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 kg, proton mass = 1.67 × 10–27 kg, 1 eV = 1.60 × 10–19 J).
Answers
Explanation:
Electron is faster; Ratio of speeds is 13.54 : 1
Mass of the electron, me = 9.11 × 10–31 kg
Mass of the proton, mp = 1.67 × 10– 27 kg
Kinetic energy of the electron, EKe = 10 keV = 104 eV
= 104 × 1.60 × 10–19
= 1.60 × 10–15 J
Kinetic energy of the proton, EKp = 100 keV = 105 eV = 1.60 × 10–14 J
For the velocity of an electron ve, its kinetic energy is given by the relation:
EKe = (1/2) mve2
∴ ve = (2EKe / m)1/2
= (2 × 1.60 × 10-15 / 9.11 × 10-31)1/2 = 5.93 × 107 m/s
For the velocity of a proton vp, its kinetic energy is given by the relation:
EKp = (1/2)mvp2
vp = (2 × 1.6 × 10-14 / 1.67 × 10-27 )1/2 = 4.38 × 106 m/s
Hence, the electron is moving faster than the proton.
The ratio of their speeds
ve / vp = 5.93 × 107 / 4.38 × 106 = 13.54 : 1
Answer:
Explanation:
Given,
mass of electron ( Me) = 9.11 × 10^-31 Kg
mass of proton ( Mp) = 1.67 × 10^-27 Kg
1 eV = 1.6 × 10^-19 J
kinetic energy of electron ( Ke) = 10KeV
= 10 × 10³ × 1.6 × 10^-19 j
= 1.6 × 10^-15 j
Let electron moves with Ve speed ,
then,
kinetic energy of electron = Ke
1/2 MeVe² = Ke
1/2 × 9.11 × 10^-31 × Ve² = 1.6 × 10^-15 j
Ve² = 3.2 × 10^-15/9.11 × 10^-31
Ve = { 3.2/9.11 × 10^(-15+31)}^1/2
Ve = 5.93 × 10^7 m/s
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again,
kinetic energy of proton (Kp) = 100 KeV
= 100 × 10³ × 1.6 × 10^-19 j
= 1.6 × 10^-14
Let speed of proton = Vp
kinetic energy = Kp
1/2MpVp² = Kp
1/2 × 1.67 × 10^-27 × Vp² = 1.6 × 10^-19 j
Vp² = 3.2 × 10^-14/1.67 × 10^-27
Vp = 4.38 × 10^6 m/s
from above explanation ,
we observed that
Ve > Vp { e.g speed of electron is greater then speed of proton }
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ratio of their speed = Ve/Vp
= 5.93 × 10^7/4.38 × 10^6
= 13.5
Hope it helps you...