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When light of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, light of wavelength 600 nm is sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

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

 

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Ans : (b)
$ \phi_1 = \large\frac{hc}{\lambda h_1} = \large\frac{hc}{300}$
$ \phi_2 = \large\frac{hc}{\lambda h_2} = \large\frac{hc}{600}$
So, $ \large\frac{ \phi_1}{ \phi_2} = \large\frac{600}{300} = \large\frac{2}{1}$

 

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

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