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Need help converting units, momentum of quark question

by DunWorry
Tags: converting, momentum, quark, units
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DunWorry
#1
Jan2-13, 06:18 AM
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 [itex]\Delta[/itex]P[itex]\Delta[/itex]X ~ [itex]\frac{\hbar}{2}[/itex]

[itex]\Delta[/itex]P ~ [itex]\frac{\hbar}{2 x 1x10^{-15}}[/itex] = 5.276x10[itex]^{-20}[/itex] kgm/s

I'm unsure how to put this into MeV/C

thanks
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mfb
#2
Jan2-13, 09:00 AM
Mentor
P: 11,570
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.
DunWorry
#3
Jan2-13, 11:27 AM
P: 40
Quote Quote by mfb View Post
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[itex]^{-20}[/itex] x (3x10[itex]^{8}[/itex])[itex]^{2}[/itex] = 4.7484x10[itex]^{-3}[/itex] Joules / c[itex]^{2}[/itex] Then to convert to eV I divided by 1.6x10[itex]^{-19}[/itex]

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

mfb
#4
Jan2-13, 12:02 PM
Mentor
P: 11,570
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.
DunWorry
#5
Jan3-13, 05:17 AM
P: 40
Hmm that seems more correct, did I do the sum incorrectly?
mfb
#6
Jan3-13, 11:05 AM
Mentor
P: 11,570
I don't see a sum, but a missing factor of 3*10^8 would explain the difference.
HallsofIvy
#7
Jan3-13, 12:52 PM
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Thanks
PF Gold
P: 39,285
Quote Quote by mfb View Post
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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