J.J. Thomson's m/e Experiment, Help

  • Thread starter SuicideSteve
  • Start date
  • Tags
    Experiment
In summary, Thomson's experimental determination of the ratio m/e of the mass to the charge of an electron used the equations m(d2x/dt2) + He(dy/dt) = Ee and m(d2y/dt2) - He(dx/dt) = 0. By substituting 1 for the constants, the resulting parametric equations for the path were found to be x = 1 - cost and y = t - sint. However, there may be an error in the given equations as the stated solution includes an "E" that is not present.
  • #1
SuicideSteve
24
0

Homework Statement



In Thomson's experimental determination of the ratio m/e of the mass to the charge of an electron, in which the electrons were subjected to an electric field of intensity E and a magnetic field of intensity H, the equations

m(d2x/dt2) + He(dy/dt) = Ee , m(d2y/dt2) - He(dx/dt) = 0 ,

were employed. If x=y=dx/dt=dy/dt=0 for t=0, show that the path is a cycloid whose parametric equations are:

x = {Em/H2e}(1 - cos([He/m]t))
y = {Em/H2e}([He/m]t - sin([He/m]t))

The Attempt at a Solution



I have solved the differential equation by substituting 1 for the constants and come out with:
x = 1 - cost
y = t - sint

My problem is I can not figure out how to end up with the constants in the results.

Any help is greatly appreciated,
Thanks.
Steve.
 
Last edited:
Physics news on Phys.org
  • #2
SuicideSteve said:

Homework Statement



In Thomson's experimental determination of the ratio m/e of the mass to the charge of an electron, in which the electrons were subjected to an electric field of intensity E and a magnetic field of intensity H, the equations

m(d2x/dt2) + He(dy/dt) , m(d2y/dt2) - He(dx/dt) = 0 ,

were employed. If x=y=dx/dt=dy/dt=0 for t=0, show that the path is a cycloid whose parametric equations are:

x = {Em/H2e}(1 - cos([He/m]t))
y = {Em/H2e}([He/m]t - sin([He/m]t))

The Attempt at a Solution



I have solved the differential equation by substituting 1 for the constants and come out with:
x = 1 - cost
y = t - sint

My problem is I can not figure out how to end up with the constants in the results.

Any help is greatly appreciated,
Thanks.
Steve.

There is something wrong with your two differential equations, since an "E" appears in the stated solution but is not present in your DEs.

RGV
 
  • #3
Ray Vickson said:
There is something wrong with your two differential equations, since an "E" appears in the stated solution but is not present in your DEs.

RGV

Sorry, fixed it.
 

FAQ: J.J. Thomson's m/e Experiment, Help

What is J.J. Thomson's m/e experiment?

J.J. Thomson's m/e experiment was a groundbreaking experiment conducted in the late 19th century that helped to discover the existence of subatomic particles, specifically the electron. It involved passing an electric current through a vacuum tube and observing the effects of a magnetic field on the particles within the tube.

Why is J.J. Thomson's m/e experiment significant?

J.J. Thomson's m/e experiment is significant because it provided evidence for the existence of subatomic particles, which was a major breakthrough in the field of physics. It also helped to disprove the previously accepted theory of the "plum pudding" model of the atom and paved the way for further discoveries about the structure of the atom.

What were the results of J.J. Thomson's m/e experiment?

The results of J.J. Thomson's m/e experiment showed that the particles within the vacuum tube were deflected by a magnetic field, indicating that they had a negative charge. This led Thomson to conclude that these particles were subatomic particles, which he called electrons. He also calculated the mass-to-charge ratio (m/e) of the electrons, which helped to further understand the structure of the atom.

How did J.J. Thomson's m/e experiment contribute to the development of modern physics?

J.J. Thomson's m/e experiment was a crucial contribution to the development of modern physics. It provided evidence for the existence of subatomic particles and helped to disprove previous theories about the structure of the atom. This experiment also opened the door for further research and discoveries in the field of particle physics, leading to a better understanding of the fundamental nature of matter.

Can J.J. Thomson's m/e experiment be replicated?

Yes, J.J. Thomson's m/e experiment can be replicated, and it has been replicated numerous times since its initial discovery. The experiment has been refined and improved upon with modern technology, but the basic principles and results remain the same. It is still a significant experiment in the field of physics and is often used as a teaching tool to demonstrate the existence and properties of subatomic particles.

Similar threads

Back
Top