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Step-by-Step Solution
Step 1: Write down the given potential energy function
The potential energy of the diatomic molecule is given as
U = \frac{\alpha}{r^{10}} \;-\; \frac{\beta}{r^{5}} \;-\; 3,
where \alpha and \beta are positive constants and r is the interatomic distance.
Step 2: Express the force as the negative gradient of potential energy
In one dimension (along the line of the two atoms), the force F can be found from
F = -\frac{dU}{dr}.
Differentiate U with respect to r :
\frac{dU}{dr} \;=\; \frac{d}{dr}\!\biggl(\frac{\alpha}{r^{10}} - \frac{\beta}{r^{5}} - 3\biggr).
Computing these derivatives term by term, we get:
\frac{d}{dr}\Bigl(\frac{\alpha}{r^{10}}\Bigr) = -10\,\frac{\alpha}{r^{11}}, \quad
\frac{d}{dr}\Bigl(\frac{\beta}{r^5}\Bigr) = -5\,\frac{\beta}{r^6}, \quad
\frac{d}{dr}\Bigl(3\Bigr) = 0.
Thus,
\frac{dU}{dr} = -10\,\frac{\alpha}{r^{11}} + 5\,\frac{\beta}{r^6}.
Therefore,
F = -\frac{dU}{dr} = -\Bigl(-10\,\frac{\alpha}{r^{11}} + 5\,\frac{\beta}{r^6}\Bigr)
= 10\,\frac{\alpha}{r^{11}} - 5\,\frac{\beta}{r^6}.
Step 3: Impose the condition for equilibrium
At equilibrium, the net force must be zero:
F = 0 \implies 10\,\frac{\alpha}{r^{11}} - 5\,\frac{\beta}{r^6} = 0.
Solve this equation for r :
10\,\frac{\alpha}{r^{11}} = 5\,\frac{\beta}{r^6}.
We can simplify this:
2\,\frac{\alpha}{r^{11}} = \frac{\beta}{r^6}.
Rearranging, we get:
\frac{2 \alpha}{\beta} = r^5.
Step 4: Solve for the equilibrium distance
Taking the fifth root of both sides gives:
r = \Bigl(\frac{2 \alpha}{\beta}\Bigr)^{\frac{1}{5}}.
This distance can be written in the form
r = \Bigl(\frac{2\alpha}{\beta}\Bigr)^{\frac{a}{b}}.
Comparing exponents, we see a = 1 and b = 5.
Step 5: State the value of a
The value of a is therefore
\boxed{1}.
