A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of $10^{12} / \mathrm{sec}$. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number $=6.02 \times 10^{23} \mathrm{gm}$ mole$^{-1}$)

(C) $7.1 \mathrm{~N} / \mathrm{m}$

(A) $5.5 \mathrm{~N} / \mathrm{m}$

(B) $6.4 \mathrm{~N} / \mathrm{m}$

(D) $2.2 \mathrm{~N} / \mathrm{m}$

JEE Main · 2018 · 08 Apr 2018 · Offline · Elasticity & SHM

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of

1 0^{12} / sec

. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number

= 6.02 \times 1 0^{23} gm

mole

^{- 1}

)

Answer: (C)

A bound atom oscillating in simple harmonic motion obeys

ω = \frac{k}{m}

, so the force constant is

k = m ω^{2}

. The angular frequency is

ω = 2 π f = 2 π \times 1 0^{12}

rad/s and the mass of one silver atom is

m = \frac{108 \times 1 0 ^{- 3}}{6.02 \times 1 0 ^{23}}

kg, so

k = m ω^{2} = \frac{108 \times 1 0 ^{- 3}}{6.02 \times 1 0 ^{23}} \times 4 π^{2} \times 1 0^{24} = \frac{108 \times 4 π ^{2} \times 10}{6.02} \approx 7.1

N/m.

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option