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Maximum velocity of the photoelectrons emitted by a metal surface is $1.2 \times 10^6 m/s$. Assuming the specific charge of the electron to be $1.8 \times 10^{11}C/kg$, the value of the stopping potential is

$\begin {array} {1 1} (a)\;2\: V & \quad (b)\;3\: V \\ (c)\;4\: V & \quad (d)\;6\: V \end {array}$


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Ans : (c)
$ \large\frac{1}{2} mv^2_{max} = eV_s$
$ \Rightarrow V_s = \large\frac{mv^2_{max}}{2e} = \large\frac{(1.2 \times 10^6)^2}{[2 \times 1.8 \times 10^{11} ) ] } 4\: V$


answered Dec 28, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1

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