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Light of wavelength $488\: nm$ is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the cut-off potential of photoelectrons is $0.38\: V$. The work function of the material from which the emitter is made is

$\begin {array} {1 1} (a)\;2.54\: eV & \quad (b)\;2.16\: eV \\ (c)\;21.6\: eV & \quad (d)\;2.26\: eV \end {array}$

 

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Ans : (b)
$\lambda = 488\: nm =488 \times 10^{-9} m$
$V_{\circ} = 0.38 V = 0.38/ (1.6 \times 10^{-19} ) eV$
So, $eV_{\circ} = hc/ \lambda – \phi_{\circ}$
$ \Rightarrow \phi_{\circ} = hc/ \lambda - eV_{\circ}$
$= (6.6 \times 10^{-34} \times 3 \times 10^8 )/ (1.6 \times 10^{-19} \times 488 \times 10^{-9} ) –
(1.6 \times 10^{-10} \times 0.38)/(1.6 \times 10^{-19} )$
$= 2.54 – 0.38 = 2.16 \: eV$

 

answered Dec 24, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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