Minimum momentum of a quark confined in a proton

[DunWorry]

Homework Statement

A proton has a diameter of about 1fm. Estimate the minimum momentum in units of MeV/C of a quark confined in a proton.

Homework Equations

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

 

[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.

 

[DunWorry]

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

 

[mfb]

Did you consider m/s -> c?
If I divide your value by 3*10⁸, I get 10⁸eV/c or ~100MeV/c.

 

[DunWorry]

Hmm that seems more correct, did I do the sum incorrectly?

 

[mfb]

I don't see a sum, but a missing factor of 3*10^8 would explain the difference.

 

[HallsofIvy]

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!