Momentum: laser beam and photons

[KL90]

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=\Deltap/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=\Deltap/(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

 

[Redbelly98]

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=\Deltap/(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

a & b look good

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

Watch the +/- signs here. Are the atoms accelerating or decelerating?

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

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.

 

[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!