Electrons are accelerated through a potential difference of 750 kV, so that their kinetic energy...

maduregimc

maduregimc

Answered

2022-01-15

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?

Answer & Explanation

encolatgehu

encolatgehu

Expert

2022-01-16Added 27 answers

Step 1
γ=11v2c2
K=mc21v2c2mc2=(γ1)mc2
E=K+mc2, (E is the total energy of a praticle)
Step 2
a) K=mc21v2c2mc2, (solving forvc)
mc21v2c2mc2=7.50×105×1.602×1019J=1.2016×1013J
m=9.1×1031kg
1v2c2=mc21.2016×1013J+mc2=0.4052
v2c2=10.4052
v=c10.4052=0.9142c
Step 3
b) k=12mv2
v=2km=2×1.2016×1013J9.1×1031kg=5.13×108m/s

otoplilp1

otoplilp1

Expert

2022-01-17Added 41 answers

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=Emc2
k=mc21v2c2mc2=7,5×105×1.6×1019J=12×1014J
mc21v2c2=12×1014J+mc2
1v2c2=mc212×1014J+mc2
but, mass of electron, m=9.1×1031kg and speed of light c=3×108ms
1v2c2=9.1×1.031(3×108)212×1014+9.1×1031(3×108)2=0.4057
1v2c2=0.40572
1v2c2=0.1646
v2c2=10.1646
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=2×12×10149.1×1031
alenahelenash

alenahelenash

Expert

2022-01-23Added 366 answers

Step 1
The formula to calculate relativistic kinetic energy of the particle is,
K=γmc2mc2
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=11v2c2
Use 11v2c2 for γ in the force equation to rewrite K,
K=11v2c2mc2mc2
rewrite the above expression in terms of v.
mc2111v2c2=K+mc2
1v2c2=mc2K+mc2
v=c1mc2K+mc2
Substitute 9.1×1031kg for m, 3×108ms1 for c and 7.50×105eV for K in the above velocity equation to find v,
v=c1(9.1×1031kg)(3×108m×s1)2(7.50×105eV×(1.6×1019JcV))+(9.1×1031kg)(3×108m×s1)2
=c10.128
=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×(1.6×1019JcV))(9.1×1031kg)
=2.64×1017m2×s2
=5.14×108m×s1

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?