Question 5

For a circular orbit, the centripetal force is provided by the central force: $\frac{mv^2}{r} = kr \Rightarrow mv^2 = kr^2 \dots (1)$ According to Bohr's quantization rule: $L = mvr = n\hbar \Rightarrow v = \frac{n\hbar}{mr} \dots (2)$ Substituting (2) into (1): $m \left( \frac{n\hbar}{mr} \right)^2 = kr^2 \Rightarrow \frac{n^2\hbar^2}{mr^2} = kr^2 \Rightarrow r^4 = \frac{n^2\hbar^2}{mk} \Rightarrow r^2 = n\hbar \sqrt{\frac{1}{mk}}$ Thus, option (A) is correct. Now, substitute $r^2$ into equation (1): $v^2 = \frac{k}{m} r^2 = \frac{k}{m} \left( n\hbar \sqrt{\frac{1}{mk}} \right) = n\hbar \sqrt{\frac{k^2}{m^2 mk}} = n\hbar \sqrt{\frac{k}{m^3}}$ Thus, option (B) is correct. From equation (1), we have: $\frac{v^2}{r^2} = \frac{k}{m} \Rightarrow \frac{v}{r} = \sqrt{\frac{k}{m}}$ Also, $\frac{L}{mr^2} = \frac{mvr}{mr^2} = \frac{v}{r} = \sqrt{\frac{k}{m}}$. Thus, option (C) is correct. Total energy $E$ is the sum of kinetic and potential energy: $E = K + V = \frac{1}{2}mv^2 + \frac{1}{2}kr^2$ Since $mv^2 = kr^2$, we have: $E = \frac{1}{2}kr^2 + \frac{1}{2}kr^2 = kr^2 = k \left( n\hbar \sqrt{\frac{1}{mk}} \right) = n\hbar \sqrt{\frac{k^2}{mk}} = n\hbar \sqrt{\frac{k}{m}}$ Thus, option (D) is incorrect as it contains a factor of $1/2$.