A particle of mass m moves around the origin | vedclass

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(A) The potential energy of the particle is given as $U = \frac{1}{2} m \omega^{2} r^{2}$.For a circular orbit,the centripetal force is provided by th

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$a \sqrt{n}$

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(A) The potential energy of the particle is given as $U = \frac{1}{2} m \omega^{2} r^{2}$.For a circular orbit,the centripetal force is provided by the gradient of the potential: $F = -\frac{dU}{dr} = -m \omega^{2} r$.Equating the magnitude of the force to the centripetal force: $m \omega^{2} r = \frac{m v^{2}}{r}$,which implies $v^{2} = \omega^{2} r^{2}$,or $v = r \omega$.Now,the angular momentum $L$ of the particle is $L = mvr = m(r \omega)r = m r^{2} \omega$.According to Bohr's quantization condition,$L = \frac{n h}{2 \pi}$.Equating the two expressions for $L$: $m r^{2} \omega = \frac{n h}{2 \pi}$.Solving for $r^{2}$: $r^{2} = \frac{n h}{2 \pi m \omega}$.Taking the square root: $r = \sqrt{n} \sqrt{\frac{h}{2 \pi m \omega}}$.Given $a = \sqrt{\frac{h}{2 \pi m \omega}}$,we get $r = a \sqrt{n}$.

$A$ particle of mass $m$ moves around the origin in a potential $\frac{1}{2} m \omega^{2} r^{2}$,where $r$ is the distance from the origin. Applying the Bohr's model in this case,the radius of the particle in its $n$th orbit in terms of $a=\sqrt{\frac{h}{2 \pi m \omega}}$ is

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