Electrons are accelerated through a potential difference of 750 kV, so that their kinetic energy is 750×105eV a) What is the ratio of the speed v of an electron having this energy to the speed of light, c? b) What would the speed be if it were computed from the principles of classical mechanics?

Question

Electrons are accelerated through a potential difference of 750 kV, so that their kinetic energy is 750×105eV  a) What is the ratio of the speed v of an electron having this energy to the speed of light, c?  b) What would the speed be if it were computed from the principles of classical mechanics?

encolatgehu · Accepted Answer

Step 1 γ=11−v2c2 K=mc21−v2c2−mc2=(γ−1)mc2 E=K+mc2, (E is the total energy of a praticle) Step 2 a) K=mc21−v2c2−mc2, (solving forvc) mc21−v2c2−mc2=7.50×105×1.602×10−19J=1.2016×10−13J m=9.1×10−31kg 1−v2c2=mc21.2016×10−13J+mc2=0.4052 v2c2=1−0.4052 v=c1−0.4052=0.9142c Step 3 b) k=12mv2 v=2km=2×1.2016×10−13J9.1×10−31kg=5.13×108m/s

otoplilp1 · Accepted Answer

Step 1  Given;   potential of the electron, V=750kV   kinetic energy of the electron, K.E=7.50×105eV   Part (A) the ratio of the speed of an electron having this energy to the speed of light   E=k+mc2   where;   E is the total energy of the electron   k is kinetic energy  k=E−mc2  k=mc21−v2c2−mc2=7,5×105×1.6×10−19J=12×10−14J  mc21−v2c2=12×10−14J+mc2  1−v2c2=mc212×10−14J+mc2  but, mass of electron, m=9.1×10−31kg and speed of light c=3×108ms  1−v2c2=9.1×1.0−31(3×108)212×10−14+9.1×10−31(3×108)2=0.4057  1−v2c2=0.40572  1−v2c2=0.1646  v2c2=1−0.1646  v2c2=0.8354  vc=0.914  v=0.914c  Step 2  b) speed of the electron when computed from principles of classical mechanics  k=12mv2  v2=2km  v=2km  v=2×12×10−149.1×10−31

alenahelenash · Accepted Answer

Step 1 The formula to calculate relativistic kinetic energy of the particle is, K=γmc2−mc2 m is mass of the electron mc2 is rest mass energy of the electron γ is the relativistic factor, c is the speed of light, The relativistic factor γ is, γ=11−v2c2 Use 11−v2c2 for γ in the force equation to rewrite K, K=11−v2c2mc2−mc2 rewrite the above expression in terms of v. mc21−11−v2c2=K+mc2 1−v2c2=mc2K+mc2 v=c1−mc2K+mc2 Substitute 9.1×10−31kg for m, 3×108ms−1 for c and 7.50×105eV for K in the above velocity equation to find v, v=c1−(9.1×10−31kg)(3×108m×s−1)2(7.50×105eV×(1.6×10−19JcV))+(9.1×10−31kg)(3×108m×s−1)2 =c1−0.128 =0.934c Therefore, the velocity of the electron is 0.934c Step 2 b) Kinetic energy of the electron K is 7.50×105eV The formula to calculate the classical kinetic energy of the electron is given by, K=12mv2 From the above non relativistic kinetic enrgy equation velocity of the electron is, 2Km=v2 v=2Km Substitute 7.50×105eV for K in the above velocity equation to find v. v=2(7.50×105eV×(1.6×10−19JcV))(9.1×10−31kg) =2.64×1017m2×s−2 =5.14×108m×s−1

Electrons are accelerated through a potential difference of 750 kV, so

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