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The minimum energy required to overcome the attractive forces between electron and the surface of Ag metal is $7.52\times 10^{-19}J$.What will be the maximum kinetic energy of electron ejected out from Ag.Which is being exposed to uv light of $\lambda=360A^{\large\circ}$?

$\begin{array}{1 1}(a)\;32.63\times 10^{-19}J&(b)\;47.68\times 10^{-19}J\\(c)\;47.68\times 10^{-12}J&(d)\;40.6\times 40.6\times 10^{-19}J\end{array}$

1 Answer

Energy absorbed =$\large\frac{hc}{\lambda}$
$\Rightarrow \large\frac{6.625\times 10^{-27}\times 3.0\times 10^{10}}{360\times 10^{-8}}$
$\Rightarrow 5.52\times 10^{-11}erg$
$\Rightarrow 5.52\times 10^{-18}J$
$\therefore E_{absorbed}$=E used in attractive forces +kinetic energy of electron
$\Rightarrow 5.52\times 10^{-18}-7.52\times 10^{-19}J$
$\Rightarrow 47.68\times 10^{-19}J$
Hence (b) is the correct answer.
answered Jan 17, 2014 by sreemathi.v

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