Atoms & Nuclei

JEE Main level, single correct, medium

81 of 239

Question 81 of 239

The acceleration of an electron in the first orbit of the hydrogen atom (n=1n=1) is :

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Answer: (D) h24π2m2r3\dfrac{{h}^{2}}{4{\pi }^{2}{m}^{2}{r}^{3}}

For H−H-atom (n=1)\left(n=1\right)

mv2r=14πϵoe2r2...(1)\dfrac{m{v}^{2}}{r}=\dfrac{1}{4\pi {ϵ}_{o}}\dfrac{{e}^{2}}{{r}^{2}}...\left(1\right)

mvr=h2π...(2)mvr=\dfrac{h}{2\pi }...\left(2\right)

On squaring (2)(2)

m2v2r2=h24π2{m}^{2}{v}^{2}{r}^{2}=\dfrac{{h}^{2}}{4{\pi }^{2}}

On dividing both sides by r3,{r}^{3},

m2v2r=h24πr3⇒v2r=h24π2r3m2\dfrac{{m}^{2}{v}^{2}}{r}=\dfrac{{h}^{2}}{4\pi {r}^{3}}\Rightarrow \dfrac{{v}^{2}}{r}=\dfrac{{h}^{2}}{4{\pi }^{2}{r}^{3}{m}^{2}}

Acceleration=h24π2r3m2=\dfrac{{h}^{2}}{4{\pi }^{2}{r}^{3}{m}^{2}}.

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