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.

\frac{36 U_{0}}{R_{0}^{2}} x^{2} - U_{0}

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

\frac{24 U_{0}}{R_{0}} x - U_{0}

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

\frac{96 U_{0}}{R_{0}^{2}} - U_{0}

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

none of these

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

Open in App

Solution

The correct option is A $\frac{36 U_{0}}{R_{0}^{2}} x^{2} - U_{0}$

$U = U_{0} [{(\frac{R_{0}}{R_{0} + x})}^{12} - 2 {(\frac{R_{0}}{R_{0} + x})}^{6}] = U_{0} ⎡ ⎢ ⎢ ⎢ ⎣ \frac{R_{0}^{12}}{R_{0}^{12}} {⎡ ⎢ ⎢ ⎢ ⎣ \frac{1}{1 + \frac{x}{R_{0}}} ⎤ ⎥ ⎥ ⎥ ⎦}^{12} ⎤ ⎥ ⎥ ⎥ ⎦ - 2 ⎡ ⎢ ⎢ ⎢ ⎣ \frac{R_{0}}{R_{0}} {⎡ ⎢ ⎢ ⎢ ⎣ \frac{1}{1 + \frac{x}{R_{0}}} ⎤ ⎥ ⎥ ⎥ ⎦}^{6} ⎤ ⎥ ⎥ ⎥ ⎦ = U_{0} ⎡ ⎢ ⎢ ⎢ ⎣ {⎡ ⎢ ⎢ ⎢ ⎣ \frac{1}{1 + \frac{x}{R_{0}}} ⎤ ⎥ ⎥ ⎥ ⎦}^{12} - 2 {⎡ ⎢ ⎢ ⎢ ⎣ \frac{1}{1 + \frac{x}{R_{0}}} ⎤ ⎥ ⎥ ⎥ ⎦}^{6} ⎤ ⎥ ⎥ ⎥ ⎦ = U_{0} [{(1 + \frac{x}{R_{0}})}^{- 12} - 2 {(1 + \frac{x}{R_{0}})}^{- 6}] = U_{0} [[1 - \frac{12 x}{R_{0}} + \frac{66 x^{2}}{R_{0}^{2}}] - 2 (1 - \frac{6 x}{R_{0}} + \frac{15 x^{2}}{R_{0}^{2}})] = \frac{36 U_{0}}{R_{0}^{2}} x^{2} - U_{0}$

Suggest Corrections

Similar questions

U_{0}

U (x) = U_{0} [{(\frac{a}{x})}^{12} - 2 {(\frac{a}{x})}^{6}], U_{0}, a > 0

m

U (x) = U_{0} (1 - sin b x)

U_{0}

b

\frac{1}{u_{0}} - \frac{1}{u_{1}} + \frac{1}{u_{2}} - \frac{1}{u_{3}} + \dots + {(- 1)}^{n} \frac{1}{u_{n}} = \frac{1}{u_{0} +} \frac{u_{0}^{2}}{u_{1} - u_{0} +} \frac{u_{1}^{2}}{u_{2} - u_{1} +} \dots \frac{u_{n - 1}^{2}}{u_{n} - u_{n - 1}}

U (x) = U_{0} [1 - cos a x]

U_{0}

Join BYJU'S Learning Program