A particle of mass m moves in circular orbits | vedclass

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(B) The potential energy is $V(r) = Fr$. The magnitude of the centripetal force is $F_c = |-\frac{dV}{dr}| = F$.For circular motion,$F = \frac{mv^2}{R

Vedclass · Accepted Answer

$B, C$

Vedclass · Answer

(B) The potential energy is $V(r) = Fr$. The magnitude of the centripetal force is $F_c = |-\frac{dV}{dr}| = F$.For circular motion,$F = \frac{mv^2}{R} \implies v^2 = \frac{FR}{m}$.Using Bohr's quantization condition,$mvr = \frac{nh}{2\pi} \implies v = \frac{nh}{2\pi mR}$.Substituting $v$ into the force equation: $F = \frac{m}{R} \left( \frac{n^2 h^2}{4 \pi^2 m^2 R^2} \right) = \frac{n^2 h^2}{4 \pi^2 m R^3}$.Thus,$R^3 = \frac{n^2 h^2}{4 \pi^2 mF} \implies R \propto n^{2/3}$.From $v = \frac{nh}{2\pi mR}$,since $R \propto n^{2/3}$,we get $v \propto \frac{n}{n^{2/3}} = n^{1/3}$. So,$(B)$ is correct.The total energy $E = K.E. + P.E. = \frac{1}{2}mv^2 + FR$.Since $mv^2 = FR$,$E = \frac{1}{2}FR + FR = \frac{3}{2}FR$.Substituting $R = \left( \frac{n^2 h^2}{4 \pi^2 mF} \right)^{1/3}$,we get $E = \frac{3}{2} F \left( \frac{n^2 h^2}{4 \pi^2 mF} \right)^{1/3} = \frac{3}{2} \left( \frac{n^2 h^2 F^3}{4 \pi^2 mF} \right)^{1/3} = \frac{3}{2} \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$. So,$(C)$ is correct.

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