Work Energy Power

JEE Main level, single correct, medium

73 of 183

Question 73 of 183

A small particle of mass mm moves in such a way that its potential energy U=12mω2r2U=\dfrac{1}{2}m{\omega }^{2}{r}^{2} where ω\omega is constant and rr is the distance of the particle from origin. Assuming Bohr’s quantization of momentum and circular orbit, the radius of nth{n}^{th} orbit will be proportional to

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Answer: (A) n\sqrt{n}

The data given is
U=12mω2r2U=\dfrac{1}{2}m{\omega }^{2}{r}^{2}
From Bohr's quantization,
mvr=nh2π⇒v2=(nh2πrm)2...(i)mvr=\dfrac{nh}{2\pi } \Rightarrow {v}^{2}={\left(\dfrac{nh}{2\pi rm}\right)}^{2}...(i)
In an orbit the value of the kinetic energy is half of the potential energy,
K=mω2r24K=\dfrac{m{\omega }^{2}{r}^{2}}{4}
Substituting the value of equation (i) in the kinetic energy,
K=12m(n2h24π2m2r2)=mω2r24⇒n2h22π2m2ω2=r4⇒r∝nK=\dfrac{1}{2}m\left(\dfrac{{n}^{2}{h}^{2}}{4{\pi }^{2}{m}^{2}{r}^{2}}\right)=\dfrac{m{\omega }^{2}{r}^{2}}{4} \Rightarrow \dfrac{{n}^{2}{h}^{2}}{2{\pi }^{2}{m}^{2}{\omega }^{2}}={r}^{4} \Rightarrow r\propto \sqrt{n}

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