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The work function for a metal is 4eV.To emit a photon-electron of zero velocity from the surface of the metal,the wavelength of incident light should be

$(a)\;2700A^{\circ}\qquad(b)\;1700A^{\circ}\qquad(c)\;5900A^{\circ}\qquad(d)\;3100A^{\circ}$

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$hV$=work function +K.E
Given K.E=0
$hV=4eV$
$4=\large\frac{12375}{\lambda}$
(where $\lambda$ is in $A^{\circ}$)
$\therefore \lambda=3100A^{\large\circ}$
Hence (d) is the correct answer.
answered Jan 10, 2014 by sreemathi.v
 

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