millikan's experiment (2 Viewers)

[{*(00)*}]

(the oildrop in the magnetic field experiment)
why does the results in milikan's experiment vary, why isn't the charge of an electron exactly -1.602 x 10^-19? and what was the thing that millikan needed to realise in his experiment in order to realise the correct results?.... thanks {*(00)*}!

 

[alcalder]

Millikan was using charged particles of oil not just single electrons. Sometimes the oil would gain a charge of 1e- and sometimes more(2e-, 5e- etc) , but he always found the charge to be a multiple of the charge on an electron.

Therefore Millikan realised that the common factor between all the charge he was measuring was that they were all a multiple of -1.602 x 10^-19.

Does that help?

 

[ianc]

Millikan's experiment is not actually in the HSC syllabus

 

[{*(00)*}]

Millikan was using charged particles of oil not just single electrons. Sometimes the oil would gain a charge of 1e- and sometimes more(2e-, 5e- etc) , but he always found the charge to be a multiple of the charge on an electron.

Therefore Millikan realised that the common factor between all the charge he was measuring was that they were all a multiple of -1.602 x 10^-19.

Does that help?

yup! thanks. but how did he know that -1.602 was a common factor? was it just trial and error?

 

[Riviet]

was it just trial and error?

It would have been for him, he would have seen a pattern eventually after doing the experiment several times.

 

[{*(00)*}]

does anyone know who found the mass of an electron?
jj thompson found the charge to mass ratio...
then millikan found the charge...
but how did he find the charge without first knowing the mass? he may have known the equation m=vd but how did he know the volume of the oil drop and the density?
and what is terminal velocity?

 

[Riviet]

what is terminal velocity?

Basically, terminal velocity is the velocity at which an object in free fall does not accelerate and falls at constant speed.

 

[alcalder]

Millikan measured the terminal velocity of the oil drop when it fell under only the influence of gravity.

Then he used an oil of known density. Fluid mechanics was then brought in.

(THis is from the old syllabus elective - Physics in Technology, so probably not relevant but it shows how he did it.)

Stokes Law
F = 6πrηv where:
F is the frictional force on particle falling with terminal velocity through a viscous fluid r is the Stokes radius of the particleη is the fluid viscosity (known), andv is the terminal velocity (measurable)
But if a particle is falling with terminal velocity then the weight force is balanced by both the buoyant force of the viscous fluid and the viscous Force.

Oil has density p (known)
Viscous Fluid it falls through has density p' (known)

weight = mg
= volume x density x g
= 4/3 π r ² p g

buoyant force = 4/3 π r ² p' g

So
weight = buoyant force + viscous force
4/3 π r ²pg = 4/3 π r ²p' g + 6πrηv

Thus fluid viscosity is
η = 2 r²g(p-p') / 9v

He knows p, p', η so he can work out radius of the oil drop.

From this he could determine the volume and hence mass of the oil drop.

You don't need to know this, but it may help in your understanding. It is proably enough to say, "using fluid mechanics and an oil of known density, he determined the radius of the oil drop and hence the mass."


BTW: If you reach terminal velocity, it certainly is. LOL