The energy-separation curve for two atoms, a distance, r, apart is: U(r)=−A/r^m+B/r^n
Derive and expression for the stiffness of the bond at the equilibrium spacing, in terms of A, B, m, n, and r0.
S=dF/dr at r=r0:
S = dF/dr
So, what's F?
If you mean dU/dr, then that would be
Am/r^(m+1) - Bn/r^(n+1)
F is dU/dr, but your answer is not right :(
To derive an expression for the stiffness of the bond at the equilibrium spacing, we need to calculate the derivative of the energy-separation curve with respect to distance, r, and evaluate it at the equilibrium spacing, r0.
Given the energy-separation curve: U(r) = -A/r^m + B/r^n
First, let's differentiate U(r) with respect to r:
dU/dr = d/dr (-A/r^m + B/r^n)
For the first term, -A/r^m, we can use the power rule of differentiation:
dU/dr = A*m/r^(m+1) + ... (derivatives of other terms)
Similarly, for the second term, B/r^n:
dU/dr = A*m/r^(m+1) - B*n/r^(n+1)
Now, we need to evaluate this expression at the equilibrium spacing, r=r0. So, substitute r=r0 into the derivative:
dU/dr |r=r0 = A*m/r0^(m+1) - B*n/r0^(n+1)
The stiffness of the bond at equilibrium, denoted as S, is defined as dF/dr at r=r0, where F represents the force. Since energy is the negative of the force, we multiply the derivative by (-1):
S = -dU/dr |r=r0
Substituting the calculated derivative into the above expression:
S = - ( A*m/r0^(m+1) - B*n/r0^(n+1) )
Thus, the final expression for the stiffness of the bond at equilibrium in terms of A, B, m, n, and r0 is:
S = B*n/r0^(n+1) - A*m/r0^(m+1)