Question

The potential energy function for either one of the two atoms in a diatomic molecule is often approximated by U(x) = −a/x12 − b/x6 where x is the distance between the atoms.

a) At what distance of separation does the potential energy have a local minimum (not at x = [infinity])

Answer 1

Answer: $x = \sqrt[6]{2a/b}$

Explanation:

If we want to find a local minimum, we can serach for the points where

U'(x) = 0

here we have that U(x) = -a/x^12 - b/x^6

then U'(x) = (-12)*-a/x^11 -(-6)*b/x^5

this must be zero.

12a/x^11 + 6b/x^5 = 0

12a/x^11 = 6b/x^5

12a = (6b/x^5)*x^11 = 6b*x^(11-5) = 6bx^6

x^6 = 12a/6b = 2a/b

$x = \sqrt[6]{2a/b}$

This is the distance at wich the potential energy has a local minimum.

Answer 2

Answer:

x = ( 2a / b ) ^( 1 / 6 )

Explanation:

Solution:-

- The The potential energy function for either one of the two atoms in a diatomic molecule is often approximated by:

                         

                            $U ( x ) = - \frac{a}{x^1^2} - \frac{b}{x^6}$

Where, x : The separation between two atoms.

- We are to find the separation where the potential energy between two atoms is minimum. For that we have to resort to the methods of calculus. The given function U ( x ) is a single variable function of separation "x" which differential over all the real numbers interval.

- We will determine the first derivative of the potential energy function U ( x ) and set it to zero to calculate the critical values of separation x.

                       $\frac{d U ( x )}{dx } = 12*\frac{a}{x^1^3} + 6*\frac{b}{x^7} \\\\\frac{d U ( x )}{dx } = 12*\frac{a*x^7}{x^1^3*x^7} + 6*\frac{b*x^1^3}{x^1^3*x^7} \\\\\frac{d U ( x )}{dx } = \frac{12ax^7 + 6bx^1^3}{x^2^0} = 0\\\\\\frac{12ax^7 + 6bx^1^3}{x^2^0} = 0\\\\12ax^7 + 6bx^1^3 = 0\\\\x^7* ( 2a + bx^6 ) = 0\\\\x^7 \neq 0 , 2a + bx^6 = 0 \\\\\\x = ( \frac{2a}{b})^\frac{1}{6}$

- The potential energy function U ( x ) has a local minima at x = ( 2a / b ) ^( 1 / 6 )