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A particle with kinetic energy $\;10\;MeV\;$ and charge $\;4e\;$ is heading towards a stationary nucleus of atomic number 100 . Calculate the distance of closest approach ?

$(a)\;5.76\times10^{-14}\;m\qquad(b)\;57.6\times10^{-14}\;m\qquad(c)\;0.576\times10^{-14}\;m\qquad(d)\;None$

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Answer : (a) $\;5.76\times10^{-14}\;m$
Explanation :
Since there is no external force
$\bigtriangleup T\;.E=0$
$\bigtriangleup K\;.E + \bigtriangleup P\;.E=0$
$ \bigtriangleup K\;.E=10\;MeV$
$\large\frac{9\times10^{9}\times100\times1.6\times10^{-19}\times4\times1.6\times10^{-19}}{r}=10^{7}\times1.6 \times10^{-19}$
$r=9\times10^2\times100\times1.6\times10^{-19}\times4$
$r=57.6\times10^{-15}\;m\;.$

 

answered Feb 4, 2014 by yamini.v
edited Aug 13, 2014 by thagee.vedartham
 

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