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When radiation falls on plate C, 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 :

$\begin {array} {1 1} (a)\;2\: eV & \quad (b)\;4\: eV \\ (c)\;8\: eV& \quad (d)\;10\: eV \end {array}$


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Ans : (b)
P.D. across the two electrodes when jockey is at the mid point = $4V$
kinetic energy of the emitted electron = $e \times 4V = 4eV$


answered Dec 30, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1

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