Question

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 the frequency of small oscillations of one Ar atom about its equilibrium position.

Answer 1

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

Step-by-step explanation:

From the given information:

The elastic potential energy can be calculated by using the formula:

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

Making K 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 frequency of the small oscillation can be determined by using the formula:

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

where;

m = mass of each atom = 1.66 × 10⁻²⁶ kg

`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}`