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When photons of energy hv fall on an aluminum plate (of work function $E_{\circ}$) photoelectrons of maximum $KE,\: K$ are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

$\begin {array} {1 1} (a)\;k+E_{\circ} & \quad (b)\;2K \\ (c)\;K & \quad (d)\;K+hv \end {array}$


1 Answer

Ans : (d)
$hv = E_{\circ}+K$
$h(2v) = E_{\circ}+K’$
$ \Rightarrow K’ = K+hv$


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

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