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? ................ $\mathrm{N/m}$ (Mole wt. of silver $= 108 $ andAvagadro number $= 6.02 \times 10^{23}$ $gm \ mole^{ -1}$ )
JEE MAIN 2018, Medium
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As we know, frequency in $SHM$
$\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}=10^{12}$ where $\mathrm{m}=\mathrm{mass}$ of one atom
Mass of one atom of silver, $=\frac{108}{\left(6.02 \times 10^{23}\right)} \times 10^{-3} \mathrm{kg}$
$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{108 \times 10^{-3}} \times 6.02 \times 10^{23}}=10^{12}$
Solving we get, spring constant, $\mathrm{K}=7.1 \mathrm{N} / \mathrm{m}$
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