Answer to Question #309916 in Physics for Almas

Question #309916

Determination of charge/mass ratio of electron using narrow fine beam tube apparatus


1
Expert's answer
2022-03-13T18:48:02-0400

The change in kinetic energy of electron when it moves through potential difference "V" is given by

"\\frac{mv^2}{2}=eV"

Hence, the speed of electron when it enters in magnetic field

"v=\\sqrt{2eV\/m}"

The electron in magnetic field moves along a circular path of radius

"r=\\frac{mv}{eB}=\\frac{m\\sqrt{2eV\/m}}{eB}=\\sqrt{(2mV)\/(eB^2)}"

Finally

"\\frac{e}{m}=\\frac{2V}{r^2B^2}"


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