The electron-volt, \(\mathrm{eV},\) is a unit of energy \((1 \mathrm{eV}=\) \(\left.1.602 \cdot 10^{-19} \mathrm{~J}, 1 \mathrm{MeV}=1.602 \cdot 10^{-13} \mathrm{~J}\right) .\) Since the unit of \(\mathrm{mo}\) mentum is an energy unit divided by a velocity unit, nuclear physicists usually specify momenta of nuclei in units of \(\mathrm{MeV} / c,\) where \(c\) is the speed of light \(\left(c=2.998 \cdot 10^{9} \mathrm{~m} / \mathrm{s}\right) .\) In the same units, the mass of a proton \(\left(1.673 \cdot 10^{-27} \mathrm{~kg}\right)\) is given as \(938.3 \mathrm{MeV} / \mathrm{c}^{2} .\) If a proton moves with a speed of \(17,400 \mathrm{~km} / \mathrm{s}\) what is its momentum in units of \(\mathrm{MeV} / \mathrm{c}\) ?

Step by step solution

01

Converting units of speed

First, we need to convert the speed of the proton from km/s to m/s. We know that 1 km = 1000 m, so 17,400 km/s can be converted to m/s as follows: \[17,400\,km/s \times \frac{1000\,m}{1\,km} = 17,400 \times 10^3\,m/s\]

02

Finding the momentum in SI units

Next, we can find the momentum of the proton in SI units by using the formula: \[p = mv\] where p is the momentum, m is the mass of the proton (1.673 * 10^-27 kg), and v is the speed we found in Step 1. \[p = (1.673 \times 10^{-27}\,kg)(17,400 \times 10^3\,m/s) = 2.90862 \times 10^{-23}\,kg\cdot m/s\]

03

Converting the momentum to MeV/c

Now, we convert the momentum to MeV/c units. We have to divide by two conversion factors: The energy conversion factor (1.602 * 10^-13 J/MeV) and the speed of light (2.998 * 10⁹ m/s). So we can calculate the momentum like this: \[p[\frac{MeV}{c}] = \frac{p_{(SI\,units)}\, [\frac{kg\,m}{s}]}{ENERGY\,CONVERSION\,FACTOR\,[\frac{J}{MeV}] \times c\,[\frac{m}{s}]}\] \[p[\frac{MeV}{c}] = \frac{2.90862 \times 10^{-23}\,\frac{kg\cdot m}{s}}{(1.602 \times 10^{-13}\,\frac{J}{MeV}) \times (2.998 \times 10^9\,\frac{m}{s})} = 6.015 \mathrm{MeV}/\mathrm{c}\] So, the momentum of the proton in units of MeV/c is approximately 6.015 MeV/c.

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