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Physics Question on Kinetic Energy

An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV\text {keV}, and the second with 100 keV\text {keV}. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = 9.11×10−319.11\times 10^{-31} kg, proton mass = 1.67×10−271.67\times 10^{-27} kg, 1 eV\text {eV} = 1.60×10−191.60\times 10^{-19} J)

Answer

Electron is faster; Ratio of speeds is 13.5413.54 : 11

Mass of the electron, mem_e = 9.11×10−319.11\times 10^{-31} kg
Mass of the proton, mpm_p = 1.67×10−271.67\times 10^{-27} kg
Kinetic energy of the electron, EkeE_{ke} = 10$$\text {keV} = 104 eV\text {eV}
= 104104 × 1.60×10−191.60\times 10^{-19}
= 1.60×10−151.60\times 10^{-15} J\text J
Kinetic energy of the proton, EkpE_{kp} = 100100 keV\text {keV} = 105105 eV = 1.60×10−141.60\times 10^{-14} J
For the velocity of an electron vev_e , its kinetic energy is given by the relation:
EkeE_{ke} = 12\frac{1}{2} mve2mv^2_e

∴ vev_e = 2×Ekem\sqrt{\frac{2\times E_{ke}}{m}} = 2×1.60×10−159.11×10−31\sqrt{\frac{2\times 1.60\times 10^{-15}}{9.11\times 10^{-31}}} = 5.93×107 m/s5.93\times 10^7\, m/s
For the velocity of a proton vp, its kinetic energy is given by the relation:

EkpE_{kp} = 12 mvp2\frac{1}{2}\,mv^2_p

vpv_p = 2×Ekpm\sqrt{\frac{2\times E_{kp}}{m}}

∴ vpv_p = 2×1.6×10−141.67×10−27\sqrt{\frac{2\times 1.6\times 10^{-14}}{1.67\times 10^{-27}}}= 4.38×106 m/s4.38\times 10^6\, m/s

Hence, the electron is moving faster than the proton.
The ratio of their speeds:

vevp\frac{v_e}{v_p}= 5.93×1074.38×106\frac{5.93\times 10^7}{4.38\times 10^6}= 13.5413.54 : 11