A particle of mass m moves in circular orbits with potential energy V(r)=𝐹𝑟, where F is a positive constant and r is its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the particle’s orbit is denoted by R and its speed and energy are denoted by v and E, respectively, then for the n^th orbit (here h is the Planck’s constant)

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$E = \frac{3}{2} {(\frac{n^{2} h^{2} F^{2}}{4 π^{2} m})}^{1 / 3}$

$E = 2 {(\frac{n^{2} h^{2} F^{2}}{4 π^{2} m})}^{1 / 3}$

$R \propto n^{2 / 3} and v \propto n^{1 / 3}$

$R \propto n^{1 / 3} and v \propto n^{2 / 3}$

answer is B, C.

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Detailed Solution

U = Fr
[Using U = Potential energy and v = velocity]

$\begin{array}{l} \Rightarrow Force = \frac{- dU}{dr} = - F \\ \Rightarrow Magnitude of force = Constant = F \\ \Rightarrow F = \frac{{mv}^{2}}{R} . . . . (1) \\ \Rightarrow mvR = \frac{nh}{2 π} . . . (2) \\ \Rightarrow F = \frac{m}{R} \times \frac{n^{2} h^{2}}{4 π^{2}} \times \frac{1}{m^{2} R^{2}} \\ \Rightarrow R = {(\frac{n^{2} h^{2}}{4 π^{2} mF})}^{1 / 3} \dots \dots (3) \end{array}$

$\Rightarrow v = \frac{nh}{2 πmR}$

$\begin{array}{l} \Rightarrow v = \frac{nh}{2 π m} {(\frac{4 π^{2} mF}{n^{2} h^{2}})}^{1 / 3} \\ \Rightarrow v = \frac{n^{1 / 3} h^{1 / 3} F^{1 / 3}}{2^{1 / 3} π^{1 / 3} m^{2 / 3}} \end{array}$

$\begin{array}{l} (B) is correct \\ \begin{aligned} \Rightarrow E & = \frac{1}{2} {mv}^{2} + U \\ = \frac{1}{2} {mv}^{2} + FR \\ \Rightarrow E & = \frac{1}{2} m (\frac{n^{2 / 3} h^{2 / 3} F^{2 / 3}}{2^{2 / 3} π^{2 / 3} m^{4 / 3}}) + F \times {(\frac{n^{2} h^{2}}{4 π^{2} mF})}^{1 / 3} \\ \Rightarrow E & = {(\frac{n^{2} h^{2} F^{2}}{4 π^{2} m})}^{1 / 3} [\frac{1}{2} + 1] \\ = \frac{3}{2} {(\frac{n^{2} h^{2} F^{2}}{4 π^{2} m})}^{1 / 3} \end{aligned} \end{array}$