Modern Physics Ques 17
- A particle of mass $m$ moves in a circular orbit in a central potential field $U(r)=\frac{1}{2} $ $k r^{2}$. If Bohr’s quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number $n$ as
(Main 2019, 12 Jan I)
(a) $r _n \propto n, E _n \propto n$
(b) $r _n \propto n^{2}, E _n \propto \frac{1}{n^{2}}$
(c) $r _n \propto \sqrt{n}, E _n \propto n$
(d) $r _n \propto \sqrt{n}, E _n \propto \frac{1}{n}$
Show Answer
Answer:
Correct Answer: 17.(c)
Solution:
Formula:
- As, for conservative fields $\mathbf{F}=-(\frac{d \mathbf{U}}{d \mathbf{r}})$
$\therefore$ Magnitude of force on particle is
$ \begin{array}{ll} \Rightarrow & F=\frac{d U}{d r}=\frac{d}{d r} (\frac{1}{2} k r^{2}) \\ \Rightarrow & F=k r \end{array} $
This force is acting like centripetal force.
$\therefore \quad \frac{m v^{2}}{r}=k r$ $\quad$ …….(i)
So, for $n^{\text {th }}$ orbit,
$ \begin{aligned} & \Rightarrow \quad m^{2} v _n^{2}=m k r _n^{2} \\ & \Rightarrow \quad \frac{n^{2} h^{2}}{4 \pi^{2} r^{2}}=m k r _n^{2} \quad [\because v _n=\frac{n h}{2 \pi m r}] \\ & \text { Therefore, } \quad r _n^{4} \propto n^{2} \\ & \Rightarrow \quad r _n^{2} \propto n \\ & \text { So, } \quad r _n \propto \sqrt{n} \end{aligned} $
Energy of particle is
$ \begin{aligned} E _n=PE+KE & =\frac{1}{2} k r _n^{2}+\frac{1}{2} m v _n^{2} \\ & =\frac{1}{2} k r _n^{2}+\frac{1}{2} k r _n^{2} \quad \text { [using Eq. (i)] } \\ & =k r _n^{2} \end{aligned} $
So, energy, $E _n \propto r _n^{2}$
$ \Rightarrow \quad E _n \propto n $
BETA
