Question

The potential energy for the interaction between two gas molecules can be represented by the function
$\varphi \left(r\right)=-[2{\left(\frac{r_o}{r}\right)}^6-{\left(\frac{r^o}{r}\right)}^{12}]$
where ${\varphi }_o$ and r${}_o$ are constants. Then (m is mass of each molecule)

• A. the force constant is ${\varphi }_o$
• B. the force constant is ${\varphi }_o$ (72 r${{}_0}^{-2}$)
• C. the angular frequency for small oscillations is $\sqrt{\frac{144{\varphi }_o}{m{r}_o^2}}$
• D. the angular frequency for small oscillations is $\sqrt{\frac{{\varphi }_o}{m{r}_o^2}}$

Answer 1

Answer: (B), (C)

the force constant is ${\varphi }_o$ (72 r${{}_0}^{-2}$)
the angular frequency for small oscillations is $\sqrt{\frac{144{\varphi }_o}{m{r}_o^2}}$

For equilibrium position, $\frac{\partial \varphi }{\partial r}=0$
$\Rightarrow -{\varphi }_0[-2r_0^6{r}^{-7}\times 6+12r_0^{12}{r}^{-13}]=0$
$\Rightarrow$ Force constant $K=\frac{{\partial }^2\varphi }{\partial r^2}$
$=-{\varphi }_0[12r_0^6\times 7\times r^{-8}-12r_0^{12}\times 13\times r^{-14}]$
$={\varphi }_0\left(72r_0^{-2}\right)$
where r = $r_o$ for equilibrium
$\omega =\sqrt{\frac{K}{\mu }}$
$\mu =$ reduced mass $=\frac{m\times m}{m+m}$
$\Rightarrow \mu =\frac{m}{2}$
$\omega =\sqrt{\frac{72{\varphi }_{o}r{{}_0}^{-2}}{\left(\frac{m}{2}\right)}}$
$=\sqrt{\frac{144{\varphi }_o}{m{r}_0^2}}$