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  • A small particle of mass m mores in such a way that the potential energy <img alt="\mathrm{v=\frac{1}{2} m \omega^2 r^2 }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bv%3D

A small particle of mass m mores in such a way that the potential energy \mathrm{v=\frac{1}{2} m \omega^2 r^2 } where \mathrm{ \omega} is constant and r is the distance of the particle from the origin. Assuming Bohr's model of quantization of angular momentum and circular orbits, that radius of \mathrm{ n^{\text {th }}} allowed orbit is proportional to

Option: 1

n


Option: 2

\sqrt{n}


Option: 3

\sqrt[3]{n}


Option: 4

n^{2 / 3}


Answers (1)

best_answer

The force at a distance r is -
\mathrm{ F=-\frac{d U}{d r}=-m \omega^2 r}
suppose the particle moves along a circle of radius r. The net force on it showed be \mathrm{\frac{m v^2}{r} } along the radius. Comparing with (i).

\mathrm{\begin{aligned} & \frac{m c^2}{r}=m \omega^2 r \\ & k=\omega \gamma \end{aligned} }
The quantization of angular momentum gives-
\mathrm{\begin{aligned} & \text { mvr }=\frac{nh}{2\pi } \\ & k=\frac{n h}{2 \pi m r}-\text { (3) } \end{aligned} }
from (2) \& (3)
\mathrm{\omega r=\frac{n h}{2 \pi m r} }
\mathrm{\begin{aligned} & r^2=\frac{n h}{2 \pi m \omega} \\ & r=\sqrt{\frac{n h}{2 \pi m \omega}} \Rightarrow r \alpha \sqrt{n} \text { Ans. } \end{aligned} }
 

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Nehul

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