Recent content by 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!

I can't seem to find that information out

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

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

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

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

I got v=c

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

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...

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

Homework Statement 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...

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