Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

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


Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App