Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 12 - Exercises and Problems - Page 222: 26

Answer

The stability location of the electron is at x = 0 & this corresponds to an unstable equilibrium.

Work Step by Step

$U(x)=ax^{2}-bx^{4}-(1)$ Equilibria occur where the potential energy has a maximum or minimum - that is, where its derivative is zero. Taking the derivative of U(x) & setting it to zero gives, $0=\frac{dU}{dx}= 2ax-4bx^{3}=2x(a-bx^{2})$ $x=0$ $\space\space\space or$ $a-2bx^{2}=0$ $x=\pm \sqrt {\frac{a}{2b}}$ ; Let's plug known values into this equation. $x=\pm\sqrt {\frac{-8\space aj/nm^{2}}{1\space aj/nm^{4}}}= \pm \sqrt {-8}\space nm$ This is not a real number. So this solution should neglect. Therefore the location of the stability is x = 0 Let's find the value of $\frac{d^{2}U}{dx^{2}},$ to find the state of equlibrium. $\frac{d^{2}U}{dx^{2}}=2a-12bx^{2}=-16\space aj/nm^{2}-12x^{2}\space aj/nm^{4}$ This value is negative for all x values ; $\frac{d^{2}U}{dx^{2}}\lt0$ So this corresponds to an unstable equilibrium
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