Q 11-13-154JEE MainJEE Main 2018 (8 Apr)Easy
A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of $10^{12}\ \text{s}^{-1}$. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver $= 108\ \text{g mol}^{-1}$ and Avogadro number $= 6.02\times10^{23}$)
Answer: (C) $7.1\ \text{N m}^{-1}$
Mass of one atom: $m = \dfrac{0.108}{6.02\times10^{23}} = 1.79\times10^{-25}$ kg.
$$k = m\omega^2 = m(2\pi f)^2 = 1.79\times10^{-25}\times(2\pi\times10^{12})^2 \approx 7.1\ \text{N m}^{-1}$$
Solution by Sreeraj P, M.Sc Physics