Find the value of $r_{0}$ corresponding to the equilibrium position of the particle; examine whether this position is steady.

r_{0} = \frac{a}{b}

, steady

No worries! We‘ve got your back. Try BYJU‘S free classes today!

r_{0} = 0

, steady

No worries! We‘ve got your back. Try BYJU‘S free classes today!

r_{0} = \frac{2 a}{b}

, steady

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

r_{0} = \frac{a}{2 b}

, steady

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $r_{0} = \frac{2 a}{b}$ , steady
We have,
$F = - \frac{d U}{d r}$
$F = 2 a r^{- 3} - b r^{- 2}$
$F = - b r^{- 2} + 2 a r^{- 3}$
$F = - \frac{b}{r^{2}} + \frac{2 a}{r^{3}}$
At equilibrium
F = 0, and r = $r_{0}$ hence
$\frac{b}{(r_{0})^{2}} - \frac{2 a}{(r_{0})^{3}} = 0$
$b r_{0} = 2 a$
$∴ r_{0} = \frac{2 a}{b}$ in steady position.

Suggest Corrections

Similar questions

The potential energy of a particle in a certain field has the form U = $\frac{a}{r^{2}}$ - $\frac{b}{r}$ , where a and b are postive constants , r is the distance from the centre of the field . Find the value of $r_{0}$ for which the paricle is in stable equilibrium

Q. The potential energy of a particle in a certain field has the form

U = \frac{a}{r^{2}} - \frac{b}{r},

where

a

and

b

are postive constants,

r

is the distance from the centre of the field.

Q. In a Vander waals interaction:

U = U_{0} [{(\frac{R_{0}}{r})}^{12} - 2 {(\frac{R_{0}}{r})}^{6}]

A small displacement x is given from equilibrium position

r = R_{0}

. Find the approximate PE function.

The wavefunction of a 2s electron is given by $ψ_{2 s} = \frac{1}{4 \sqrt{2} π} (\frac{1}{a_{0}})^{\frac{3}{2}} (2 - \frac{r}{a_{0}}) e^{\frac{- r}{a_{0}}}$
It has a node at $r = r_{0}$ . The value of $\frac{r_{0}}{a_{0}}$ is___

\frac{d r}{d t} = - r t, r (0) = r_{0}

Join BYJU'S Learning Program

Find the value of r0 corresponding to the equilibrium position of the particle; examine whether this position is steady.

The correct option is B r0=2ab, steadyWe have,F=−dUdrF=2ar−3−br−2F=−br−2+2ar−3F=−br2+2ar3At equilibrium F = 0, and r = r0 henceb(r0)2−2a(r0)3=0br0=2a∴r0=2ab in steady position.

Find the value of $r_{0}$ corresponding to the equilibrium position of the particle; examine whether this position is steady.