# Calculate the recoil momentum and kinetic energy of the atom

1. Sep 26, 2015

### tauristar

1. The problem statement, all variables and given/known data
Suppose an atom of iron at rest emits an X-ray photon of energy 6.4 keV. Calculate the “recoil” momentum and kinetic energy of the atom. (Hint: Do you expect to need classical or relativistic kinetic energy for the atom? Is the kinetic energy likely to be much smaller than the atom’s rest energy?)

2. Relevant equations
Where E is the energy of the photon (in this case 6.4 KeV), Ei is the initial energy of the atom (I believe it is zero), Ef is the final energy of the atom and K is the "recoil" kinetic energy of the atom.
E = (Ei − Ef) − K

3. The attempt at a solution
Since the atom starts at rest Ei =0 and so I get
6.4 Kev=(0-Ef)-K
6.4 Kev = -Ef-K

Not sure where to go from here

2. Sep 26, 2015

### PietKuip

3. Sep 26, 2015

### tauristar

I tried going that route using P=E/c so P=6.4 kev/c now what...?

4. Sep 26, 2015

### PietKuip

So you know the momentum of the atom. Why not try to calculate its velocity?

5. Sep 26, 2015

### tauristar

Not sure what to use. pc/E=v/c --> v=pc^2/E so then v= (6.4 kev/c)c^2/(6.4 kev) --> v=c I feel like i'm going around in circles I have no clue what to do and I've been working on this for two days I just don't get this stuff. I feel like I'm missing a big piece of the puzzle and I just don't know where to find it.

6. Sep 26, 2015

### PietKuip

I repeat my hint: read the question. Consider classical mechanics.

7. Sep 26, 2015

### tauristar

classical mechanics Ki+Ui=Kf-Uf so (1/2) m vi^2 + mgyi = (1/2) m vf^2 + mgyf
initial Ki=0 the Ui and Uf are zero so that doesn't leave me with anything useful

8. Sep 26, 2015

### PietKuip

Ok, go back one step: you know the momentum of the atom. Why not calculate its velocity? (That is what I wrote in message #4.)

9. Sep 26, 2015

### tauristar

I got v=c

10. Sep 26, 2015

### Staff: Mentor

You are overthinking this. Forget energy. What is the momentum of a particle in classical mechanics?

11. Sep 26, 2015

### tauristar

p=mv
if I plug in what i found for p then I get v= 6.88e28 eV/kg

12. Sep 26, 2015

### tauristar

6.4 keV = 9.3E-26 kg v
v= (6.4 keV)/(9.3E-26 kg) = 6.88E28 ev/kg

13. Sep 26, 2015

### Staff: Mentor

6.4 keV is not the momentum.

14. Sep 26, 2015

### tauristar

I thought it was p=E/c and E (energy of the photon) is 6.4 keV which would give me p=6.4 keV/c

15. Sep 26, 2015

### PietKuip

So convert to SI units. Or imperial, or whatever you prefer.

16. Sep 26, 2015

### tauristar

(6400 eV/c)(c/1eV)(5.34e-28 kg m/s)= 3.42e-24 kg m /s

17. Sep 26, 2015

### PietKuip

18. Sep 26, 2015

### tauristar

19. Sep 26, 2015

### Staff: Mentor

You know p=mv.

You know p=3.42e-24 kg m /s
You know m.

What is v?

20. Sep 26, 2015

### tauristar

so the velocity is (3.42e-24 kg m /s)(1/9.27e-27)= 36.7 m/s

I know i close but not sure how to finish it off

oh i just need to convert!

Last edited by a moderator: Sep 26, 2015