \times 10^3 \times 6 \\ & n=\frac{P}{E_{\text{photon }}}=\frac{39.6}{3.9756 \times 10^{-19}} \approx 9.96 \times 10^{19} \\ & I=n \times e=9.96 \times 10^{19} \times 1.6 \times 10^{-1 \xi} \end{aligned}$$) Conclusion: The correct choices are (4) (\mathrm{a}, \mathrm{b}, \& \mathrm{c}).