Question

The potential energy (U) of diatomic molecule is a function dependent on \[r\] (interatomic distance) as \[U = \dfrac{\alpha }{{{r^{10}}}} - \dfrac{\beta }{{{r^5}}} - 3\], where \[\alpha \] and \[\beta \] are positive constants. The equilibrium distance between two atoms will be \[{\left( {\dfrac{{2\alpha }}{\beta }} \right)^{\dfrac{a}{b}}}\] where \[a\]is equal to:

Answer 1

Hint: Here, the potential energy of a diatomic molecule is given and which is the function of \[r\]. Now, let us understand what is asked in the question, we have been given the equilibrium distance between two atoms and asked to find out \[a\], so we have to compare equilibrium distance and then calculate \[a\]. Equilibrium occurs when all the net forces acting on atoms will become zero.

Formula used:
\[F = - \dfrac{{dU}}{{dr}}\]

Complete answer:
First of all let us draw a figure and understand what is the given data showing us:
Given data: potential Energy \[U = \dfrac{\alpha }{{{r^{10}}}} - \dfrac{\beta }{{{r^5}}} - 3\]
Equilibrium distance between two items will be, \[{\left( {\dfrac{{2\alpha }}{\beta }} \right)^{\dfrac{a}{b}}}\]

As we know that, equilibrium occurs only when net forces become zero. Whenever there is equilibrium between the two atoms, the force becomes zero on both the atoms, in this condition we know the relation between force and potential energy is that the negative work done is equal to the force exerted on the atoms.
Thus, we have
\[F = - \dfrac{{dU}}{{dr}}\]
(since, potential energy is the function of\[r\] )
\[F = - \left[ {\dfrac{d}{{dr}}(\alpha {r^{ - 10}} - \beta {r^{ - 5}} - 3)} \right]\]
\[F = - \left( {\alpha ( - 10){r^{ - 11}} - \beta ( - 5){r^{ - 6}} - 0} \right)\]
\[F = \dfrac{{10\alpha }}{{{r^{11}}}} - \dfrac{{5\beta }}{{{r^6}}}\]

For equilibrium force must be zero that is why,
\[\dfrac{{10\alpha }}{{{r^{11}}}} - \dfrac{{5\beta }}{{{r^6}}} = 0\]
\[\dfrac{{10\alpha }}{{{r^{11}}}} = \dfrac{{5\beta }}{{{r^6}}}\]

On solving the above equation we get
\[2\alpha = \beta {r^5}\]
\[{r^5} = \dfrac{{2\alpha }}{\beta }\]
\[r = {\left( {\dfrac{{2\alpha }}{\beta }} \right)^{\dfrac{1}{5}}}\]

Now on equating the given equilibrium distance and the calculated one we get
\[{\left( {\dfrac{{2\alpha }}{\beta }} \right)^{\dfrac{a}{b}}} = {\left( {\dfrac{{2\alpha }}{\beta }} \right)^{\dfrac{1}{5}}}\]
Now, on comparing both the sides we get
\[a = 1,b = 5\]

Therefore, the answer to this question is \[a = 1\]

Note: Here, we have used the concept of equilibrium and the work done by the atoms in diatomic molecule, where the diatoms are apart such that the net force on them becomes zero. This is the most important part of this question that we had to understand.