Photons of energy 5 eV are incident on cathode. Electrons reaching the anode have kinetic energies varying from 6 eV to 8 eV. Then which of the following is true?
Answer & explanation
Correct answer: option 1
Maximum kinetic enegy, $K_{max} = (5 − \phi)eV$
When these electrons are accelerated through 5 V, they will reach the anode with maximum energy =
$(5 − \phi + 5)eV$
$∴10 − \phi = 8$ or $\phi = 2eV$
Current is less than saturation current because even if slowest electron reaches the plate it will have 5 eV energy at the anode, but there it is given that the minimum energy is 6 eV.