Bohr model is applied to a particle of mass m | vedclass

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(A) According to the Bohr quantization condition,the angular momentum is given by $mvr = \frac{nh}{2 \pi}$.Therefore,$vr = \frac{nh}{2 \pi m} \dots (i

Vedclass · Accepted Answer

$\frac{nhqB}{4 \pi m}$

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(A) According to the Bohr quantization condition,the angular momentum is given by $mvr = \frac{nh}{2 \pi}$.Therefore,$vr = \frac{nh}{2 \pi m} \dots (i)$.For a particle moving in a circular path under a magnetic field,the magnetic force provides the centripetal force: $qvB = \frac{mv^2}{r}$.This simplifies to $mv = qBr$,or $v = \frac{qBr}{m} \dots (ii)$.Substituting $v$ from $(ii)$ into $(i)$,we get $(\frac{qBr}{m})r = \frac{nh}{2 \pi m}$,which implies $r^2 = \frac{nh}{2 \pi qB}$.The kinetic energy $E$ is given by $E = \frac{1}{2}mv^2$. Since $mv = qBr$,we have $E = \frac{1}{2}m(\frac{qBr}{m})^2 = \frac{q^2 B^2 r^2}{2m}$.Substituting $r^2$ into the energy expression: $E = \frac{q^2 B^2}{2m} 	imes \frac{nh}{2 \pi qB} = \frac{nhqB}{4 \pi m}$.

Bohr model is applied to a particle of mass $m$ and charge $q$ moving in a plane under the influence of a transverse magnetic field $B$. The energy of the charged particle in the $n^{\text{th}}$ level will be $[h = \text{Planck's constant}]$

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