Question

(c) Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0= 1.68×10-21 J and R0= 3.82×10-10 m. Find the frequency of small oscillations of one Ar atom about its equilibrium position.

Answer 1

Step-by-step explanation:

Answer:

`\mathbf{f_o =1.87 * 10^(11) \ Hz}`

Step-by-step explanation:

The formula for calculating the elastic potential energy is:

`U_o = (1)/(2)kR_o^2`

By rearrangement and using (K) as the subject;

`K = (2 U_o)/(R_o^2)`

`k = (2* 1.68 * 10^(-21))/((3.82* 10^(-10))^2)`

k = 2.3 × 10⁻² N/m

Now; the formula used to calculate the frequency of the small oscillation is:

`f_o = (1)/(2 \pi)\sqrt{(k)/(m)}`

where;

m = mass of each atom

assuming

m = 1.66 × 10⁻²⁶ kg

Then:

`f_o = (1)/(2 \pi)\sqrt{(2.3 * 10^(-2) N/m)/(1.66 * 10^(-26) \ kg)}`

`\mathbf{f_o =1.87 * 10^(11) \ Hz}`