JEE PYQ: Atomic Physics Question 65
Question 65 - 2019 (12 Jan Shift 2)
A particle of mass m moves in a circular orbit in a central potential field $U(r) = \frac{1}{2}kr^2$. If Bohr’s quantization conditions are applied, radii of possible orbits and energy levels vary with quantum number $n$ as:
(1) $r_n \propto \sqrt{n}$, $E_n \propto n$
(2) $r_n \propto \sqrt{n}$, $E_n \propto \frac{1}{n}$
(3) $r_n \propto n$, $E_n \propto n$
(4) $r_n \propto n^2$, $E_n \propto \frac{1}{n^2}$
Show Answer
Answer: (1)
Solution
Let force of attraction towards the centre be $F$. $F = \frac{dU}{dr} = kr = \frac{mv^2}{r}$ (centripetal force). $mvr = \frac{nh}{2\pi}$ [Bohr’s quantization rule]. So $\frac{m^2v^2}{m} = kr^2 \Rightarrow \left(\frac{nh}{2\pi}\right)^2 \frac{1}{m} = kr^2$. $r^2 \propto n \Rightarrow r \propto \sqrt{n}$. Also $E = \frac{1}{2}kr^2 + \frac{1}{2}mv^2 = \frac{1}{2}kr^2 + \frac{1}{2}(kr^2) = kr^2 \propto n$.
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