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

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

## 1 Answer

Ans : (b)
Work function, $\phi = \large\frac{hc}{\lambda}$
$\Rightarrow \phi \sim \large\frac{1}{\lambda}$
Since, $\large\frac{\phi_1}{\phi_2} = \large\frac{\lambda_1}{\lambda_2 }= \large\frac{600}{300}=\large\frac{2}{1}$

answered Dec 27, 2013
edited Mar 13, 2014

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