Force and acceleration on sodium atom if laser bundle shines on atom

[jennyjones]

Homework Statement

I want to calculate the average force and acceleration on an sodium atom, if a laser bundle shines on it
which has I>>>> I(sat). I(sat) is the saturation intensity of the 589nm 3s-3p transition in sodium.
3p lives 16 ns

Homework Equations

c = 3 * 10 ^ 8
m sodium = 3.8 * 10^-26 kg
planck = 6.6* 10 ^-34 Js
tua = 16 ns

The Attempt at a Solution

F = - h(bar)/(2*λ*(tau)) ?

not sure which formula to use


thanx,

jenny

 

[haruspex]

Not an area I know anything about, but the formula you quoted looks almost right to me.
The h/λ part gives you the momentum of each photon. 1/tau, where tau is the remanence time in the excited state, gives you the max rate of excitation. The reradiation is in a random direction, so you should only count the momentum of the absorbed photons. Force equals rate of change of momentum, so momentum per excitation multiplied by rate of excitation events.

What I don't understand is the constant factor of 1/4π. Maybe someone who knows this area will comment.

 

[BvU]

Hello Jenny,

Could you reveal a bit more of the context of your post ?
I am intrigued by the I >> Isat. I take it the laser is tuned to the 589 nm, but the spectral width isn't indicated: is it sharper than the D-line or much broader?

When you irradiate Na atoms, you get resonance fluorescence (in all directions) and you also get stimulated emission (in the same direction as the incoming photon). When the intensity in increased, the latter shortens the residence time in the upper level (power broadening, page 90). Hope this link works, otherwise: Laser Spectroscopy: Vol. 1: Basic Principles
By Wolfgang Demtröder

I seem to recall that with a broadly tuned laser you get line broadening and with a very sharply tune laser you can get line splitting, but I'm not that sure - long time ago.

However, momentum conservation goes: Cohen-Tannoudji and Dalibard -- real experts in this field, the former a Nobel laureate -- mention a recoil of the atom of around 3 cm/s on p 16 . ( $p = \hbar{\bf k} = h/\lambda$. No $1/(4\pi)$. The $1/\tau$ comes in as the number of photons that can be absorbed per second, as Haruspex already wrote).

CT & D work out radiation pressure on p 17; more or less what you are asking.

 

[jennyjones]

thanks so much for the help!

i think i know how to solve the problem now!

F = dp/dt = - h/λτ
F = - 6.6* 10 ^-34/(589 * 10^-9 * 16 * 10^-9) = -7 * 10^-20

a = F/m = -7 *10^-20/ 3.8 * 10^-26 kg = -1.8 *10^6