Question #ffc16

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 = "329 N/m"# or #"kg/s"^2#, so recall that the angular frequency is:

#omega = sqrt(k/mu)#

where #mu = (m_1m_2)/(m_1+m_2) = m/2# for #"Cl"_2#, #m# is the mass in #"kg"# (you must divide by #6.022 xx 10^(23)# to get its molar mass from #"kg/mol"# to #"kg"# if you use that), and #omega# is the angular frequency in #"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")))#

#= color(blue)(5.613 xx 10^(-21))# #color(blue)("J")#

or

#color(blue)("0.03504 eV")#

or

#color(blue)("282.59 cm"^(-1))#

A reference value from here is #"279.8756 cm"^(-1)# using only the simple harmonic oscillator approximation in the equation on pg. 392 in Table 2.