In J.J. Thomson's experiment the specific charge of an electron $(e / m)$ is related to potential difference $(V)$ between the cathode and the anode as

\frac{e}{m} α V

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\frac{e}{m} α \frac{1}{V}

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\frac{e}{m} α \frac{1}{\sqrt{V}}

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\frac{e}{m} α V^{2}

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Solution

The correct option is B $\frac{e}{m} α \frac{1}{V}$
In J.J. Thomson's experiment when sufficient amount of potential (V) is applied between the two electrodes electrons emitted from the cathode accelerate with velocity v then
$e V = \frac{m v^{2}}{2}$
or, $\frac{e}{m} = \frac{v^{2}}{2 V}$ .
So the specific charge of an electron (e/m) is related to potential difference (V) between the cathode and the anode as
$\frac{e}{m} \propto \frac{1}{V}$ .

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In J.J. Thomson's experiment the specific charge of an electron (e/m) is related to potential difference (V) between the cathode and the anode as

In J.J. Thomson's experiment the specific charge of an electron $(e / m)$ is related to potential difference $(V)$ between the cathode and the anode as