Give step-by-step solution with explanation and final answer:Problem 4 (10 points): A crystal is a periodic lattice of positively and negatively charged ions. The
goal of this problem is to show that

Question

Give step-by-step solution with explanation and final answer:Problem 4 (10 points): A crystal is a periodic lattice of positively and negatively charged ions. The
goal of this problem is to show that the potential energy per ion is negative. This is one condition for a
crystal to be a stable configuration: otherwise, it would be energetically favorable for the ions to be infinitely
far apart, which would disintegrate the crystal. Although in general a crystal can have a complicated 3D
arrangement of fons, here we will consider a 1D crystal as a simpler example.
Consider an infinite 1D crystal of alternating ions with charges + and —c, separated by distance d:

ds

—O—2— 02020202
fe —¢ te —c He —¢ +e —c +c ec
2 eli il ie2

In the figure above, we have labeled some of the charges by i, i=1, etc. The fotal potential energy of the
infinite lattice is

0-5 (£0) 1)

LE
where Uy is the potential energy between charges i and j. The term in parentheses, denoted Up, given by

Uo=Y Uy. ®)

f=

is the same for every charge i and is the potential encrgy per charge. Show that Up = —e2In(2)/(47cod).
Hint: You should encounter an infinite sum of the form 32 ,(~1)"*/n. You

VIPSolutions · Accepted Answer

Let's solve the problem step by step:

### 1. **Potential Energy Between Two Charges**

The potential energy between two charges $ +e $ and $ -e $ separated by distance $ r $ is:
$$
U_{ij} = \frac{q_i q_j}{4\pi \epsilon_0 r_{ij}}
$$
In our 1D lattice, charges alternate between $ +e $ and $ -e $, and are separated by distance $ d $.

### 2. **Labeling the Charges**

Let the charge at position $ i $ be:
$$
q_i = (-1)^i e
$$
The charge at $ j $ is $ q_j = (-1)^j e $.

The distance between charges $ i $ and $ j $ is $ |i-j|d $.

### 3. **Potential Energy per Charge (U₀)**

For a given charge (say at $ i = 0 $), the potential energy due to all charges to its right is:
$$
U_0 = \sum_{j=1}^{\infty} U_{0j}
$$
where
$$
U_{0j} = \frac{q_0 q_j}{4\pi \epsilon_0 |0-j|d}
$$
Plugging in $ q_0 = e $, $ q_j = (-1)^j e $, and $ |0-j|=j $:
$$
U_{0j} = \frac{e \cdot (-1)^j e}{4\pi \epsilon_0 j d} = \frac{(-1)^j e^2}{4\pi \epsilon_0 j d}
$$

### 4. **Sum for Potential Energy per Charge**

So,
$$
U_0 = \sum_{j=1}^{\infty} \frac{(-1)^j e^2}{4\pi \epsilon_0 j d}
$$
$$
U_0 = \frac{e^2}{4\pi \epsilon_0 d} \sum_{j=1}^{\infty} \frac{(-1)^j}{j}
$$

### 5. **Evaluating the Infinite Sum**

The sum
$$
\sum_{j=1}^{\infty} \frac{(-1)^j}{j}
$$
is a known alternating series. In particular,
$$
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \ln(2)
$$
So,
$$
\sum_{j=1}^{\infty} \frac{(-1)^j}{j} = -\ln(2)
$$

### 6. **Final Expression for $ U_0 $**

Plug this result in:
$$
U_0 = \frac{e^2}{4\pi \epsilon_0 d} \left[ -\ln(2) \right]
$$
$$
U_0 = -\frac{e^2 \ln(2)}{4\pi \epsilon_0 d}
$$

### 7. **Conclusion**

- The potential energy per charge is **negative**.
- This negative value shows that the lattice is stable: the ions prefer to stay together rather than fly apart.

---

## **Final Answer**

$$
\boxed{U_0 = -\frac{e^2 \ln(2)}{4\pi \epsilon_0 d}}
$$

This shows the potential energy per ion is negative, as required.

---

### **Summary of Steps**

1. Wrote out the expression for the pairwise potential energy.
2. Expressed the total energy per charge as an infinite sum.
3. Used the known result for the alternating series.
4. Arrived at the negative potential energy per charge.

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