Question

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of ${10}^{12}{s}^{-1}$. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = $108amu$ and $N_A=6.02x{10}^{23}gmo{l}^{-1}$)

• A. $2.2N/m$
• B. $5.5N/m$
• C. $6.4N/m$
• D. $7.1N/m$

Answer 1

The correct option is D

$7.1N/m$

Step 1. Given data

Frequency,$\nu ={10}^{12}{s}^{-1}$

Molar mass of silver atom =$108amu=108g=108\times {10}^{-3}kg$

Step 2.Finding the force constant

Frequency of SHM is, $\nu =\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

Squaring both side and then rearranging, we get

$k=4{\pi }^2{\nu }^{2}m$

Mass of one silver atom, $m=\frac{molarmass}{AvagadroNumber}=\frac{108\times {10}^{-3}}{6.02x{10}^{23}}=17.9\times {10}^{-26}kg$

$k=4\times {(3.14)}^2\times {\left({10}^{12}\right)}^2\times 17.9\times {10}^{-26}=7.06N/m\approx 7.1N/m$

Hence, the correct option is D