The work function of caesium (Cs) metal is 2 eV. When monochromatic radiation beam of intensity (60 kWm ⁻² ) and frequency (6 10¹⁴ ~Hz ), is falling normally on Cs plate of (6.6 ~cm ^2 ) in a photocell then (a) Energy of a photoelectron may be (8 10⁻²⁰ ~J ) (b) Stopping potential is 0.48 V (c) Saturation current is 16 A Correct choices are

Step-by-step solution

Given data: - Work function of Caesium (( )=2 eV = ) (2 1.6 10⁻¹⁹ ~J =3.2 10⁻¹⁹ ~J ) - Frequency of radiation ((f)=6 10¹⁴ ) Hz - Area of the plate = (6.6, ~cm ^2=6.6 10⁻⁴, ~m ^2 ) - Intensity of radiation = (60, ~kW / m ^2=60 10^3, ~W / m ^2 ) Energy of incident photons The energy of each photon is given by: (E=h f ) (E=6.626 10⁻³⁴ 6 10¹⁴=3.97 € ) Maximum kinetic energy of ejected electrons The maximum kinetic energy of the ejected electrons is given by the photoelectric equation: (K_ =E_ photon - ) (K_ = 3.9756 10