Answer to Question #309927 in Physics for Almas

Question #309927

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


1
Expert's answer
2022-03-13T18:48:16-0400

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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