Velocity of Lithium Atom Powered by 1mW Laser Beam

[stunner5000pt]

A lithium atom (mass doesn't matter) is being pushed from rest by a laser beam whose power is 1mW. Find the velocity of this lithium atom as a function of time
So here goes what i have done
v(t) = \frac{1}{2} at^2

now F = ma = \frac{\Delta p}{\Delta t}
so then a = \frac{\Delta p}{\Delta t m}
and then subbing into the first equation

v(t) = \frac{1}{2} \frac{\Delta p}{m \Delta t} t^2

v(t) = \frac{\Delta p t}{2m}
The radiation pressure becasue of the lithium absorbing (right?) is given by
\Delta p = \frac{\Delta U}{c}

so then v(t) = \frac{\Delta U}{c} \frac{t}{2m}
and \Delta U = P t = 0.001t P = 1mW
v(t) = \frac{0.001t^2}{2mc}

am i right in assuming that the lithium atom will absorb jump to higher level and then go back down? Because i do know that the lifetime of the lithium atom in the ecited state is 27ns

If this is not clear to you you might want to have a look at the b part of the question posted in thsi thread
https://www.physicsforums.com/showthread.php?t=64390

thanks in advance for your help!

 

[James R]

This is only a matter of momentum conservation between the incident photons and the atoms. You don't need to worry about what is going on inside the atom, in terms of energy level transitions etc.

 

[stunner5000pt]

This is only a matter of momentum conservation between the incident photons and the atoms. You don't need to worry about what is going on inside the atom, in terms of energy level transitions etc.

So the photons are absorbed by the Lithium atoms, and everything up there is OK?

 

[stunner5000pt]

can someone tell me if this is right so far or have I made a mistake along the way??

Seems my threads get ignored a lot thee days...

 

[vincentchan]

the whole question is wrong... you can't calculate how fast the atom runs, since you don't know what percentage of 1mW the atom absorbs... the atom can absorbs 10%of light or 0.000000001% or 0%.

second, because of conservation of momentum, the atom cannot absorbs 100% of light, even if you ignore quantum mechanic...

can you tell me what level of physics are you in.. (high school, college lower division, or upper division) I just don't want to over complicate your problem...

if you are a high school kid, or taking a lower division class in college, just use the conservation of energy, ignore the momentum part...
if you are doing a upper division physics course, you should know about compton scattering and stuff like that... the atom will not go stright and it depend on the photon's wave length and stuff... make this problem extremely complicated

 

[stunner5000pt]

I'm currently in the second year of physics ... err... no, no compton scattering or photoelectric effect needs to be included in this case (although i have studied that and i know how that would complicate matters)

Take a look at the other thread which is linked in post #1 part (b) asks for an expression of the doppler shift and since the doppler shift in the Y direction would be dependant on the atom's velocity in the Y direction that the expression which is needed

ALso my prof tells us to assume tht the atom will absorb a photon, then return to ground, absorb another, return to ground and so on all the while getting deflected and gaining velocity in the Y hat direction.

Alll i have to do is assume that is this is like a force pushing an object ointo a parabolic(?) tracjectory. Just like James R said.

So give this info can you tell me if my first post's workout is good or are there problem,s??

 

[vincentchan]

ok, than I'll use the energy point of view to do this problem
the gain of energy is dE/dt = 10mW
the total energy is zero (assume it started at rest)

since dE/dt =constant, E=10t, the kinetic energy of the atom is E=1/2mv^2, therefore,
1/2mv^2=10t
v= sqrt(20t/m)... do the proper unit convertion

this method seems like too simple for a 2nd year college kids, but, since you didn't tell me the wavelength of the photon, this is the best one can do for this problem, BTW, what course/level are you taking?

 

[stunner5000pt]

Don't you have to include radiation pressure in this calculation??

The course is caled Optics and Spectra Physics 2060 and we use a text Halliday Resnick Krane Volume 2 (pretty similar to HRW if you've come across that)

i have to sleep now (EST is pretty late actaulyl) but i hope you'll see what level i am on and help me at that level.

 

[vincentchan]

Don't you have to include radiation pressure in this calculation??

if you assume the atom absorbs all the 10mW energy, my method is the most straight forward... why do you want to do it by radiation pressure?
by the way, you make a mistake here:
v(t) = \frac{1}{2} \frac{\Delta p}{m \Delta t} t^2 = \frac{\Delta p t}{2m}
the delta t at the bottom can't cancel the t at the top, imagine \frac{\Delta p}{m \Delta t} is dp/dt, and you will see why they can't equal..
sure you can do it by radiation presure, but you don't have the energy intensity (and the cross section area of atom)...
you should know the best what your prof expect you do answer.. I have no idea your textbook and stuff... but i think this is a basic physics class, right? how many physics class you have taken in college?

 

[vincentchan]

one more small thing... if you do it by radiatioin pressure, you will have you momentum conserve, but the energy will not be conserve... the answer will not be same as mine... That's okay, because either way will not yeild the correct answer anyway, as i said, the exact solution is very very complicate...
summery:
consevation of energy: do it in my way
consevation of momentum: do it by radiation pressure
both are wrong anyway..

 

[stunner5000pt]

one more small thing... if you do it by radiatioin pressure, you will have you momentum conserve, but the energy will not be conserve... the answer will not be same as mine... That's okay, because either way will not yeild the correct answer anyway, as i said, the exact solution is very very complicate...
summery:
consevation of energy: do it in my way
consevation of momentum: do it by radiation pressure
both are wrong anyway..

well in any case i do need a momentum because in the end i need an expression for the velocity of the lithium atom

i do understand that this would be wrong otherwise but since I'm not at that level (yet) ill have to stick with this

 

[Doc Al]

ALso my prof tells us to assume tht the atom will absorb a photon, then return to ground, absorb another, return to ground and so on all the while getting deflected and gaining velocity in the Y hat direction.

That's what I would assume as well. (Assume a resonant laser beam.) When the atom absorbs a photon, it will gain the photon's momentum (h \nu / c). You can figure out the change in the atom's velocity with each absorption. (Assume that when the atom emits the photon, it does so randomly. So the average change in momentum due to spontaneous decay is zero.) How many collisions does it undergo? Use the lifetime of the excited state: assume the atom will immediately absorb a new photon after each spontaneous emission. So you can find the rate of change of the atom's speed.