Question

The potential energy of a pair of hydrogen atoms separated by a large distance x is given by u(x)=−c6/x6, where c6 is a positive constant. is this force attractive or repulsive?

Answer 1

(-6C₆) / x⁷ newton; it is an attractive force.  

Further explanation

Given:

The potential energy of a pair of hydrogen atoms separated by a large distance x is given by $\boxed{ \ U(x) = - \frac{C_6}{x^6} \ }$, where C₆ is a positive constant.  

Question:

• What is the force that one atom exerts on the other?  
• Is this force attractive or repulsive?

The Process:

Let us rewrite the potential energy of the pair of the hydrogen atom:

$\boxed{ \ U(x) = - \frac{C_6}{x^6} \ }$

The component of a conservative force in a particular direction, equals the negative of the derivative of the corresponding potential energy, properly concerning a displacement in that direction.

For one-dimensional motion, say along the x-axis, the relationship between force and potential energy is as follows:

$\boxed{ \ \overrightarrow{F} = F_x \hat{i} = - \frac{\delta U}{\delta x} \hat{i} \ }$

Let us determine the force.

$\boxed{ \ F_x \hat{i} = - \frac{\delta}{\delta x} \Big( - \frac{C_6}{x^6} \Big) \ \hat{i} \ }$

$\boxed{ \ F_x \hat{i} = - (- C_6) \frac{\delta}{\delta x} \Big( \frac{1}{x^6} \Big) \ \hat{i} \ }$

$\boxed{ \ F_x \hat{i} = C_6 \frac{\delta}{\delta x} (x^{-6}) \ \hat{i}} \ }$

$\boxed{ \ F_x \hat{i} = (C_6)(-6)x^{-7} \ \hat{i} \ }$

Thus, we get: $\boxed{\boxed{ \ F_x = - \frac{6C_6}{x^7} \ \hat{i} \ }}$

From the data problem above, we know that C₆ is a positive constant. Hence, Fₓ is positive when x is negative and vice versa. Thus Fₓ is always directed toward the origin. Besides, a negative sign means that an attractive force occurs.

In essence, the directions are toward the origin, since this is the potential energy for a restoring force.

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Notes:

We know that the difference of potential energy from point 1 to point 2 as the negative of the work done:  

$\boxed{ \ \Delta U_{12} = U_2 - U_1 = - W_{12} \ }$

The work done by the given force as the particle moves from coordinate x to (x + d x) in one dimension is  

$\boxed{ \ dW = \overrightarrow{F} \cdot d \vec{r} \ } \rightarrow \boxed{ \ \overrightarrow{F} = \frac{dW}{d \vec{r}} \ }$

Therefore, for motion along a straight line, a conservative force  is the negative derivative of its associated potential energy function, i.e.,

$\boxed{ \ \overrightarrow{F} = F_x \hat{i} = - \frac{\delta U}{\delta x} \hat{i} \ }$

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Conservative force : force that does work independent of path.

Learn more

1. The derivatives of the composite function  brainly.com/question/6013189  
2. About vector components brainly.com/question/1600633  
3. What is the approximate potential energy (PE) due to the gravity of the mass with respect to the ground? brainly.com/question/39225

Answer 2

We need to see what happens to the energy depending on their distance. If the atoms get closer, we have that x gets smaller and their energy smaller; 1/x^6 is getting bigger but accounting for the minus sign, u gets smaller. If their distance increases, x gets bigger and similarly u increases. The forces act in a way to decrease potential energy, in every case of power. In this case, since potential energy is reduced when distance is decreased, we have that the force is attractive.