A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of

Q: Why do we model the atom as a spring-mass system?

In solid-state physics, the interatomic forces near the equilibrium position can be approximated by Hooke's Law ($F = -kx$). This means small vibrations of atoms behave exactly like a mass on a spring, allowing us to use standard SHM equations.

Q: Why did we divide by Avogadro's number?

The given molecular weight (108 g/mol) is the mass of one *mole* of silver atoms. Since the formula requires the mass of a *single* oscillating atom, we must divide the molar mass by the number of atoms in a mole, which is Avogadro's number.

Q: Why must the molar mass be converted to kg?

To get the force constant $k$ in standard SI units (Newtons per meter, where $1 \text{ N} = 1 \text{ kg}\cdot\text{m/s}^2$), all input variables must be in SI units. Therefore, grams must be converted to kilograms.

Q: Is the force constant $k$ the same as the spring constant?

Yes, mathematically they are identical in this context. The term 'force constant' is often preferred in chemistry and atomic physics to describe the stiffness of a chemical bond, while 'spring constant' is used for macroscopic mechanical springs.

Q: How do I handle the calculation without a calculator?

Group the powers of 10 together and the significant digits together. Use approximations like $\pi^2 \approx 10$ or $9.87$. For example, $108 / 6.02$ is roughly $108 / 6 = 18$. Then $4 \times 10 \times 18 = 720$. Adjusting for the powers of 10 ($10^{24} \times 10^{-3} / 10^{23} = 10^{-2}$) gives $720 \times 10^{-2} = 7.2$, which is very close to the exact answer 7.1.

Question

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of $10^{12}$ per second. What is the force constant of the bonds connecting one atom with the other? (Take, molecular weight of silver = 108 and Avogadro number = $6.02 	imes 10^{23} 	ext{ g mol}^{-1}$)

Options:

(A) 6.4 N/m

(B) 7.1 N/m

(C) 2.2 N/m

(D) 5.5 N/m

Sigma Percentile · Accepted Answer

### Modeling the Atomic Lattice

Imagine the solid lattice of silver. Each atom is held firmly in place by chemical bonds with its neighboring atoms. When thermal energy is present, these atoms don't just sit still; they vibrate around their equilibrium positions. We can elegantly model this single silver atom as a small mass attached to a spring, oscillating back and forth. In this analogy, the spring represents the restoring force of the chemical bond, and the spring constant $k$ represents the "stiffness" or force constant of that bond.

### The Master Equation

For any simple harmonic oscillator, the time period $T$ is given by the classic formula:

$$T = 2\pi \sqrt{\frac{m}{k}}$$

Since frequency $f$ is the reciprocal of the time period ($f = \frac{1}{T}$), we can express the frequency as:

$$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$$

Our goal is to find the force constant $k$. Let's isolate $k$ by squaring both sides of the equation:

$$f^2 = \frac{1}{4\pi^2} \frac{k}{m}$$

$$\Rightarrow k = 4\pi^2 f^2 m$$

### Finding the Mass of a Single Atom

Before we can plug numbers into our master equation, we need the mass $m$ of a *single* silver atom in standard SI units (kilograms). We are given the molecular weight (molar mass) of silver as $M = 108 	ext{ g/mol}$.

First, we must convert this to kilograms:

$$M = 108 	imes 10^{-3} 	ext{ kg/mol}$$

This is the mass of one mole of silver atoms. To find the mass of just one atom, we divide by Avogadro's number $N_A$:

$$m = \frac{M}{N_A} = \frac{108 	imes 10^{-3}}{6.02 	imes 10^{23}} 	ext{ kg}$$

### Final Calculation

Now, let's carefully substitute all our known values into the rearranged formula for $k$:

$$k = 4 \pi^2 (10^{12})^2 \left( \frac{108 	imes 10^{-3}}{6.02 	imes 10^{23}} ight)$$

Don't get intimidated by the large numbers! Let's break it down. We can approximate $\pi^2 \approx 9.87$. Evaluating the powers of 10 gives us $(10^{12})^2 = 10^{24}$.

$$k \approx 4 	imes 9.87 	imes 10^{24} 	imes \frac{108 	imes 10^{-3}}{6.02 	imes 10^{23}}$$

$$k \approx 39.48 	imes 10^{24} 	imes 17.94 	imes 10^{-26}$$

$$k \approx 708.27 	imes 10^{-2} 	ext{ N/m}$$

$$k \approx 7.08 	ext{ N/m}$$

Rounding to one decimal place, we get **$7.1 	ext{ N/m}$**. This perfectly matches option (b). It is truly fascinating how macroscopic properties like molar mass and Avogadro's number allow us to calculate the mechanical stiffness of a single atomic bond!

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Visualized Solution

$M o d e l o f A t o m$

$F r e q u e n cy F or m u l a$

$M a ss o f O n e A t o m$

$R e a r r an g in g f or k$

$C a l c u l a t i o n$

$C o n c l u s i o n$

The Sigma Insight: Simple Harmonic Motion (SHM)

Modeling the Atomic Lattice

The Master Equation

Finding the Mass of a Single Atom

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Model of Atom

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Frequency Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass of One Atom

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rearranging for k

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion