JJ Thompson Experiment (1 Viewer)

[shredinator]

Can someone who is good at explaining things please explain the theory behind JJ Thompsons experiment to measure the charge to mass ratio of electrons? I simply do not get it.

Thanks alot.

 

[willC]

Ok, hope this helps...

There are two physical steps to Thompson's experiment, the rest is purely mathematical

THE EXPERIMENTAL STEPS

Step 1: Place an electric and a magnetic field on a cathode ray tube so that their force on the electrons in the cathode ray balance each other. When the forces are balanced the beam will flow in a straight line (as it won't be deflected by the magnetic or electric fields-they are balanced)

Step 2: Remove the electric field. The electrons passing through the magnetic field, without the electric field to balance its force, will be deflected.

IMPORTANT!: An electron passing through a magnetic field that is threading perpendicular to its direction of motion will be deflected in a circle (if you can't really understand this draw a diagram then use the right hand rule to determine the force on the electron as it moves through the field).

So now the cathode ray will move in a circular path. The next step is simply to measure the radius of this circle.

SO...HOW DO YOU CALCULATE THE CHARGE TO MASS RATIO?

1. Before the electric field is removed the forces on the electrons are balanced

Therefore-

Force (due to electric field)=Force (due to magnetic field)

inserting the formulas for force due to electric and magnetic fields:

Force (due to electric field)=Eq
and Force (due to magnetic field)=qvBsinθ

Therefore:

Eq=qvBsinθ

but as the electrons are moving through at 90 degrees to the magnet field θ=90 so sinθ=1. Therefore...

Eq=qvB

cancel the q's

E=vB OR v=E/B (save this fact for later)

2. After the electric field has been removed the electrons will deflect toward the negative plate

This is the really cool part....

As the electron is moving in a circular path (as explained above), its force can be expressed in terms of centripedal force

F=(mv^2)/r

so, as only the magnetic field is acting, this centripedal force is provided by the magnetic force on the electron

qvBsinθ=(mv^2)/r

but the magnetic field is always 90 degrees to the electron's direction of motion so θ=90 and sinθ=1 (just like above)

qvB=(mv^2)/r

times both sides by r

rqvB=mv^2

divide through by v^2

rqB/v=m

divide through by q

rB/v=m/q

flip both sides because we want charge on mass ratio (q=charge, m=mass)

v/rB=q/m

Now! From above (1) we learnt that v=E/B. Substituting that in...

q/m=(E/B)/Br

q/m=E/(B^2)r

so the charge on mass ratio is equal to the elctric field strength divided by the radius times the magnetic field strength squared...

We can measure the magnetic field strength...

We can measure the electric field strength...

And we can measure the radius of deflection!

SO...we can work out the charge to mass ratio!!!

This was calculated to be a very very small number. It proved that the charge on an electron was constant and that and electron must be a very very very small particle-in fact one of the elementry particles.


Thats the best I can do. Sorry about the messy maths, if you cant understand it i'll try and post some diagrams and neater equations.

 

[Forbidden.]

... if you cant understand it i'll try and post some diagrams and neater equations.

Experiment with the [ sup][ /sup] and [ sub][ /sub] tags (WITHOUT the spaces).

 

[shredinator]

Will C - you, my friend, are a legend.

Couldn't have been explained better. Cheers!

Oh and the beatles rock.

 

[willC]

No worries! I enjoyed it.

Thanks for the hints, too, Forbidden...

Yes...Beatles rock.

 

[dwatt]

While the charge/mass ratio is of course very small, it's probably important to say that most other charge/mass ratios are much smaller. In fact, from memory, the charge/mass ratio of the electron was considered quite large (relatively).

A proton has a much smaller charge/mass ratio. So this showed that electrons have a tiny mass.

 

[helper]

While the charge/mass ratio is of course very small, it's probably important to say that most other charge/mass ratios are much smaller. In fact, from memory, the charge/mass ratio of the electron was considered quite large (relatively).

A proton has a much smaller charge/mass ratio. So this showed that electrons have a tiny mass.

The charge mass ratio of an electron is very large, not just relatively.

1.7 x 10 ¹¹

 

[checheville]

YOUR MY SUPERMAN WILLC! thanxx a million

and can you plz tell me one thingg...

if you stuff up your 2 assessments in beginning of YR12 ( but have 2 assessments left ) and trials and then the real HSC but you do really good in all the otherrss

>>>>> jzz those 2 friggin assessments that you stuffed upp!<<<<

have you ruined your chance of getting a UAI of 77+??????

plz reply ASAP! and this goes to anyone elsee out there that can helppp.

 

[Forbidden.]

YOUR MY SUPERMAN WILLC! thanxx a million

and can you plz tell me one thingg...

if you stuff up your 2 assessments in beginning of YR12 ( but have 2 assessments left ) and trials and then the real HSC but you do really good in all the otherrss

>>>>> jzz those 2 friggin assessments that you stuffed upp!<<<<

have you ruined your chance of getting a UAI of 77+??????

plz reply ASAP! and this goes to anyone elsee out there that can helppp.

I barely passed a few but I still made it through.

 

[zahmad]

Do you need to know the mathematical part of this dot point?

 

[Felix Jones]

The charge mass ratio of an electron is very large, not just relatively.

1.7 x 10 ¹¹

no..it to the power of (-11)....