A small particle of mass m moves in such a way that its potential energy U=1/2mω^2r^2 where ω is constant and r is the distance of the particle from…

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A small particle of mass m moves in such a way that its potential energy U=1/2mω^2r^2 where ω is constant and r is the distance of the particle from origin. Assuming Bohr’s quantization of momentum and circular orbit, the radius of n^th orbit will be proportional to

MARKS Content Team · Accepted Answer

Correct answer: (A) √n. The data given is U=1/2mω^2r^2 From Bohr's quantization, mvr=nh/(2π)⇒v^2=nh2πrm^2 ...(i) In an orbit the value of the kinetic energy is half of the potential energy, K=(mω^2r^2)/4 Substituting the value of equation (i) in the kinetic energy, K=1/2m(n^2h^2)/(4π^2m^2r^2)=(mω^2r^2)/4⇒(n^2h^2)/(2π^2m^2ω^2)=r^4⇒r∝√n

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

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A small particle of mass m moves in such a way that its potential energy U=12mω2r2 where ω is constant and r is the distance of the particle from origin. Assuming Bohr’s quantization of momentum and circular orbit, the radius of nth orbit will be proportional to

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