Question

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of ${10^{12}}{\sec ^{ - 1}}$. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver $ = 108$ and Avogadro’s number$ = 6.02 \times {10^{23}}gm/mole$ )
(A) $2.2\dfrac{N}{m}$
(B) $5.5\dfrac{N}{m}$
(C) $6.4\dfrac{N}{m}$
(D) $7.1\dfrac{N}{m}$

Answer 1

Hint: We can easily calculate the value of the force constant if we know the relation between the frequency, mass and force constant for a body in simple harmonic motion. As in the question force constant of a bond connecting one atom of silver to another is asked, we will use the mass of one silver atom for calculating the force constant.

Formula Used:
 For simple harmonic motion,
$T = 2\pi \sqrt {\dfrac{m}{k}} $
Where $T$ is the time period of the simple harmonic oscillation, $m$ is the mass of the body oscillating in simple harmonic motion and $k$ is the force constant of the bonds connecting one atom with the other.

Complete step by step answer:
We know that for a simple harmonic motion,
 $T = 2\pi \sqrt {\dfrac{m}{k}} $
Where $T$ is the time period of the simple harmonic oscillation, $m$ is the mass of one silver atom oscillating in simple harmonic motion and $k$ is the force constant of the bonds connecting one silver atom with the other.
In the question frequency ($f$ ) is given as${10^{12}}{\sec ^{ - 1}}$. But we need a time period for the equation. So,
$T = \dfrac{1}{f}$
$ \Rightarrow T = \dfrac{1}{{{{10}^{12}}}} = {10^{ - 12}}\sec $
Also in the question mass of one mole of silver atom is given. We need to calculate the mass of one atom of silver. So mass of one silver atom ( $m$ ) is,
$m = \dfrac{M}{{{N_A}}}$
Where $M$is the mass of one mole of silver atom and ${N_A}$ is the Avogadro’s number i.e.$6.02 \times {10^{23}}gm/mole$
$ \Rightarrow m = \dfrac{{108}}{{6.02 \times {{10}^{23}}}} \times {10^{ - 3}}kg$
Substituting the values of $T$ and $m$in the equation of time period for simple harmonic motion, we get
${10^{ - 12}} = 2\pi \sqrt {\dfrac{{108 \times {{10}^{ - 3}}}}{{k \times 6.02 \times {{10}^{23}}}}} $
$ \Rightarrow k = 7.1\dfrac{N}{m}$

Hence, the correct option is (D).

Note: In the above question we should be careful with the values because molar mass is given but we need to find a force constant for a single atom. Also we should be careful about the units of all the individual quantities as the options are given in $\dfrac{N}{m}$ and to calculate in Newton the mass should be kilogram and not in grams.