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?
2.2N/m
5.5N/m
7.1N/m
6.4N.m
Solution:
Time period of SHM is given by: T = 2π√(m/k) Frequency = 1/(2π√(m/k)) = 10¹² where m = mass of one atom = 108/(6.02 × 10²³ × 10⁻³ ) kg 1/(2π√(m/k)) = 10¹² k = 7.1 N/m