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Q. A monochromatic ray of photons of energy $5eV$ are incident on cathode. Electrons reaching the anode have kinetic energies varying from $6eV$ to $8eV$ . Choose the current option.
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NTA AbhyasNTA Abhyas 2022

Solution:

$\left(KE\right)_{max}=\left(\right.5-\phi\left.\right)eV$
when these electrons are accelerated through $5 \text{V}$ , they will reach the anode with maximum energy $=\left(\right.5 \, -\phi \, + \, 5\left.\right)eV$
$10 \, - \, \phi \, = \, 8 \, \phi \, = \, 2eV \, $
Current is less than saturation current because if slowest electron also reached the plate it would have $5 \, eV$ energy at the anode, but there it is given that the minimum energy is $6 \, eV$ .