What Are the Quantum Limits of Cooling Sodium Atoms in an Atom Trap?

  • Thread starter Thread starter natalie.*
  • Start date Start date
  • Tags Tags
    Atom Quantum
AI Thread Summary
Physicists have successfully cooled sodium atoms in an atom trap to approximately 1 nK, raising questions about the quantum limits of such cooling. The discussion involves estimating the smallest range of speeds for sodium atoms confined in a 1-mm box, utilizing the Heisenberg Uncertainty Principle. The calculated range of speeds indicates that even when atoms are cooled, they will still exhibit a distribution of speeds. The root-mean-square speed (v_rms) is estimated to be half of the maximum speed derived from the uncertainty principle, which is then used to calculate the corresponding temperature. The analysis highlights the interplay between quantum mechanics and thermodynamic limits in atomic cooling.
natalie.*
Messages
6
Reaction score
0

Homework Statement


Physicists use laser beams to create an "atom trap" in which atoms are confined within a spherical region of space with a diameter of about 1 mm. The scientists have been able to cool the atoms in an atom trap to a temperature of approximately 1 nK, which is extremely close to absolute zero, but it would be interesting to know if this temeprature is close to any limit set by quantum physics. We can explore this issue with a 1-D model of a sodium atom in a 1-mm-long box.

a) Estimate the smallest range of speeds you might find for a sodium atom in this box.

b) Even if we do our best to bring a group of sodium atoms to rest, individual atoms will have speeds within the range you found in part a. Because there's a distribution of speeds, suppose we estimate that the root-mean-square spped v_rms of the atoms in the trap is half the value you found in part a. Use this v_rms to estimate the temperature of the atoms when they've been cooled to the limit set by the uncertainty principle.

HINT: use the equation:

v_rms = Sqrt[(3(k_B)T)/m], where k_B = 1.38 x 10^-23 J/K is Boltzmann's constant.

Homework Equations


For a particle in a box, the allowed energies are:
E_{n}=\frac{1}{2}mv_{n}^{2}=n^2\frac{h^2}{8mL^2}
which means that the allowed velocities are:
v_{n}=n\frac{h}{2L}

Heisenberg Uncertainty Principle
\Delta x \Delta p_{x}\geq\frac{h}{2}

The Attempt at a Solution


Using the uncertainty principle I found that \Delta v=\frac{h}{2 \Delta x m} where I think that \Delta x=L, the length of the box. I don't know what this range is around though? The answer for a) has to be entered as the lower speed and then the higher speed.
 
Physics news on Phys.org
For anyone else googling this problem or anything:

I ended up figuring it out. The delta v you can find with the Heisenberg uncertainty principle is around 0 m/s, so:
-0.5*(delta v) <= v <= 0.5*(delta v)

Then your v_rms=0.25*(delta v) and you sub that into the equation given to find T (in Kelvin). Remember that the mass of a sodium atom is the mass in amu=g/mol divided by Avogadro's number divided by 1000 (SI units is kg not g).
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top