# LASER Operating At A Certain Frequency

Q. A LASER is operating at a frequency f = 6.1*10^14 Hz.
A)Calculate the momentum of the photon emitted by the laser.
B)A helium atom flies towards the laser at a speed of v = 3.5 m/s.During one laser pulse the gold atom absorbs on average 5 photons.Find the speed the helium atom after one laser pulse.Treat the interaction between photons and the helium atom as an inelastic collision.Use m=4.0u for mass of the helium atom.

ATTEMPT: wavelength λ= c/f = 3×10^8/6.1×10^14
≈492 nm.
⇒Total momenta of 5 photons in 1 pulse =5×h/λ =5×6.63×10^-34/4.92×10^-7
= 5×1.35×10^-27
= 6.75×10^-27 kg.m/s
Again, like in the previous question i posted, i can't figure out how to find the speed of the helium atom.
(Here i can't divide the momenta of 5 photons by the mass of the He atom)
BTW the mass of the He atom will probably be :-
M(He)=4×1.6605×10^-27
= 6.64×10^-27 kg
That's as far i can get.I guess what i must do is deal with the conservation of linear momentum but i am not sure.Any suggestions?I really must complete these questions in time.

BvU
Homework Helper
It's all you've got and it holds. No big deal!

But the one who gave these questions isn't going to hold! Haven't you got a soluti-----I mean hint?

BvU
Homework Helper
Treat it as a fully ineastic collision: add up the momentum of the photons and the He, to get the momentum after the absorption. Then divide by the He mass to get its velocity.

You mean pphotons/MHe=6.75×10-27/6.64×10-27=1.01 m/s.But the answer is 2.5 m/s

Found it! it should be done like this:
pHe before collision=Mu=6.64×10-27 × 3.5
= 2.324×10-26 kg.m/s

So,apparently, Mv=pHe-pphotons
=6.75×10-27-2.324×10-26
=1.65×10-26kg.m/s
and v = Mv/M=1.65×10-26/6.64×10-27
= 2.5 m/s
YAY

BvU