Potential energy in a field is given by `U=-A/(r^6)+B/(r^11)` then find the value of r for equilibrium position.

Options

A. `r=((6A)/(11B))^(1/5)`
B. `r=((11B)/(6A))^(1/5)`
C. `r=((6B)/(11A))^(1/5)`
D. `r=((11A)/(6B))^(1/5)`

Answer

To find the value of \( r \) for the equilibrium position given the potential energy function \( U = -\frac{A}{r^6} + \frac{B}{r^{11}} \), we will follow these steps:

### Step 1: Understand the relationship between potential energy and force
The force \( F \) is related to the potential energy \( U \) by the equation:
\[ F = -\frac{dU}{dr} \]
At equilibrium, the force must be zero:
\[ F = 0 \implies -\frac{dU}{dr} = 0 \implies \frac{dU}{dr} = 0 \]

### Step 2: Differentiate the potential energy with respect to \( r \)
We need to differentiate \( U \) with respect to \( r \):
\[ U = -\frac{A}{r^6} + \frac{B}{r^{11}} \]
Using the power rule for differentiation:
\[ \frac{dU}{dr} = \frac{d}{dr}\left(-\frac{A}{r^6}\right) + \frac{d}{dr}\left(\frac{B}{r^{11}}\right) \]
\[ = 6A r^{-7} - 11B r^{-12} \]

### Step 3: Set the derivative equal to zero for equilibrium
Now, we set the derivative equal to zero:
\[ 6A r^{-7} - 11B r^{-12} = 0 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ 6A r^{-7} = 11B r^{-12} \]
Multiplying both sides by \( r^{12} \) to eliminate the negative powers:
\[ 6A r^5 = 11B \]

### Step 5: Solve for \( r \)
Now, we can solve for \( r^5 \):
\[ r^5 = \frac{11B}{6A} \]
Taking the fifth root of both sides gives:
\[ r = \left(\frac{11B}{6A}\right)^{\frac{1}{5}} \]

### Final Answer
Thus, the value of \( r \) for the equilibrium position is:
\[ r = \left(\frac{11B}{6A}\right)^{\frac{1}{5}} \]