A proton has a radius of about 1 fm (10^−15 m). Use dimensional analysis with the speed of light c=3×10^8 m/s, the reduced Plank constant ℏ=1.1×10^−34 Js to estimate the kinetic energy of quarks inside a proton. Recall the conversion between Joules and electron volts, 1 eV=1.6×10^−19 J, and that MeV = mega electron volts, i.e. 1 MeV=10^6 eV.

Note that the point of this problem, and similar ones, is not just to do some conversion of numbers. The point is that once you know one scale (in this case a length scale---the size of the proton), it encodes a lot of information and characterizes the typical scales in the system (in this case the length scale also sets the energy scale).

so we have :

$$R=1fm(10^{-15}m)$$

$$c = 3*10^8m/s$$

$$h{bar}=1.1*10^{-34}J*s$$

$$V=1eV=1.6*10^{-19}J$$

$$1MeV=10^6eV$$

so we need a answer in MeV:

first we get a value in J*m:

$$a=c*h{bar}=3*10^8m/s*1.1*10^{-34}J*s=3.3e-26J*m$$

Now we need it to be in joules only:

$$b = a/R=\frac{3.3e-26J*m}{1*10^{-15}m}=3.3e-11J$$

this value to eV:

$$D = b/V=\frac{3.3e-11J}{1.6e-19J}=206250000eV$$

Finally this to MeV:

$$ANS=D/1e6=\frac{206250000eV}{1e6eV}=206.25MeV!!$$

Note that the point of this problem, and similar ones, is not just to do some conversion of numbers. The point is that once you know one scale (in this case a length scale---the size of the proton), it encodes a lot of information and characterizes the typical scales in the system (in this case the length scale also sets the energy scale).

so we have :

$$R=1fm(10^{-15}m)$$

$$c = 3*10^8m/s$$

$$h{bar}=1.1*10^{-34}J*s$$

$$V=1eV=1.6*10^{-19}J$$

$$1MeV=10^6eV$$

so we need a answer in MeV:

first we get a value in J*m:

$$a=c*h{bar}=3*10^8m/s*1.1*10^{-34}J*s=3.3e-26J*m$$

Now we need it to be in joules only:

$$b = a/R=\frac{3.3e-26J*m}{1*10^{-15}m}=3.3e-11J$$

this value to eV:

$$D = b/V=\frac{3.3e-11J}{1.6e-19J}=206250000eV$$

Finally this to MeV:

$$ANS=D/1e6=\frac{206250000eV}{1e6eV}=206.25MeV!!$$