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The vapours of Hg absorb some electrons accelerate by a potential difference of 4.5volts as a result of which light is emitted .If the full energy of single incident electron is supposed to be converted into light emitted by single Hg atom,find the wave number $\big(\large\frac{1}{\lambda}\big)$ of the light.

$\begin{array}{1 1}(a)\;2.63\times 10^6m^{-1}&(b)\;4.62\times 10^6m^{-1}\\(c)\;3.63\times 10^6m^{-1}&(d)\;3.63\times 10^2m^{-1}\end{array}$

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Energy of an accelerated electron =$Q.V=1.602\times 10^{-19}\times 4.5$
$\Rightarrow 7.209\times 10^{-19}J$
This energy is completely converted into light.
$\large\frac{hc}{\lambda}$$=7.209\times 10^{-19}$
$\large\frac{6.625\times 10^{-34}\times 3.0\times 10^8}{\lambda}$$=7.209\times 10^{-19}$
$\large\frac{1}{\lambda}$=wave no =$3.63\times 10^6m^{-1}$
Hence (c) is the correct answer.
answered Jan 16, 2014 by sreemathi.v

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