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Calculate the velocity of electron ejected from platinum surface when radiation of 200nm fall on it. Work function of platinum is 5eV. $(1eV=1.6\times 10^{-19}J$)

$\begin{array}{1 1}(a)\;6.54\times 10^5m/s&(b)\;5.5\times 10^5m/s\\(c)\;6.54\times 10^2m/s&(d)\;4.63\times 10^3m/s\end{array}$

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Using Einstein's photoelectric equation, $E_1=KE+W$
$E_1=h$$\large\frac{c}{\lambda}=\frac{(6.63\times 10^{-34})\times(3\times 10^8)}{200\times 10^{-9}}$
Given, $W=5eV=5\times (1.6\times 10^{-19})J = 8.0\times 10^{-19}J$
$KE=E_1-W= 9.945\times 10^{-19}-8.0\times 10^{-19} = 1.945\times 10^{-19}J$
Now, Kinetic energy of the electron, $KE=\large\frac{1}{2}$$mv^2$
$\Rightarrow v=\sqrt{\large\frac{2KE}{m}}$
$\Rightarrow v=\sqrt{\large\frac{2(1.195\times 10^{-19})}{9.1\times 10^{-31}}}$
$\Rightarrow 6.54\times 10^2m/s$
Hence (c) is the correct answer.
answered Jan 9, 2014 by sreemathi.v
edited Mar 19, 2014 by mosymeow_1

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