When other light falls on the anode plate, the ammeter reading remains zero till jockey is moved from the end P to the middle point of the wire PQ. Therefore, the deflection is recorded in the ammeter. The maximum kinetic energy of the emitted electron is

16 e V

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13.3 e V

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4 e V

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10 e V

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Solution

The correct option is B $13.3 e V$
When the jockey is in the middle of the potentiometer wire, the thevenin voltage becomes:
$V = \frac{20 \times 4}{20 + 4} = \frac{80}{6} N o w, \frac{1}{2} m v_{m a x}^{2} = {K E}_{m a x} = q V = e V = 13.3 e V$

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When other light falls on the anode plate, the ammeter reading remains zero till jockey is moved from the end P to the middle point of the wire PQ. Therefore, the deflection is recorded in the ammeter. The maximum kinetic energy of the emitted electron is

The correct option is B 13.3eVWhen the jockey is in the middle of the potentiometer wire, the thevenin voltage becomes:V=20×420+4=806Now,12mv2max=KEmax=qV=eV=13.3eV

The correct option is B $13.3 e V$
When the jockey is in the middle of the potentiometer wire, the thevenin voltage becomes:
$V = \frac{20 \times 4}{20 + 4} = \frac{80}{6} N o w, \frac{1}{2} m v_{m a x}^{2} = {K E}_{m a x} = q V = e V = 13.3 e V$