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Maximum velocity of the photoelectron emitted by a metal is 1.8 x 10⁶ ms^-1. Take, the value of specific
charge of the electron is 1. 8 x 10¹¹ C kg^-1, then the stopping potential in volt is [KCET 2013]

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answer is C.

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Detailed Solution

Given, $v = 1.8 \times 10^{6} {ms}^{- 1}$

$\frac{e}{m} = 1.8 \times 10^{11} {Ckg}^{- 1}$

We have, $e V_{0} = \frac{1}{2} m v^{2} \Rightarrow V_{0} \frac{e}{m} = \frac{v^{2}}{2}$

$\Rightarrow V_{0} \times 1.8 \times 10^{11} = \frac{1.8 \times 1.8 \times {(10^{6})}^{2}}{2} \Rightarrow V_{0} = 9 V$

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