A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/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 × 1023 gm mole–1)

Simple Harmonic Motion PYQ — JEE Main 2018 · Shift 0 · Q55 with Solution

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Correct answer: C

For a particle of mass $m$ executing simple harmonic motion,

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

So,

k = 4\pi^2 f^2 m

Given:

Hence mass of one atom,

m = \frac{0.108}{6.02\times 10^{23}}\ \text{kg}

m \approx 1.79\times 10^{-25}\ \text{kg}

Given frequency,

f = 10^{12}\ \text{s}^{-1}

Therefore,

k = 4\pi^2 (10^{12})^2 (1.79\times 10^{-25})

Using $4\pi^2 \approx 39.48$ ,

k \approx 39.48 \times 10^{24} \times 1.79\times 10^{-25}

k \approx 39.48 \times 1.79 \times 10^{-1}

k \approx 70.7 \times 10^{-1}

k \approx 7.1\ \text{N/m}

The calculated force constant is

\boxed{7.1\ \text{N/m}}

So the correct option is C.