An electron and a proton are detected in a cosmic ray experiment,… - Vidyadip

Question

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

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Answer

Electron is faster, Ratio of speeds is 13.54: 1 Mass of the electron, $m_e=9.11 \times 10^{-31} \mathrm{~kg}$ Mass of the proton, $m_p=$
$1.67 \times 10^{-27} \mathrm{~kg}$ Kinetic energy of the electron, $\mathrm{E}_{\mathrm{Ke}}=10 \mathrm{keV}=10^4 \mathrm{eV}=10^4 \times 1.60 \times 10^{-19}=1.60 \times 10^{-15} \mathrm{~J}$ Kinetic energy of the proton, $\mathrm{E}_{\mathrm{Kp}}=100 \mathrm{keV}=10^5 \mathrm{eV}=1.60 \times 10^{-14} \mathrm{~J}$ For the velocity of an electron $\mathrm{v}_{\mathrm{e}}$ its kinetic energy is given by the relation, $\mathrm{E}_{\mathrm{ke}}=\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^2 \therefore \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \times \mathrm{E}_{\mathrm{kr}}}{\mathrm{m}}}=\sqrt{\frac{2 \times 1.60 \times 10^{-10}}{9.11 \times 10^{-31}}}=5.93 \times 10^7 \mathrm{~m} / \mathrm{s}$ For the velocity of a proton $\mathrm{v}_{\mathrm{p}}$ its kinetic energy is given by the relation, $\mathrm{E}_{\mathrm{kp}}=\frac{1}{2} \mathrm{mv}_{\mathrm{p}}^2 \mathrm{v}_{\mathrm{p}}=\sqrt{\frac{2 \times \mathrm{E}_{\mathrm{kp}}}{\mathrm{m}}} \therefore \mathrm{v}_{\mathrm{p}}=\sqrt{\frac{2 \times 1.6 \times 10^{-14}}{1.67 \times 10^{-27}}}$
$=4.38 \times 10^6 \mathrm{~m} / \mathrm{s}$ Hence, the electron is moving faster than the proton. The ratio of their speeds, $\frac{v_c}{v_p}=\frac{5.93 \times 10^7}{4.38 \times 10^6}=13.54: 1$

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