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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2018
  • Medium
  • MCQ

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 \(= 6.02 \times 10^{23}\) \(gm \ mole^{ -1}\) )

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Answer & Solution

Correct Answer

(A) \(7.1\) 

Step-by-step Solution

Detailed explanation

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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