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JEE Main 20179 AprilOnline

Exam: JEE Main
Year: 2017
Date: 9 April
Mode: Online
Subject: Physics
Chapter: Atomic Physics

JEE Main 2017 · 9 April · Online · Physics · Atomic Physics

The acceleration of an electron in the first orbit of the hydrogen...

The acceleration of an electron in the first orbit of the hydrogen atom (

n = 1

) is :

Options

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

Correct answer

D. $\dfrac{{h}^{2}}{4{\pi}^{2}{m}^{2}{r}^{3}}$

Step-by-step solution

For

H -

atom \left(n = 1\right)

<br /><br />

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

mvr =

\dfrac{h}{2\pi}

. . . \left(2\right)

On squaring

( 2 )

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

On dividing both sides by

{r}^{3},

\dfrac{{m}^{2}{v}^{2}}{r} = \dfrac{{h}^{2}}{4\pi{r}^{3}} ⇒ \dfrac{{v}^{2}}{r} = \dfrac{{h}^{2}}{4{\pi}^{2}{r}^{3}{m}^{2}}

Acceleration

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

.

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