Specific charge of electron

APPARATUS • Kent \(e/m\) Experimental Apparatus Model TG-13 • GW Laboratory DC Power Supply Model GPS-1850 • Heathkit Regulated Power Supply Model PS-4 • Triplett Multimeter Model 4000 • Simpson Digital Multimeter Model 464 • Bell 620 Gaussmeter with Hall Probe • Magnetic compass INTRODUCTION Measuring separately the electric charge (\(e\)) and the rest mass (\(m\)) of an electron is a difficult task because both quantities are extremely small (\(e\) = 1.60217733×10 -19 coulombs, \(m\) = 9.1093897×10 -31 kilograms). Fortunately, the ratio of these two fundamental constants can be determined easily and precisely from the radius of curvature of an electron beam traveling in a known magnetic field. An electron beam of a specified energy, and therefore a specified speed, may be produced conveniently in an \(e/m\) apparatus. The central piece of this apparatus is an evacuated electron-beam bulb with a special anode. A known current flows through a pair of Helmholtz coils and produces a magnetic field. The trajectory of the speeding electrons moving through the magnetic field is made visible by a small amount of mercury vapor. THEORY An electron moving in a uniform magnetic field travels in a helical path around the field lines. The electron's equation of motion is given by the Lorentz relation. If there is no electric field, then this relation can be written as \begin. • Compute the ratio \(e/m\) using the “accepted” values provided in the “Introduction” section of this experim...

Experimentally, we tried to find the specific charge, $\frac$ from the equation. Is there a way in which we can use the measurements $d$, $V_a$, $V_p$ and $l$ to calculate the specific charge? Or is it necessary to use a magnetic field and electric field together to equate the two forces and then find the specific charge? The simple answer is no, and if you think about this right, the reason is simple. Take your experiment and assume any value you like for e and m. Now repeat the experiment for a new m (M), where M = 4m. Particle velocity will be v/2, rather than v. This means that the particle will spend twice as long in deflecting region, but it will deflect at a rate 1/4 as great. So when you calculate d/2, the factors cancel, and the deflection remains the same.

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Why is the JJ Thomson experiment for specific charge seen as such strong evidence for the electron being smaller than the hydrogen ion? He showed that the specific charge of an electron was over 1000x greater than a hydrogen ion. This means that the electron is either much lighter(as known today), or more charged. Since the idea that atoms were indivisible was so ingrained at the time, what evidence made it clear to rule out the latter? As only this leads to the conclusion that it is smaller than the atom. $\begingroup$ @JonCuster I think this evidence is circular because it wasn't known at the time, although these things are obvious now. The answer seems to be that they didnt know for sure, but it was just a reasonable assumption and no one was really arguing otherwise, and it was not until Millikan oil drop that it was known. $\endgroup$