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IPhO 1991, Теория и експеримент — Задача 3. Cooling Atoms by laser

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THEORETICAL PROBLEMS · 10 т.

Условие

To study the properties of isolated atoms with a high degree of precision they must be kept almost at rest for a length of time. A method has recently been developed to do this. It is called “laser cooling” and is illustrated by the problem below.

In a vacuum chamber a well collimated beam of Na $^{23}$ atoms (coming from the evaporation of a sample at $10^3$ K) is illuminated head-on with a high intensity laser beam (fig. 3.1). The frequency of laser is chosen so there will be resonant absorption of a photon by those atoms whose velocity is $v_0$ . When the light is absorbed, these atoms are exited to the first energy level, which has a mean value $E$ above the ground state and uncertainty of $\Gamma$ (fig. 3.2).

Vacuum chamber (rectangle) with a sodium oven labelled Na and 10³ K at the left end, a collimator, and a wavy laser beam entering from the right and travelling towards the atoms. — Figure 3.1

Energy-level diagram: ground level 0 and a hatched excited band between E−Γ/2 and E+Γ/2; an incoming photon hν and a vertical arrow of height E from the ground level into the band. — Figure 3.2

For this process, the atom’s decrease in velocity $\Delta v_1$ is given by $\Delta v_1 = v_1 - v_0$ . Light is then emitted by the atom as it returns to its ground state. The atom’s velocity changes by $\Delta v' = v'_1 - v_1$ and its direction of motion changes by an angle $\varphi$ (fig. 3.3).

An Na23 atom moving right with velocity v1; after emission it moves along v'1 deflected upward by angle φ from the dashed original direction, while the emitted photon (wavy arrow) leaves downward at angle θ. — Figure 3.3

This sequence of absorption and emission takes place many times until the velocity of the atoms has decreased by a given amount $\Delta v$ such that resonant absorption of light at frequency $v$ no longer occurs. It is then necessary to change the frequency of laser so as to maintain resonant absorption. The atoms moving at the new velocity are further slowed down until some of them have a velocity close to zero.

As first approximation we may ignore any atomic interaction processes apart from the spontaneous absorption and emission light described above.

Furthermore, we may assume the laser to be so intense that the atoms spend practically no time in the ground state.

Questions

a) Find the laser frequency needed ensure the resonant absorption of the light by those atoms whose kinetic energy of the atoms inside the region behind the collimator. Also find the reduction in the velocity of these atoms, $\Delta v_1$ , after the absorption process. [3 т.]

b) Light of the frequency calculated in question a) is absorbed by atoms which velocities lie within a range $\Delta v_0$ . Calculate this velocity range. [1,5 т.]

c) When an atom emits light, its direction of motion changes by $\varphi$ from initial direction. Calculate $\varphi$ . [1,5 т.]

d) Find the maximum possible velocity decrease $\Delta v$ for a given frequency. [1 т.]

e) What is the approximate number $N$ of absorption-emission events necessary to reduce the velocity of an atom from is initial value $v_o$ -found in question a) above- almost to zero? Assume the atom travels in a straight line. [1 т.]

f) Find the time t that the process in question e takes. Calculate the distance $\Delta S$ an atom travels in this time. [2 т.]

Data

$E = 3{,}36\cdot 10^{-19}\ \mathrm{J}$

$\Gamma = 7{,}0\cdot 10^{-27}\ \mathrm{J}$

$c = 3\cdot 10^{8}\ \mathrm{ms^{-1}}$

$m_p = 1{,}67\cdot 10^{-27}\ \mathrm{kg}$

$h = 6{,}62\cdot 10^{-34}\ \mathrm{Js}$

$k = 1{,}38\cdot 10^{-23}\ \mathrm{JK^{-1}}$

where c is speed of light, h is Planck’s constant, k is the Boltzmann constant, and mp is the mass of proton.

Отговори

Решение

Question a:

The velocity $v_o$ of the atoms whose kinetic energy is the mean of the atoms on issuing from the collimator is given is given by:

\frac{1}{2}mv_o^2 = \frac{3}{2}kT \Rightarrow v_o = \sqrt{\frac{3kT}{m}} \qquad (1)

v_o = \sqrt{\frac{3\cdot 1{,}38\cdot 10^{-23}\cdot 10^3}{23\cdot 1{,}67\cdot 10^{-27}}}\ \mathrm{m/s}

$v_o \approx 1{,}04\cdot 10^3$ m/s because:

m \approx 23\ m_p \qquad (2)

Since this velocity is much smaller than c, $v_o \ll c$ , we may disregard relativistic effects.

Light is made up of photons with energy $h\nu$ and momentum $h\nu/c$ .

In the reference system of the laboratory, the energy and momentum conservation laws applied to the absorption process imply that:

\frac{1}{2}mv_o^2 + h\nu = \frac{1}{2}mv_1^2 + E;\ mv_o - \frac{h\nu}{c} = mv_1 \Rightarrow \Delta v_1 = v_1 - v_o = \frac{-h\nu}{mc}

\frac{1}{2}m(v_1^2 - v_o^2) = h\nu - E \Rightarrow \frac{1}{2}m(v_1 + v_o)(v_1 - v_o) = h\nu - E

$h\nu/c \ll mv_o$ . Then $v_1 \approx v_o$ and this implies $mv_o\Delta v_1 = h\nu = E$ , where we assume that $v_1 + v_o \approx 2v_o$

Combining these expressions:

\nu = \frac{\frac{E}{h}}{1 + \frac{v_o}{c}} \qquad (3)

and:

\Delta v_1 = -\frac{E}{mc}\frac{1}{1 + \frac{v_o}{c}} \qquad (4)

And substituting the numerical values:

$\nu \approx 5{,}0\cdot 10^{14}$ Hz $\qquad \Delta v_1 \approx -3{,}0\cdot 10^{-2}$ m/s

If we had analyzed the problem in the reference system that moves with regard to the laboratory at a velocity $v_o$ , we would have that:

\frac{1}{2}m(v_1 - v_2)^2 + E = h\nu

Where $\nu = \dfrac{\nu'}{1 + \frac{v_o}{c}}$ is the frequency of the photons in the laboratory system. Disregarding $\Delta v_1^2$ we get the same two equations above.

The approximations are justifiable because:

-\frac{|\Delta v_1|}{v_o} \sim 10^{-4}

Then $v_1 + v_o = 2v_o - \Delta v_1 \approx 2v_o$

Question b:

For a fixed $\nu$ :

v_o = c\left(\frac{E}{h\nu} - 1\right) \qquad (5)

if E has an uncertainty $\Gamma$ , $v_o$ would have an uncertainty:

\Delta v_o = \frac{c\Gamma}{h\nu} = \frac{c\Gamma\left(1 + \frac{v_o}{c}\right)}{E} \approx \frac{c\Gamma}{E} = 6{,}25\ \mathrm{m/s} \qquad (6)

so the photons are absorbed by the atoms which velocities are in the interval $\left(v_o - \frac{\Delta v_o}{2}, v_o + \frac{\Delta v_o}{2}\right)$

Question c:

The energy and momentum conservation laws imply that:

\frac{1}{2}mv_1^2 + E = \frac{1}{2}mv'^2_1 + h\nu'

( $\nu'$ – is the frequency of emitted photon)

mv_1 = mv'_1\cos\varphi + \frac{h\nu'}{c}\cos\theta

0 = mv'_1\sin\varphi - \frac{h\nu'}{c}\sin\theta

The deviation $\varphi$ of the atom will be greatest when $\theta = \frac{\pi}{2}$ , then:

mv_1 = mv'_1\cos\varphi_m;\ \frac{h\nu'}{c} = mv'_1\sin\varphi_m \Rightarrow \tan\varphi_m = \frac{h\nu'}{mv_1c}

since $\nu' \approx \nu$ :

\tan\varphi_m \approx \frac{E}{mv_1c} \qquad (7)

\varphi_m = \mathrm{arctg}\frac{E}{mvc} \Rightarrow \varphi_m \approx 5\cdot 10^{-5}\ \mathrm{rad} \qquad (8)

Question d:

As the velocity of the atoms decreases, the frequency needed for resonant absorption increases according to:

\nu = \frac{\frac{E}{h}}{1 + \frac{v_o}{c}}

When the velocity is $v_o = \Delta v$ , absorption will still be possible in the lower part of the level if:

h\nu = \frac{E - \frac{\Gamma}{2}}{1 + \frac{v_o - \Delta v}{c}} = \frac{E}{1 + \frac{v_o}{c}} \Rightarrow \Delta v = \frac{c\Gamma}{2E}\left(1 + \frac{v_o}{c}\right) \qquad (9)

$\Delta v = 3{,}12$ m/s

Question e:

If each absorption-emission event varies the velocity as $\Delta v_1 \approx \frac{E}{mc}$ , decreasing velocity from $v_o$ to almost zero would require N events, where:

N = \frac{v_o}{|\Delta v_1|} \approx \frac{mcv_o}{E} \Rightarrow N \approx 3{,}56\cdot 10^4

Question f:

If absorption is instantaneous, the elapsed time is determined by the spontaneous emission. The atom remains in the excited state for a certain time, $\tau = \frac{h}{\Gamma}$ , then:

\Delta t = N\tau = \frac{Nh}{\Gamma} = \frac{mchv_o}{\Gamma E} \Rightarrow \Delta t \approx 3{,}37\cdot 10^{-9}\ \mathrm{s}

The distance covered in that time is $\Delta S = v_o\Delta t/2$ . Assuming that the motion is uniformly slowed down:

\Delta S = \frac{1}{2}mchv_o^2\Gamma E \Rightarrow \Delta S \approx 1{,}75\ \mathrm{m}

Marking Code

a)Finding	$v_o$	1 pt	Total 3 pt
“	$\nu$	1 pt
“	$\Delta v_1$	1 pt
b) “	$\Delta v_o$	1,5 pt	Total 1,5 pt
c) “	$\varphi_m$	1,5 pt	Total 1,5 pt
d) “	$\Delta v$	1 pt	Total 1 pt
e) “	N	1 pt	Total 1 pt
f) “	$\Delta t$	1 pt	Total 2 pt
“	$\Delta S$	1 pt
			Overall total 10 pts

We suggest in all cases: 0,75 for the formula; 0,25 for the numeral operations.

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