Question ffc16

Mar 24, 2017

For a simple harmonic oscillator in one dimension, we have the following Hamiltonian:

hatH = -ℏ^2/(2mu)(del^2)/(delx^2) + 1/2 kx^2 = -ℏ^2/(2mu)(del^2)/(delx^2) + 1/2 muomega^2x^2

and the following energy (note that ℏomega = h/(2pi)omega = hnu):

E_upsilon = ℏomega(upsilon+1/2) = hnu(upsilon+1/2)

where $\upsilon$ is the vibrational quantum number, $\nu$ is the fundamental frequency, and ℏ = h/(2pi) is the reduced Planck's constant.

The zero-point energy is the energy at the bottom of the potential energy well, specifically at

E_0 = hnu(0+1/2) = 1/2hnu = 1/2ℏomega. You were given that $k = \text{329 N/m}$ or ${\text{kg/s}}^{2}$, so recall that the angular frequency is:

$\omega = \sqrt{\frac{k}{\mu}}$

where $\mu = \frac{{m}_{1} {m}_{2}}{{m}_{1} + {m}_{2}} = \frac{m}{2}$ for ${\text{Cl}}_{2}$, $m$ is the mass in $\text{kg}$ (you must divide by $6.022 \times {10}^{23}$ to get its molar mass from $\text{kg/mol}$ to $\text{kg}$ if you use that), and $\omega$ is the angular frequency in ${\text{s}}^{- 1}$.

Thus:

color(blue)(E_0) = 1/2ℏomega = 1/2*1/(2pi)*hsqrt(k/mu)

= 1/(4pi)(6.626 xx 10^(-34) "J"cdot"s")sqrt("329 kg/s"^2/("0.03496885 kg/mol"/2 xx "1 mol"/(6.022 xx 10^(23) "molecules")))#

$= \textcolor{b l u e}{5.613 \times {10}^{- 21}}$ $\textcolor{b l u e}{\text{J}}$

or

$\textcolor{b l u e}{\text{0.03504 eV}}$

or

$\textcolor{b l u e}{{\text{282.59 cm}}^{- 1}}$

A reference value from here is ${\text{279.8756 cm}}^{- 1}$ using only the simple harmonic oscillator approximation in the equation on pg. 392 in Table 2.