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# A small particle of mass m moves in such a way that the potential energy $\;u=ar^2\;$ where a is a constant and r is the distance of particle from origin. Assuming Bohr's model of quantization of angular momentum and circular orbits. Then the radium of $\;n^{th}\;$ allowed orbit is

$(a)\;r=(\large\frac{n^2 h^2}{82 a m \pi^{2}})^{\large\frac{1}{4}}\qquad(b)\;r=(\large\frac{n^2 h^2}{4 a m \pi^{2}})^{\large\frac{1}{4}}\qquad(c)\;r=(\large\frac{n^2 h^2}{8 a m \pi^{2}})^{\large\frac{1}{4}}\qquad(d)\;r=(\large\frac{n^2 h^2}{16 a m \pi^{2}})^{\large\frac{1}{4}}$

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Answer : (c) $\;r=(\large\frac{n^2 h^2}{8 a m \pi^{2}})^{\large\frac{1}{4}}$
Explanation :
The force at a distance r is ,
$F=-\large\frac{dV}{dr}=-2ar$
Suppose r be the radius of $\;n^{th}\;$ orbit . Then the necessary centripetal force is provided by above force
Thus ,
$\large\frac{mV^2}{r}=2ar-----> (1)$
Further , the quantization of angular momentum gives $\;mvr=\large\frac{nh}{2 \pi}-----(2)$
Solving equations (1) & (2) for r we get
$r=(\large\frac{n^2 h^2}{8 a m \pi^{2}})^{\large\frac{1}{4}}\;.$
answered Feb 22, 2014 by

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