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A silver atom in a solid oscillates in s...

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of `10^(12)s^(-1)`. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of Silver = 108 and Avogadro number = `6.02xx10^(23)g*mol^(-1)`)

A

2.2 N/m

B

5.5 N/m

C

6.4 N/m

D

7.1 N/m

Text Solution

Verified by Experts

The correct Answer is:
D
Two atoms with their bond executing SHM be compared as vibration of a spring.
Now, `T=2pisqrt(m/k)` [k is spring constant]
Thus, `f=1/T`
or, `f=1/(2pi)sqrt(k/m)or,10^(12)=1/(2pi)sqrt(k/m)`
or, `k=4pi^(2)mxx10^(24)=(4xxpi^(2)xx108xx10^(-3))/(6.02xx10^(23))xx10^(24)`
= `7.08N//m~~7.1N//m`
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