# Momentum: laser beam and photons

1. Oct 31, 2009

### KL90

1. The problem statement, all variables and given/known data
A rubidium atom travelling at the speed of sound absorbs photons from an oncoming laser beam Each photon can be viewed as a tiny ping pong ball having momentum 7.79 x 10-28 Nt/sec. The atom absorbs a photon which is then reradiated in any direction. Hence, on average each photon absorption /reemission reduced the atom's momentum by a photon momentum. (Use 87Rb and speed of sound = 330 m/sec)
a) How many photons must an atom absorb and reradiate in order to be stopped?

b) What is the deceleration experienced my the atom if it can absorb and reradiate a photon every 2.5 x 10-8 sec?

c) What distance does it take to stop an atom?

d) How many photons/sec are required to stop a beam of 109 atoms/sec?

2. Relevant equations
F=ma=$$\Delta$$p/time
x = x0 + u0t + 1/2at2

3. The attempt at a solution

a) # photons = atom momentum / photon momentum
= (87 x 1.67x10-24/1000) (330m/sec)/ 7.79 x 10-28
= 61545 photons

b) a= F/atom mass=$$\Delta$$p/(time x Rb mass)=photon momentum/(time x Rb mass)
= 7.79 x 10-28 / [2.5 x 10-8 x (87x1.67x10-24/1000)]
= 2.1 x 105 m/sec2

c) x = x0 + u0t + 1/2at2
t = #photons absorbes x time to absorb 1 photon
= 61545 x 2.5 x 10-8
= 1.54 x 10-3 sec

x = 330t + 1/2 (2.1 x 105) t2
= 0.76m

d) Unsure --> attmept --> 109 / 61545
= 16248 photons/sec

2. Nov 1, 2009

### Redbelly98

Staff Emeritus
a & b look good
Watch the +/- signs here. Are the atoms accelerating or decelerating?

Hmmm, no. Think in terms of units.

You'd need 61545 photons per Rb atom to stop them all.
The beam has 109 Rb atoms per second.

3. Nov 1, 2009

### KL90

c) x = 330t - 1/2 (2.1 x 105) t2
since it is decelerating, the acceleration would be negative.

d) 109 Rbatoms/sec x 61545 photons/Rb atom = xphotons/sec

Thanks a lot!