# Need help converting units, momentum of quark question

by DunWorry
Tags: converting, momentum, quark, units
 P: 40 1. The problem statement, all variables and given/known data A proton has a diameter of about 1fm. Estimate the minimum momentum in units of MeV/C of a quark confined in a proton. 2. Relevant equations 3. The attempt at a solution I used the uncertainty relation $\Delta$P$\Delta$X ~ $\frac{\hbar}{2}$ $\Delta$P ~ $\frac{\hbar}{2 x 1x10^{-15}}$ = 5.276x10$^{-20}$ kgm/s I'm unsure how to put this into MeV/C thanks
 Mentor P: 9,601 You can convert kg in J/c^2 and that to MeV/c^2. Isolate a factor of c (based on m/s) and use this to cancel one c in the denominator. Alternatively, use ℏ given in MeV/c * length.
P: 40
 Quote by mfb You can convert kg in J/c^2 and that to MeV/c^2. Isolate a factor of c (based on m/s) and use this to cancel one c in the denominator. Alternatively, use ℏ given in MeV/c * length.
Hmmmm I tried 5.276x10$^{-20}$ x (3x10$^{8}$)$^{2}$ = 4.7484x10$^{-3}$ Joules / c$^{2}$ Then to convert to eV I divided by 1.6x10$^{-19}$

And I ended up getting 2.96775x10$^{16}$eV / C. the answer should be around 200 MeV/C

Mentor
P: 9,601

## Need help converting units, momentum of quark question

Did you consider m/s -> c?
If I divide your value by 3*108, I get 108eV/c or ~100MeV/c.
 P: 40 Hmm that seems more correct, did I do the sum incorrectly?
 Mentor P: 9,601 I don't see a sum, but a missing factor of 3*10^8 would explain the difference.
PF Patron
Thanks
Emeritus
P: 38,406
 Quote by mfb I don't see a sum, but a missing factor of 3*10^8 would explain the difference.
I suspect that Dunworry is British (or learned "British" English) and they, for some strange reason, use the word "sum" to refer to any arithmetic calcluation!

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