Q133 / 180 · COMEDK 2026 (9 May Shift 2) · Physics
Force constant of interatomic bond, in a certain element, is $7.1\ Nm^{-1}$. If the atom oscillates in SHM in a certain direction, what is its frequency?
Given: Mole weight of the given element is 108g and Avagadro's number $= 6.023 \times 10^{23}\ g\ mol^{-1}$
Given: Mole weight of the given element is 108g and Avagadro's number $= 6.023 \times 10^{23}\ g\ mol^{-1}$
- A$6.667 \times 10^{12}\ s^{-1}$
- B$0.005 \times 10^{12}\ s^{-1}$
- C$3.45 \times 10^{22}\ s^{-1}$
- Correct answerD$1 \times 10^{12}\ s^{-1}$
Solution
Mass of a single atom is given by $m = \dfrac{M}{N_A}$
Substituting the given values:
$m = \dfrac{108 \times 10^{-3}}{6.023 \times 10^{23}} \approx 1.793 \times 10^{-25}\ kg$
The frequency of oscillation of the atom in simple harmonic motion is given by:
$\nu = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}$
Substituting $k = 7.1\ N\ m^{-1}$ and $m = 1.793 \times 10^{-25}\ kg$:
$\nu = \dfrac{1}{2\pi} \sqrt{\dfrac{7.1}{1.793 \times 10^{-25}}}$
$\nu = \dfrac{1}{2\pi} \sqrt{39.6 \times 10^{24}}$
$\nu = \dfrac{6.29 \times 10^{12}}{2\pi} \approx 1 \times 10^{12}\ s^{-1}$
Answer: $1 \times 10^{12}\ s^{-1}$
Substituting the given values:
$m = \dfrac{108 \times 10^{-3}}{6.023 \times 10^{23}} \approx 1.793 \times 10^{-25}\ kg$
The frequency of oscillation of the atom in simple harmonic motion is given by:
$\nu = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}$
Substituting $k = 7.1\ N\ m^{-1}$ and $m = 1.793 \times 10^{-25}\ kg$:
$\nu = \dfrac{1}{2\pi} \sqrt{\dfrac{7.1}{1.793 \times 10^{-25}}}$
$\nu = \dfrac{1}{2\pi} \sqrt{39.6 \times 10^{24}}$
$\nu = \dfrac{6.29 \times 10^{12}}{2\pi} \approx 1 \times 10^{12}\ s^{-1}$
Answer: $1 \times 10^{12}\ s^{-1}$
Hence, the correct answer is option: 'D'.