Oscillations

JEE Main level, single correct, medium

136 of 194

Question 136 of 194

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012s−1{10}^{12}{s}^{-1} . What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver,=108gmol−1=108g{mol}^{-1} and Avogadro number =6.02×1023=6.02\times {10}^{23})

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Answer: (C) 7.1Nm−17.1N{m}^{-1}

Given, molecular weight of silver (M)=108g\left(M\right)=108g and Avogadro's number(NA)=6.02×1023\left({N}_{A}\right)=6.02\times {10}^{23},
frequency (f)=1012s\left(f\right)={10}^{12}s.
Mass of single silver atom will be given by,
m=MN=1086.02×1023=1.79×10−22g=1.79×10−25kgm=\dfrac{M}{N}=\dfrac{108}{6.02\times {10}^{23}}=1.79\times {10}^{-22}g=1.79\times {10}^{-25}kgThe bonds between the atoms can be considered as spring and then using the spring analog, the time period of the oscillation of the atoms is given by,
T=1f=2πmkT=\dfrac{1}{f}=2\pi \sqrt{\dfrac{m}{k}}; where kk is spring constant.
11012=2π(1.79×10−22)×10−3k\dfrac{1}{{10}^{12}}=2\pi \sqrt{\dfrac{\left(1.79\times {10}^{-22}\right)\times {10}^{-3}}{k}}
k=7.1Nm−1k=7.1N{m}^{-1}.

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