Q4.9 Force and potential energy (K)

  1. A particle experiences a conservative force F, described by a potential energy U(x). Write down an equation defining the form of the force F(x), and express it in words.
  2. The force which one atom exerts on another, a distance r away, can often be expressed through a potential energy function U(r) with the qualitative form shown.
    Sketch the implied form of F(r). Identify the point of mechanical equilibrium, both in the sketch of U(r) given, and your own sketch of F(r).
    Graphic - No title - lj2
  3. For a pair of Argon atoms the function U(r) is well described by the form
    \[ U(r)= 4\varepsilon\left[ \left(\frac{\sigma}{r}\right) ^{12} -\left(\frac{\sigma}{r}\right)^{6}\right] \]
    with σ=3.4×10-10 m and ε=1.67×10-21 J.
    1. Convince yourself that this function has the shape shown in the sketch.
    2. Find the value of the separation r at which the two atoms will be in mechanical equilibrium, and the value of U(r) at this point, expressing the latter in electron volts (eV). (1eV=1.6×10-19J)
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Hint

If a function has a minimum, what happens to its gradient at that point?
 

Solution

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Solution

  1. The form of the force, in terms of the potential energy is (Key Point 3.11)

    \[ F(r) = - \frac{d U(r)}{dr} \]

    In words: The force at r is the (negative) of the gradient of potential energy at r

  2. The force looks like

    Graphic - No title - ljfrc2

    Note: that the zero of the force occurs at the minimum of the potential energy, which is the point of mechanical equilibrium. Both force and potential energy are small at large atomic separations (and attractive), and very large (repulsive) at small separations.

  3. Lennard-Jones model for Argon
    1. The potential energy is the sum of two terms. The first (r-12) term is always positive and diverges at r=0 and decays at large r. The second (r-6) term diverges at r=0, but less strongly than the first, since it’s a smaller power. Therefore at short range the function must be positive. At larger r the function must decay to zero (since both terms do), but from below, since r-6>r-12 when r>1. Therefore, there must be a minimum somewhere in between.
    2. If the particle feels no force, it is in mechanical equilibrium. This happens when the gradient of the potential energy is zero, at r=rmin. Therefore we have to solve

      \[ \frac{dU(r)}{dr} = 0 \]
      Where,
      \[ U(r)= 4\varepsilon\left[ \left(\frac{\sigma}{r}\right) ^{12} -\left(\frac{\sigma}{r}\right)^{6}\right] \]
      Differentiating:
      \[ \frac{dU(r)}{dr} = - \left[ 12 \left(\frac{\sigma^{12}}{r^{13}} \right) - 6\left(\frac{\sigma^{6}}{r^{7}} \right) \right] = 0 \]
      Which we can rewrite as
      \[ - \left[ \frac{12}{\sigma} \left(\frac{\sigma}{r} \right)^{13} - \frac{6}{\sigma} \left(\frac{\sigma}{r} \right)^{7} \right] = 0 \]
      Taking out a factor of $\frac{6}{\sigma} \left(\frac{\sigma}{r} \right)^{7}$ 
      \[ - \frac{6}{\sigma} \left(\frac{\sigma}{r} \right)^{7} \left[ 2 \left(\frac{\sigma}{r} \right)^{6} - 1 \right] = 0 \]
      Cancelling all common prefactors, we are left with

      \[ 2 \left( \frac{\sigma}{r} \right)^6 = 1 \]

      or

      r=rmin=21/6σ

      Substituting into the equation for U(r),

      \[ U(r)= 4\varepsilon\left[ \left(\frac{\sigma}{r}\right) ^{12} -\left(\frac{\sigma}{r}\right)^{6}\right] = 4\varepsilon\left[ \left(\frac{\sigma}{2^{1/6}\sigma}\right) ^{12} -\left(\frac{\sigma}{2^{1/6}\sigma}\right)^{6}\right] = 4\varepsilon\left[ \frac{1}{4} -\frac{1}{2}\right] = - \varepsilon \]
      So we find that the value of the potential at the minimum is -ε. In electron volts, this is 0.01 eV. The electron volt energy scale is defined by the work done by a force to move an electron (a basic unit of charge) through a potential of 1 volt. This is as it should be: if the energies that rip electrons out of atoms are of the order of 10 eV: the potential energy should be much less than this, at its deepest, certainly for noble gases such as Argon.