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KL90
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Homework Statement
A rubidium atom traveling 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?
Homework Equations
F=ma=[tex]\Delta[/tex]p/time
x = x0 + u0t + 1/2at2
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=[tex]\Delta[/tex]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