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Step-by-Step Solution
Step 1: Identify the Given Data
• Frequency of oscillation, $f = 10^{12} \text{ s}^{-1}$
• Molar mass of silver, $M = 108 \text{ g mol}^{-1}$
• Avogadro number, $N_A = 6.02 \times 10^{23} \text{ mol}^{-1}$
• We need to find the force constant $k$ of the bond.
Step 2: Calculate the Mass of One Silver Atom
In SI units, 108 g of Ag corresponds to $6.02 \times 10^{23}$ atoms. Hence, the mass of one silver atom is:
$\displaystyle m = \frac{108 \times 10^{-3}\,\text{kg}}{6.02 \times 10^{23}}$
Step 3: Recall the Relation for the Frequency of a Simple Harmonic Oscillator
For simple harmonic motion, the frequency $f$ is given by:
$\displaystyle f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$
Rewriting and squaring both sides leads to:
$\displaystyle f^2 = \frac{1}{4\pi^2} \frac{k}{m}$
Thus,
$\displaystyle k = m\, f^2 \times 4 \pi^2$
Step 4: Substitute the Numerical Values
1. Substitute $m$ as computed and $f = 10^{12}$ into the formula for $k$:
$\displaystyle k = \left(\frac{108 \times 10^{-3}}{6.02 \times 10^{23}}\right) \times \left(10^{12}\right)^2 \times 4\pi^2$
Step 5: Simplify to Obtain the Force Constant
On performing the multiplication and unit conversions carefully, we get:
$\displaystyle k \approx 7.1 \,\text{N/m}$
Final Answer
The force constant of the bonds connecting one silver atom to another is approximately $7.1\text{ N/m}$.
