The energy-separation curve for two atoms, a distance, r, apart is: U(r)=−A/r^m+B/r^n
Derive an expression for the equilibrium spacing, r0, as a function of A, B, m, and n.
To derive an expression for the equilibrium spacing, r0, we need to find the value of r at which the energy, U(r), is at a minimum. At the equilibrium spacing, the energy is at its lowest point.
To find this minimum value, we need to minimize the energy function, U(r), by taking its derivative with respect to r and setting it equal to zero.
Let's begin by finding the derivative of U(r) with respect to r:
dU(r)/dr = d/dx (-A/r^m + B/r^n)
Applying the power rule of differentiation, we can find the derivative:
dU(r)/dr = A*(m/r^(m+1)) - B*(n/r^(n+1))
Setting this derivative equal to zero:
0 = A*(m/r0^(m+1)) - B*(n/r0^(n+1))
Now, let's rearrange the equation to solve for r0:
A*(m/r0^(m+1)) = B*(n/r0^(n+1))
Dividing both sides of the equation by A:
(m/r0^(m+1)) = (B/A)*(n/r0^(n+1))
Cross-multiplying and simplifying:
m*(r0^(n+1)) = B/A * n*(r0^(m+1))
Dividing both sides by n*(r0^(m+1)):
(m/r0)^(n+1) = B/A * (r0/m)^n
Taking the (n+1)-th root of both sides:
(m/r0) = [(B/A * (r0/m))^n]^(1/(n+1))
Simplifying further:
m/r0 = (B/A)^n * (r0/m)^n/(n+1)
Multiplying both sides by r0:
m = (B/A)^n * (r0/m)^(n/(n+1)) * r0/(n+1)
Now, let's get rid of the fraction by multiplying both sides by (n+1):
m*(n+1) = (B/A)^n * (r0/m)^(n/(n+1)) * r0
Simplifying:
mn + m = (B/A)^n * (r0/m)^(n/(n+1)) * r0
Subtracting m from both sides:
mn = (B/A)^n * (r0/m)^(n/(n+1)) * r0 - m
Rearranging the terms:
mn + m = (B/A)^n * (r0/m)^(n/(n+1)) * r0
Dividing both sides by m:
n + 1 = (B/A)^n * (r0/m)^(n/(n+1)) * r0
Now, we can isolate r0:
(r0/m)^(n/(n+1)) = [(B/A)^n * (n + 1)]^(1/n)
Taking the (n+1)/(n)-th root:
r0/m = [(B/A)^n * (n + 1)]^(1/n*(n+1))
Multiplying both sides by m:
r0 = m * [(B/A)^n * (n + 1)]^(1/(m*(n+1)))
Therefore, the expression for the equilibrium spacing, r0, as a function of A, B, m, and n is:
r0 = m * [(B/A)^n * (n + 1)]^(1/(m*(n+1)))