Question

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of $10^{12} / \sec$. What is the force constant of the bonds connecting one atom with the other ? ( Mole wt. of silver $= 108 $ and Avogadro number $= 6.02 \times 10^{23} \, gm \, mole^{-1} $ )

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

Answer 1

Answer: (B)

$K x=m a $
$\Rightarrow a=( K / m ) x$
$T=2 \pi \sqrt{\frac{m}{K}}$
$f=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{K}{m}}=10^{12}$
$=\frac{1}{4 \pi^{2}} \times \frac{K}{m}=10^{24}$
$K=4 \pi^{2} m \times 10^{24}=\frac{4 \times 10 \times 108 \times 10^{-3}}{6.02 \times 10^{23}} \times 10^{24}$
$=7.1 \,N / m$