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

Answer:

$\mathbf{f_o =1.87 \times 10^{11} \ Hz}$

Explanation:

From the given information:

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

$U_o = \dfrac{1}{2}kR_o^2$

Making K the subject;

$K = \dfrac{2 U_o}{R_o^2}$

$k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}$

k = 2.3 × 10⁻² N/m

Now; the frequency of the small oscillation can be determined by using the formula:

$f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}$

where;

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

$f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}$

$\mathbf{f_o =1.87 \times 10^{11} \ Hz}$