Atoms & Nuclei

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Question 76 of 239

A particle of mass mm moves in a circular orbit in a central potential field U(r)=12kr2.U\left(r\right)=\dfrac{1}{2}k{r}^{2}. If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number nn as:

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Answer: (B) rn∝n,En∝n{r}_{n}\propto \sqrt{n}, {E}_{n}\propto n

Fr=−dUdr=−kr{F}_{r}=\dfrac{-dU}{dr}=-kr

For circular motion

∣Fr∣=kr=mv2r⇒kr2=mv2|{F}_{r}|=kr=\dfrac{m{v}^{2}}{r}\Rightarrow k{r}^{2}=m{v}^{2} ........(i)

Bohr's quantization ⇒mvr=nh2π\Rightarrow mvr=\dfrac{nh}{2\pi } ....(ii)

From (i) and (ii)

m2v2m=kr2\dfrac{{m}^{2}{v}^{2}}{m}=k{r}^{2}

⇒1m(nh2πr)2=kr2⇒n2h24π2mk=r4\Rightarrow \dfrac{1}{m}{\left(\dfrac{nh}{2\pi r}\right)}^{2}=k{r}^{2}\Rightarrow \dfrac{{n}^{2}{h}^{2}}{4{\pi }^{2}mk}={r}^{4}

⇒r=(h24π2mk)1/4n1/2\Rightarrow r={\left(\dfrac{{h}^{2}}{4{\pi }^{2}mk}\right)}^{1/4} {n}^{1/2}

r∝nr\propto \sqrt{n}

From equation (i) U∝nU\propto \sqrt{n}

KE=12mv2KE=\dfrac{1}{2}m{v}^{2}PE=12kr2PE=\dfrac{1}{2}k{r}^{2}

E=K+U=12mv2+12kr2=kr2∝nE=K+U=\dfrac{1}{2}m{v}^{2}+\dfrac{1}{2}k{r}^{2}=k{r}^{2}\propto n

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