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The maximum kinetic energy of photoelectrons ejected from a metal,when it is irradiated with radiation of frequency $2\times 10^{14}s^{-1}$ is $6.63\times 10^{-20}J$.The threshold frequency of metal is

$\begin{array}{1 1}(a)\;2\times 10^{14}s^{-1}&(b)\;3\times 10^{14}s^{-1}\\(c)\;1\times 10^{-14}s^{-1}&(d)\;1\times 10^{14}s^{-1}\end{array}$

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Absorbed energy =Threshold energy + Kinetic energy of photoelectrons
$hv=hv_0+KE$
$hv_0=hv-KE$
$6.626\times 10^{-34}\times v_0=6.626\times 10^{-34}\times 2\times 10^{-14}-6.63\times 10^{-20}$
$V_0=\large\frac{1.3252\times 10^{-19}-6.63\times 10^{-20}}{6.626\times 10^{-34}}$
$V_0=9.99\times 10^{13}$
$\Rightarrow 10^{14}s^{-1}$
Hence (d) is the correct answer.
answered Jan 17, 2014 by sreemathi.v
 

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