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Work, Power and Energy

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Question : 12 of 30
Marks: +1, -0
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 =9.11×10−31= 9.11 \times 10^{-31} kg, proton mass =1.67×10−27= 1.67 \times 10^{-27} kg, 1 eV=1.60×10−191\ \mathrm{eV} = 1.60 \times 10^{-19} J.)
Solution:  
Mass of electron me=9.11×10−31 kgm_{e}=9.11\times 10^{-31}\ \mathrm{kg}
Velocity of electron =ve=v_{e}
Mass of proton mp=1.6×10−27 kgm_{p}=1.6\times 10^{-27}\ \mathrm{kg}
Velocity of proton =vp=v_{p}
Kinetic energy of electron, 12meve2=10 KeV\frac{1}{2} m_{e} v_{e}^{2}=10\ \mathrm{KeV} ...(i)
Kinetic energy of proton, 12mpvp2=100 KeV\frac{1}{2} m_{p} v_{p}^{2}=100\ \mathrm{KeV} ...(ii)
Divide (i) by (ii),
12meve212mpvp2=10×1.6×10−16 J100×1.6×10−16 J\frac{\frac{1}{2} m_{e} v_{e}^{2}}{\frac{1}{2} m_{p} v_{p}^{2}}=\frac{10\times 1.6\times 10^{-16}\ \mathrm{J}}{100\times 1.6\times 10^{-16}\ \mathrm{J}}
⇒meve2mpvp2=110\Rightarrow\frac{m_{e} v_{e}^{2}}{m_{p} v_{p}^{2}}=\frac{1}{10}
vevp\frac{v_{e}}{v_{p}} =110mpme=1.67×10−2710×9.11×10−31=\sqrt{\frac{1}{10}\frac{m_{p}}{m_{e}}}=\sqrt{\frac{1.67\times 10^{-27}}{10\times 9.11\times 10^{-31}}} =13.54=13.54
Therefore, electron moves faster than proton.
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