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A small particle of mass m move in such a way the potential energy `(U = (1)/(2) m^(2) omega^(2) r^(2))` when a is a constant and r is the distance of the particle from the origin Assuming Bohr's model of quantization of angular momentum and circular orbits , show that radius of the nth allowed orbit is proportional to in

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To solve the problem, we need to show that the radius of the nth allowed orbit is proportional to \( n \). We start with the given potential energy and use the principles of circular motion and Bohr's quantization condition. ### Step-by-Step Solution: 1. **Given Potential Energy**: The potential energy of the particle is given by: \[ U = \frac{1}{2} m \omega^2 r^2 \] 2. **Calculate the Force**: The force acting on the particle can be derived from the potential energy using: \[ F = -\frac{dU}{dr} \] We differentiate \( U \): \[ F = -\frac{d}{dr} \left( \frac{1}{2} m \omega^2 r^2 \right) = -m \omega^2 r \] 3. **Centripetal Force**: For a particle moving in a circular orbit, the centripetal force required to keep the particle in circular motion is given by: \[ F_c = \frac{mv^2}{r} \] Setting the magnitudes equal, we have: \[ m \omega^2 r = \frac{mv^2}{r} \] 4. **Express Velocity**: Rearranging the above equation gives: \[ v^2 = \omega^2 r^2 \] Taking the square root: \[ v = \omega r \] 5. **Bohr's Quantization Condition**: According to Bohr's model, the angular momentum \( L \) is quantized: \[ L = mvr = n \frac{h}{2\pi} \] Substituting \( v \) from the previous step: \[ L = m(\omega r)r = m \omega r^2 \] Setting this equal to the quantization condition: \[ m \omega r^2 = n \frac{h}{2\pi} \] 6. **Solve for Radius \( r \)**: Rearranging gives: \[ r^2 = \frac{n h}{2\pi m \omega} \] Taking the square root: \[ r = \sqrt{\frac{n h}{2\pi m \omega}} \] 7. **Proportionality**: Since \( \frac{h}{2\pi m \omega} \) is a constant, we can express the radius \( r \) in terms of \( n \): \[ r \propto \sqrt{n} \] ### Conclusion: Thus, we have shown that the radius of the nth allowed orbit is proportional to \( \sqrt{n} \): \[ r_n \propto \sqrt{n} \]

To solve the problem, we need to show that the radius of the nth allowed orbit is proportional to \( n \). We start with the given potential energy and use the principles of circular motion and Bohr's quantization condition. ### Step-by-Step Solution: 1. **Given Potential Energy**: The potential energy of the particle is given by: \[ U = \frac{1}{2} m \omega^2 r^2 ...
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