A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of $$10^{12}$$ s$$^{-1}$$. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 g mol$$^{-1}$$ and Avogadro number = $$6.02 \times 10^{23}$$)

The motion of an atom in the crystal can be regarded as that of a linear simple harmonic oscillator. For such an oscillator the angular frequency $$\omega$$, the force constant (or spring constant) $$k$$ and the mass of the oscillator $$m$$ are related by the well-known formula

$$\omega \;=\;\sqrt{\dfrac{k}{m}}\;.$$

Re-arranging we obtain the force constant in terms of mass and angular frequency:

$$k \;=\; m\,\omega^{2}\;.$$

The ordinary frequency of oscillation is given as $$\nu = 10^{12}\ \text{s}^{-1}$$, so the angular frequency is

$$\omega \;=\; 2\pi\nu = 2\pi \times 10^{12}\ \text{s}^{-1} = 6.283 \times 10^{12}\ \text{s}^{-1}\;.$$

Next we calculate the mass of a single silver atom. The molar mass of silver is 108 g mol$$^{-1}$$, i.e.

$$M = 108\ \text{g mol}^{-1} = 0.108\ \text{kg mol}^{-1}\;.$$

Dividing by Avogadro’s number $$N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$$ gives

$$m = \dfrac{M}{N_A} = \dfrac{0.108\ \text{kg}}{6.02 \times 10^{23}} = 1.794 \times 10^{-25}\ \text{kg}\;.$$

Now we need $$\omega^{2}$$:

$$\omega^{2} = (2\pi\nu)^{2} = (6.283 \times 10^{12})^{2} = 39.48 \times 10^{24}\ \text{s}^{-2}\;.$$

Substituting $$m$$ and $$\omega^{2}$$ into $$k = m\omega^{2}$$:

$$k = (1.794 \times 10^{-25}\ \text{kg}) \times (39.48 \times 10^{24}\ \text{s}^{-2}) = (1.794 \times 39.48)\times 10^{-25+24}\ \text{N m}^{-1}\;.$$

$$k = 70.8 \times 10^{-1}\ \text{N m}^{-1} = 7.08\ \text{N m}^{-1}\;.$$

Rounding to one decimal place, we have $$k \approx 7.1\ \text{N m}^{-1}$$.

Hence, the correct answer is Option C.