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Question: In the Vander Waals interaction U = U<sub>0</sub>\(\left\lbrack \left( \frac{R_{0}}{r} \right)^{12}...

In the Vander Waals interaction

U = U0[(R0r)12−2(R0r)6]\left\lbrack \left( \frac{R_{0}}{r} \right)^{12} - 2\left( \frac{R_{0}}{r} \right)^{6} \right\rbrack.

A small displacement x is given from equilibrium position r = R0. Find the approximate PE function.

A

36U0R02x2−U0\frac{36U_{0}}{R_{0}^{2}}x^{2} - U_{0}

B

24U0R0x−U0\frac{24U_{0}}{R_{0}}x - U_{0}

C

96U0R02−U0\frac{96U_{0}}{R_{0}^{2}} - U_{0}

D

None of these

Answer

36U0R02x2−U0\frac{36U_{0}}{R_{0}^{2}}x^{2} - U_{0}

Explanation

Solution

U = U0 [(R0R0+x)12−2[R0R0+x]6]\left\lbrack \left( \frac{R_{0}}{R_{0} + x} \right)^{12} - 2\left\lbrack \frac{R_{0}}{R_{0} + x} \right\rbrack^{6} \right\rbrack

= U0 [R012R012[11+xR0]12]−2[11+xR0]6\left\lbrack \frac{R_{0}^{12}}{R_{0}^{12}}\left\lbrack \frac{1}{1 + \frac{x}{R_{0}}} \right\rbrack^{12} \right\rbrack - 2\left\lbrack \frac{1}{1 + \frac{x}{R_{0}}} \right\rbrack^{6}

= U0[(1+xR0)−12−2[1+xR0]−6]U_{0}\left\lbrack \left( 1 + \frac{x}{R_{0}} \right)^{- 12} - 2\left\lbrack 1 + \frac{x}{R_{0}} \right\rbrack^{- 6} \right\rbrack

= U0[[1−12xR0+66x2R02]−2(1−6xR0+15x2R02)]U_{0}\left\lbrack \left\lbrack 1 - \frac{12x}{R_{0}} + \frac{66x^{2}}{R_{0}^{2}} \right\rbrack - 2\left( 1 - \frac{6x}{R_{0}} + \frac{15x^{2}}{R_{0}^{2}} \right) \right\rbrack

= z36U0R02x2−U0\frac{36U_{0}}{R_{0}^{2}}x^{2} - U_{0}.