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Relation of the stopping potential $V_{\circ}$ of a metal and the maximum velocity $v$ of the photoelectrons

$\begin {array} {1 1} (a)\;V_{\circ} \sim \large\frac{1}{v^2} & \quad (b)\;V_{\circ} \sim  v^2 \\ (c)\;V_{\circ}  \sim  v & \quad (d)\;V_{\circ} \sim  \large\frac{1}{v} \end {array}$

 

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Ans : (b)
$ \lambda= \large\frac{h}{mv}\: \: \: and\: \: \: \lambda=\large\frac{h}{ \sqrt{2meV_{\circ}}}$
where, $ \lambda $ = incident radiation wavelength
So, $mv_{max} = \sqrt{meV_{\circ}}$
$ \Rightarrow v^2 \sim V_{\circ}$

 

answered Dec 28, 2013 by thanvigandhi_1
edited Oct 30, 2014 by thagee.vedartham
 

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