JEE Main 2018 — Oscillations Question with Solution

Given, molecular weight of silver $\left(M\right)=108 \mathrm{g}$ and Avogadro's number$\left(N_A\right)=6.02\times {10}^{23}$,
frequency $\left(f\right)={10}^{12} \mathrm{s}$.

Mass of single silver atom will be given by,

$m=\frac{M}{N}=\frac{108}{6.02\times {10}^{23}}=1.79\times {10}^{-22} \mathrm{g}=1.79\times {10}^{-25} \mathrm{kg}$

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,

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

$\frac{1}{{10}^{12}}=2\mathrm{\pi }\sqrt{\frac{\left(1.79\times {10}^{-22}\right)\times {10}^{-3}}{k}}$

$k=7.1 \mathrm{N} {\mathrm{m}}^{-1}$.