A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of `10^(12)//sec`. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number `=6.02xx10^(23)" gm "mole^(-1))

Options

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

Answer

D

Explanation

Angular frequency can be written as : `omega = 2piv`
` = 2pi xx 10^(12)`
`omega = sqrt((k)/(m)) rArr k = momega^(2)`
`rArr k = ((108)/(6.02 xx 10^(23)) xx 10^(-3)) xx (2pi xx 10^(12))^(2)`
= 7.1 N/m.