Question

An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy $10keV$, and the second with $100keV$. Which is faster the electron or the proton? Obtain the ratio of their speeds. ($m_e=9.1\times {10}^{-31}kg,m_p=1.67\times {10}^{-27}kg,1eV=1.6\times {10}^{-19}J$)

Answer 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 is

$E_e=10keV={10}^{4}eV$

$E_e={10}^4\times 1.6\times {10}^{-19}=1.60\times {10}^{-15}J$

Kinetic energy of the proton

$E_p=100keV={10}^{5}eV=1.60\times {10}^{-14}J$

For the velocity of an electron $v_e$, its kinetic energy is given by the relation:

$E_e=\frac{1}{2}m{v}_e^2\Rightarrow v_e=\frac{2\times 1.6\times {10}^{-15}}{9.11\times {10}^{-31}}=5.93\times {10}^{7}m/s$

For the velocity of a proton $v_p$, its kinetic energy is given by the relation:

$E_p=\frac{1}{2}m{v}_p^2$

$v_p=4.38\times {10}^{6}m/s$
As the velocity of electron is more than that of proton, so electron is faster than proton.
The ratio of their speeds is $\frac{v_e}{v_p}=\frac{5.93\times {10}^7}{4.38\times {10}^6}=\frac{13.54}{1}$