\(\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}=10^{12}\) where \(\mathrm{m}=\mathrm{mass}\) of one atom

Mass of one atom of silver, \(=\frac{108}{\left(6.02 \times 10^{23}\right)} \times 10^{-3} \mathrm{kg}\)

\(\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{108 \times 10^{-3}} \times 6.02 \times 10^{23}}=10^{12}\)

Solving we get, spring constant, \(\mathrm{K}=7.1 \mathrm{N} / \mathrm{m}\)