Question

Potential energy as a function of r is given by \(U = \frac{A}{r^{10}}-\frac{B}{r^5}\), where \(r\) is the interatomic distance, A and B are positive constants. The equilibrium distance between the two atoms will be:

• A. \((\frac{A}{B})^{\frac{1}{5}}\)
• B. \((\frac{B}{A})^{\frac{1}{5}}\)
• C. \((\frac{2A}{B})^{\frac{1}{5}}\)
• D. \((\frac{2B}{A})^{\frac{1}{5}}\)

Answer 1

The Correct Option is C

To find the equilibrium distance between two atoms, we must first understand the concept of potential energy in the context of atomic interactions. The potential energy function given is:

U = \frac{A}{r^{10}} - \frac{B}{r^5}

where r is the interatomic distance, and A and B are positive constants. For the equilibrium position, the force experienced by the atoms must be zero. The force between the atoms is given by the negative gradient of the potential energy:

F = -\frac{dU}{dr}

Setting this force to zero will give the equilibrium condition:

\begin{align*} \frac{dU}{dr} &= \frac{d}{dr}\left(\frac{A}{r^{10}} - \frac{B}{r^5}\right) \\ &= \frac{d}{dr}(Ar^{-10}) - \frac{d}{dr}(Br^{-5}) \\ &= -10Ar^{-11} + 5Br^{-6} = 0 \end{align*}

We solve this equation for r:

-10Ar^{-11} + 5Br^{-6} = 0

Rearranging terms:

10Ar^{-11} = 5Br^{-6}

Divide both sides by 5:

2Ar^{-11} = Br^{-6}

Multiply both sides by r^{11}:

2A = Br^5

Solve for r^5:

r^5 = \frac{2A}{B}

Thus, the equilibrium distance r between the atoms is:

r = \left(\frac{2A}{B}\right)^{\frac{1}{5}}

This matches the correct option: \left(\frac{2A}{B}\right)^{\frac{1}{5}}.

Conclusion: The equilibrium distance between the two atoms is \left(\frac{2A}{B}\right)^{\frac{1}{5}}.