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Monochromatic radiation of wavelength $640.2\: nm$ from a neon lamp irradiates photosensitive material made of cesium on tungsten. The stopping voltage is measured to be $0.54\: V$. The source is replaced by an iron source and its $427.2\: nm$ light irradiates then same photocell. Predict the new stopping voltage.

$\begin {array} {1 1} (a)\;2.29\: eV & \quad (b)\;1.39\: eV \\ (c)\;1.5\: eV & \quad (d)\;2.4\: eV \end {array}$

 

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Ans : (c)
$ \lambda = 640.2 \: nm \: \: \: \: \: \: \: V_{\circ} = 0.54\: V$
Let $ \phi_{\circ}$ be work function and $v$ be frequency of emitted light.
So, $eV_{\circ} = hv - \phi_{\circ}$
$ \Rightarrow \phi_{\circ} = \large\frac{hc}{\lambda} - eV_{\circ}$
$=\large\frac{ 6.6 \times 10^{-34} \times 3 \times 10^8}{(640.2 \times 10^{-9} ) – 1.6 \times 10^{-19} \times 0.54}$
$= 3.093 \times 10^{-19} – 0.864 \times10^{-19}$
$= \large\frac{2.229 \times10^{-19} J}{ 1.6 \times 10^{-19}}$
$= 1.39 \: eV$
$ \lambda’ = 427.2\: nm$
Let $V_{\circ}’$ be new stopping potential.
$eVo’ =\large\frac{ hc}{\lambda’} - \phi_{\circ}$
$= \large\frac{6.6 \times 10^{-34} \times 3 \times 10^8 }{ (427.2 \times10^{-9} ) – 2.229 \times10^{-19}}$
$= 4.63 \times 10^{-19} – 2.229 \times 10^{-19}$
$=\large\frac{ 2.401 \times 10^{-19} J}{ 1.6 \times 10^{-19}}$
$= 1.5\: eV$

 

answered Dec 25, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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