JEE Main 2017 — Physics Modern Physics

2017mcqmedium

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

Official previous-year question

Held on 9 Apr 2017 · Verified 6 Jul 2026.

Options

A
$\frac{{h}^{2}}{4\pi {m}^{2}{r}^{3}}$
B
$\frac{{h}^{2}}{{\pi }^{2}{m}^{2}{r}^{3}}$
C
$\frac{{h}^{2}}{8{\pi }^{2}{m}^{2}{r}^{3}}$
D
$\frac{{h}^{2}}{4{\pi }^{2}{m}^{2}{r}^{3}}$

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Solution

For $H-$ atom $(n=1)$

$\frac{m{v}^{2}}{r}=\frac{1}{4\pi {\epsilon }_{o}}\frac{{e}^{2}}{{r}^{2}}...(1)$

$mvr=\frac{h}{2\pi }...(2)$

On squaring $(2)$

${m}^{2}{v}^{2}{r}^{2}=\frac{{h}^{2}}{4{\pi }^{2}}$

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

$\frac{{m}^{2}{v}^{2}}{r}=\frac{{h}^{2}}{4\pi {r}^{3}} \Rightarrow \frac{{v}^{2}}{r}=\frac{{h}^{2}}{4{\pi }^{2}{r}^{3}{m}^{2}}$

Acceleration $= \frac{{h}^{2}}{4{\pi }^{2}{r}^{3}{m}^{2}}$ .

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Solution

For $H-$ atom $(n=1)$

$\frac{m{v}^{2}}{r}=\frac{1}{4\pi {\epsilon }_{o}}\frac{{e}^{2}}{{r}^{2}}...(1)$

$mvr=\frac{h}{2\pi }...(2)$

On squaring $(2)$

${m}^{2}{v}^{2}{r}^{2}=\frac{{h}^{2}}{4{\pi }^{2}}$

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

$\frac{{m}^{2}{v}^{2}}{r}=\frac{{h}^{2}}{4\pi {r}^{3}} \Rightarrow \frac{{v}^{2}}{r}=\frac{{h}^{2}}{4{\pi }^{2}{r}^{3}{m}^{2}}$

Acceleration $= \frac{{h}^{2}}{4{\pi }^{2}{r}^{3}{m}^{2}}$ .

JEE Main 2017 — Physics Modern Physics

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JEE Main Modern Physics in other years

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JEE Main 2017 — Physics Modern Physics

Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics