This question is about laser cooling, which is a quick and efficient way of cooling ions down to ver — Physical Chemistry — Kinetics Chemistry Question
Laser Cooling
This question is about laser cooling, which is a quick and efficient way of cooling ions down to very cold temperatures. The mean kinetic energy of a molecule is related to its temperature by B=E k T3 2 , where kB is the Boltzmann constant.
Calcium atoms leak out of an oven at 600 °C. Calculate the mean kinetic energy of the calcium atoms and hence the rms momentum and rms speed of a 40Ca atom, whose relative isotopic mass is 39.96.
Model Answer
E = 3/2 kT = 1.81×10−20 J
p = \sqrt{2 mE} = 4.90×10−23 kg m s−1
v = p / m = 738 m s−1
The atoms drift into an ion trap where they are photoionised and trapped. While in this trap they are bombarded with laser light of wavelength 396.96 nm. Calculate the frequency, energy and momentum of a photon with this wavelength.
Model Answer
ν = c / λ = 7.5522×1014 Hz
E = h v = 5.0042×10−19 J
p = h / λ = 1.6692×10−27 kg m s−1
The ions go through an optical cycle repeatedly. Ions absorb a photon from the laser when they are moving in the opposite direction to the light (this is achieved using the Doppler Effect) and then re-emit a photon in a random direction. The net effect of this procedure is to slow the ion down slightly. Calculate the change in mean momentum and speed at each cycle and the number of photons that would need to be absorbed to bring the ion approximately to rest. (In practice this process was found to reduce the temperature to about 0.5 mK.)
Model Answer
At each cycle the mean momentum of the ion is reduced by the momentum of the photon it has absorbed. The re-emission is isotropic and has no effect on the mean momentum.
∆patom = −1.6692×10−27 kg m s−1
∆vatom = patom / m = −2.5156×10−2 m s−1
To slow the ion to rest therefore takes approximately 2.93×104 photons.
Write down the ground electronic configuration of the Ca+ ion, and calculate the orbital and spin angular momentum of the unpaired electron.
Model Answer
Ca+: 1s2 2s2 2p6 3s2 3p6 4s1.
l = 0, hence \sqrt{l(l+1)}\hbar = 0
s = ½, hence \sqrt{s(s+1)}\hbar = \sqrt{3}/2 \hbar
In the excited configuration involved in the laser cooling transition the unpaired electron has been excited into the lowest available p orbital. Calculate the orbital and spin angular momentum of the unpaired electron.
Model Answer
For an electron in a p orbital, l = 1, hence \sqrt{l(l+1)}\hbar = \sqrt{2}\hbar
s = ½, hence \sqrt{s(s+1)}\hbar = \sqrt{3}/2 \hbar
In this excited state the electron experiences a magnetic field because of its own orbital motion around the charged nucleus. The spin of the electron can line up either parallel or antiparallel to this field, and the two states have slightly different energies. The resultant quantum number, j, for the total electronic angular momentum takes values from l - s to l + s in integer steps. Calculate the possible values of j.
Model Answer
j = 1/2 (antiparallel)
tj = 3/2 (parallel)
The laser cooling transition is to the lower of these two levels, the transition from the ground state to the higher level has a wavelength 393.48 nm. Calculate the energy difference between the two levels resulting from the excited configuration.
Model Answer
The first transition was calculated in 16.2: E = hv = 5.0042×10−19 J
The second transition is E = hc / λ = 5.0484×10−19 J
The energy difference is ∆E = 4.43×10−21 J