Orbits of a particle moving in a circle are s | vedclass

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(C) According to the condition given,the circumference of the orbit is an integer multiple of the de-Broglie wavelength:$2 \pi r = n \lambda$Since $\l

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$n^{1/2}$

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(C) According to the condition given,the circumference of the orbit is an integer multiple of the de-Broglie wavelength:$2 \pi r = n \lambda$Since $\lambda = \frac{h}{p} = \frac{h}{mv}$,we have:$2 \pi r = \frac{nh}{mv} \implies mvr = \frac{nh}{2 \pi}$For a charged particle moving in a magnetic field $B$,the magnetic Lorentz force provides the centripetal force:$qvB = \frac{mv^2}{r} \implies mv = qBr$Substituting $mv = qBr$ into the quantization condition:$(qBr)r = \frac{nh}{2 \pi} \implies qBr^2 = \frac{nh}{2 \pi}$Solving for $r^2$:$r^2 = \frac{nh}{2 \pi qB}$Thus,$r = \sqrt{\frac{nh}{2 \pi qB}}$,which implies $r \propto n^{1/2}$.

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field,the radius of the $n^{th}$ orbital will therefore be proportional to

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