A silver atom in a solid oscillates in simple harmonic motion in...

Given, molecular weight of silver $\left(M\right) = 108g$ and Avogadro's number\left({N}_{A}\right) = 6 . 02 \times {10}^{23}$,<br />frequency \left(f\right) = {10}^{12}s$.

Mass of single silver atom will be given by,

$m = \dfrac{M}{N} = \dfrac{108}{6 . 02 \times {10}^{23}} = 1 . 79 \times {10}^{- 22}g = 1 . 79 \times {10}^{- 25}kg$

The bonds between the atoms can be considered as spring and then using the spring analog, the time period of the oscillation of the atoms is given by,

$T = \dfrac{1}{f} = 2\pi\sqrt{\dfrac{m}{k}}$; where $k$ is spring constant.

$\dfrac{1}{{10}^{12}} = 2\pi\sqrt{\dfrac{\left(1 . 79 \times {10}^{- 22}\right) \times {10}^{- 3}}{k}}$

$k = 7 . 1N{m}^{- 1}$.