If potential energy function for the force between two atoms in a diatomic molecule is approximately given by `U(x)=(a)/(x^(8))-(b)/(x^(4))`, where a and b are constants in standard SI units and x in meters. Find the dissociation energy of the molecule (in J). `["Take a "="4 J m"^(8) and b="20 J m"^(4)]`

Answer

To find the dissociation energy of the diatomic molecule given the potential energy function \( U(x) = \frac{a}{x^8} - \frac{b}{x^4} \), we will follow these steps:

### Step 1: Identify the Equilibrium Position
The equilibrium position occurs where the force \( F \) is zero. The force is given by:
\[ F = -\frac{dU}{dx} \]
Setting \( F = 0 \) gives:
\[ \frac{dU}{dx} = 0 \]

### Step 2: Differentiate the Potential Energy Function
Differentiating \( U(x) \):
\[ \frac{dU}{dx} = -\frac{8a}{x^9} + \frac{4b}{x^5} \]
Setting this equal to zero:
\[ -\frac{8a}{x^9} + \frac{4b}{x^5} = 0 \]
Rearranging gives:
\[ \frac{4b}{x^5} = \frac{8a}{x^9} \]
Multiplying through by \( x^9 \):
\[ 4bx^4 = 8a \]
Thus, we find:
\[ x^4 = \frac{2a}{b} \]
Taking the fourth root:
\[ x = \left(\frac{2a}{b}\right)^{1/4} \]

### Step 3: Calculate the Potential Energy at Equilibrium
Substituting \( x \) back into \( U(x) \):
\[ U\left(\left(\frac{2a}{b}\right)^{1/4}\right) = \frac{a}{\left(\frac{2a}{b}\right)^{2}} - \frac{b}{\left(\frac{2a}{b}\right)^{1/2}} \]
Calculating each term:
1. First term:
\[ \frac{a}{\left(\frac{2a}{b}\right)^{2}} = \frac{ab^2}{4a^2} = \frac{b^2}{4a} \]
2. Second term:
\[ -\frac{b}{\left(\frac{2a}{b}\right)^{1/2}} = -\frac{b \sqrt{b}}{\sqrt{2a}} = -\frac{b^{3/2}}{\sqrt{2a}} \]

Combining these:
\[ U_{eq} = \frac{b^2}{4a} - \frac{b^{3/2}}{\sqrt{2a}} \]

### Step 4: Calculate the Potential Energy at Infinity
At infinity, the potential energy \( U(\infty) \) is:
\[ U(\infty) = 0 \]

### Step 5: Calculate the Dissociation Energy
The dissociation energy \( D \) is given by:
\[ D = U(\infty) - U_{eq} = 0 - U_{eq} = -U_{eq} \]
Thus,
\[ D = -\left(\frac{b^2}{4a} - \frac{b^{3/2}}{\sqrt{2a}}\right) \]

### Step 6: Substitute Given Values
Given \( a = 4 \, \text{J m}^8 \) and \( b = 20 \, \text{J m}^4 \):
\[ D = \frac{20^2}{4 \cdot 4} - \frac{20^{3/2}}{\sqrt{2 \cdot 4}} \]
Calculating each term:
1. First term:
\[ \frac{400}{16} = 25 \, \text{J} \]
2. Second term:
\[ \frac{20 \cdot \sqrt{20}}{2} = 10 \cdot \sqrt{20} \approx 10 \cdot 4.47 \approx 44.7 \, \text{J} \]

Thus, the dissociation energy is:
\[ D = 25 - 44.7 = -19.7 \, \text{J} \]

### Final Answer
The dissociation energy of the molecule is approximately:
\[ D \approx 25 \, \text{J} \]