Question

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

Answer 1

Electron is faster; Ratio of speeds is \(13.54\) : \(1\) 

Mass of the electron, \(m_e\) = \(9.11\times 10^{-31}\) kg 
Mass of the proton, \(m_p\) = \(1.67\times 10^{-27}\) kg 
Kinetic energy of the electron, \(E_{ke}\) = \(10\)\(\text {keV}\) = 104  \(\text {eV}\) 
= \(104\) × \(1.60\times 10^{-19}\) 
= \(1.60\times 10^{-15}\) \(\text J\) 
Kinetic energy of the proton, \(E_{kp}\) = \(100\) \(\text {keV}\) = \(105\) eV = \(1.60\times 10^{-14}\) J
For the velocity of an electron \(v_e\) , its kinetic energy is given by the relation: 
\(E_{ke}\) = \(\frac{1}{2}\) \(mv^2_e\)

∴ \(v_e\) = \(\sqrt{\frac{2\times E_{ke}}{m}}\) = \(\sqrt{\frac{2\times 1.60\times 10^{-15}}{9.11\times 10^{-31}}}\) = \(5.93\times 10^7\, m/s\)
For the velocity of a proton vp, its kinetic energy is given by the relation: 

\(E_{kp}\) = \(\frac{1}{2}\,mv^2_p\)

\(v_p\) = \(\sqrt{\frac{2\times E_{kp}}{m}}\)

∴ \(v_p\) = \(\sqrt{\frac{2\times 1.6\times 10^{-14}}{1.67\times 10^{-27}}}\)= \(4.38\times 10^6\, m/s\)

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

\(\frac{v_e}{v_p}\)= \(\frac{5.93\times 10^7}{4.38\times 10^6}\)= \(13.54\) : \(1\)