A particle of mass m is moving in a circular orbit under the influence of the central force $ f(r) = −kr$ , corresponding to the potential energy $v(r) = \frac{kr^2}{2}$ , where $k$ is a positive force constant and $ r$ is the radial distance from the origin. According to the Bohr’s quantization rule, the angular momentum of the particle is given by $L= nℏ$ , where $ ℏ = ℎ/(2pi), ℎ$ is the Planck’s constant, and $n$ a positive integer. If $v$ and $E$ are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?

Question

A particle of mass m is moving in a circular orbit under the influence of the central force $ f(r) = −kr$ , corresponding to the potential energy $v(r) = \frac{kr^2}{2}$ , where  $k$  is a positive force constant and  $ r$  is the radial distance from the origin. According to the Bohr’s quantization rule, the angular momentum of the particle is given by  $L= nℏ$ , where $ ℏ = ℎ/(2pi), ℎ$  is the Planck’s constant, and  $n$  a positive integer. If  $v$  and  $E$  are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?

cdquestions Admin · Accepted Answer

The correct option is (A): $r^2$ = $nh\sqrt{\frac{1}{mk}}$ , (B): $v^2$ = $nh\sqrt{\frac{k}{m^3}}$ and (C): $\frac{L}{mr^2}$ = $\sqrt{\frac{k}{m}}$

Answer

$r^2$=$nh\sqrt{\frac{1}{mk}}$

Answer

$v^2$=$nh\sqrt{\frac{k}{m^3}}$

Answer

$\frac{L}{mr^2}$=$\sqrt{\frac{k}{m}}$

Answer

$E$=$\frac{nh}{2}\sqrt{\frac{k}{m}}$

The Correct Option is A, B, C

Solution and Explanation

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