(a) Find the rest energy of a proton in electron volts
Question : (a) Find the rest energy of a proton in electron volts
E_{R} =m_{p} c^{2}=\left(1.67 \times 10^{-27} \mathrm{~kg}\right)\left(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2}
=\left(1.50 \times 10^{-10} \mathrm{~J}\right)\left(1.00 \mathrm{eV} / 1.60 \times 10^{-19} \mathrm{~J}\right)
=938 \mathrm{MeV}
(b) If the total energy of a proton is three times its rest energy, with what speed is the proton moving?
Solution Equation 39.25 gives
E=3 m_{p} c^{2} =\frac{m_{p} c^{2}}{\sqrt{1-\frac{u^{2}}{c^{2}}}}{\sqrt{1-\frac{u^{2}}{c^{2}}}}
Solving for u gives
1-\frac{u^{2}}{c^{2}} =\frac{1}{9} \frac{u^{2}}{c^{2}} =\frac{8}{9} u=\frac{\sqrt{8}}{3} c=2.88 \times 10^{8} \mathrm{~m} / \mathrm{s}
(c) Determine the kinetic energy of the proton in electron volts.
Solution From Equation 39.24
K=E-m_{p} c^{2}=3 m_{p} c^{2}-m_{p} c^{2}=2 m_{p} c^{2}Because m_{p} c^{2}=938 \mathrm{MeV}, K=1880 \mathrm{MeV}
(d) What is the proton’s momentum?
Solution We can use Equation 39.26 to calculate the momentum with E=3 m_{p} c^{2}
E^{2} =p^{2} c^{2}+\left(m_{p} c^{2}\right)^{2}=\left(3 m_{p} c^{2}\right)^{2} p^{2} c^{2} =9\left(m_{p} c^{2}\right)^{2}-\left(m_{p} c^{2}\right)^{2}=8\left(m_{p} c^{2}\right)^{2} \phi =\sqrt{8} \frac{m_{p} c^{2}}{c}=\sqrt{8} \frac{(938 \mathrm{MeV})}{c}=2650 \mathrm{MeV} / cThe unit of momentum is written \mathrm{MeV} / c for convenience.