Question

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 Avagadro number = 6.02 × 10²³ gm mole^-1)

विकल्प

• 6.4 N/m

• 7.1 N/m

• 2.2 N/m

• 5.5 N/m

Answer 1

7.1 N/m

Explanation:

As we know, frequency in SHM.

f = `1/(2pi)sqrt(k/m) = 10^12`

where m = mass of one atom

Mass of one atom of silver = `108/((6.02 xx 10^23)) xx 10^-3` kg

`1/(2pi)sqrt(k/(108 xx 10^-3) xx 6.02 xx 10^23) = 10^12`

Solving we get, spring constant,

K = 7.1 N/m