• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

URGENT - J J Thomsons experiment (1 Viewer)

Mr_Kap

Well-Known Member
Joined
Mar 24, 2015
Messages
1,127
Gender
Male
HSC
2015
Ok. I was looking through my textbook and other people's hsc resources and I found that there were two ways to perform thomson's experiment?

One method is that "he calculated the radius of the curved electron beam when deflected by the magnetic field only"

While the other method says "By equating the potential energy and kinetic energy of the electrons at the
cathode and anode respectively, and substituting the potential difference across the tube and the
velocity of the electrons, he was able to calculate the charge-to-mass ratio of the electron."


which one do the markers prefer?
Which one did you get taught at school?
Which one do i use?
 

anomalousdecay

Premium Member
Joined
Jan 26, 2013
Messages
5,766
Gender
Male
HSC
2013
When calculating the charge to mass ratio, you need to look at the potential energy and kinetic energy of the electrons, while also keeping in mind the radius of curvature of the path they followed.

Using:



you can find the charge to mass ratio as:

 

Kaido

be.
Joined
Jul 7, 2014
Messages
798
Gender
Male
HSC
2015
Both of the methods you described are basically the same, the latter is more generalised and should not be used in an exam situation.
Follow anomaly's method above for ALL exam questions asking for this and you may also want to equate the ELECTRIC with MAGNETIC
 

Mr_Kap

Well-Known Member
Joined
Mar 24, 2015
Messages
1,127
Gender
Male
HSC
2015
Both of the methods you described are basically the same, the latter is more generalised and should not be used in an exam situation.
Follow anomaly's method above for ALL exam questions asking for this and you may also want to equate the ELECTRIC with MAGNETIC
Alright, thx. I know both ways so i just wanted to know which way to use...thx for the clarification.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top