Verified Solution • 2025–26 Session moderate

QJEEM18Ph69e6: JEE Main 2018 Physics - Class 11 JEEM18Ph69e6 Solution: A silver atom in a solid oscillates in simple harmonic motion…

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)

Multiple Choice Options

  • A 5.5 N/m
  • B 6.4 N/m
  • C 7.1 N/m
  • D 2.2 N/m

Step-by-Step Verified Solution

6.02 ×\times 1023 atoms of silver = 108 gm

1   atoms   of   silver   =   108×10−36.02×1023{{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}  kg

For a harmonic oscillator

f = 12π{1 \over {2\pi }} km\sqrt {{k \over m}}

Where  k = force constant

⇒\Rightarrow    \,\,\, f2 = 14π2{1 \over {4{\pi ^2}}} (km)\left( {{k \over m}} \right)

⇒\Rightarrow    \,\,\, k = mf2 ×\times 4 π\pi 2

Given,

f = 1012

m = 108×10−36.02×1023{{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}

∴   \therefore\,\,\, k = 108×10−36.02×1023{{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}} ×\times 1012 ×\times 4 π\pi 2

= 7.1 N/m

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