If one were to apply the Bohr model to a part | vedclass

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(C) The magnetic force provides the necessary centripetal force for circular motion:$q v B = \frac{m v^2}{r} \implies q B = \frac{m v}{r} \implies r =

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$n\left( \frac{hqB}{4\pi m} \right)$

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(C) The magnetic force provides the necessary centripetal force for circular motion:$q v B = \frac{m v^2}{r} \implies q B = \frac{m v}{r} \implies r = \frac{m v}{q B}$ ..... $(i)$According to Bohr's quantization condition for angular momentum:$m v r = \frac{n h}{2 \pi}$ ..... $(ii)$Substitute $r$ from equation $(i)$ into equation $(ii)$:$m v \left( \frac{m v}{q B} \right) = \frac{n h}{2 \pi}$$m^2 v^2 = \frac{n h q B}{2 \pi}$The kinetic energy $E$ is given by $E = \frac{1}{2} m v^2 = \frac{m^2 v^2}{2 m}$.Substituting the value of $m^2 v^2$:$E = \frac{1}{2 m} \left( \frac{n h q B}{2 \pi} \right) = n \left( \frac{h q B}{4 \pi m} \right)$.

If one were to apply the Bohr model to a particle of mass $m$ and charge $q$ moving in a plane under the influence of a magnetic field $B$,the energy of the charged particle in the $n^{th}$ level will be

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