JEE Mains · Physics · STD 11 - 13. oscillations
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 \(= 6.02 \times 10^{23}\) \(gm \ mole^{ -1}\) )
- A \(7.1\)
- B \(2.2 \)
- C \(5.5 \)
- D \(6.4 \)
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, \(=\frac{108}{\left(6.02 \times 10^{23}\right)} \times 10^{-3} \mathrm{kg}\)…
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