Single Choice

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

A$$\dfrac{e}{m}\alpha V$$
B$$\dfrac{e}{m}\alpha \dfrac{1}{V}$$
Correct Answer
C$$\dfrac{e}{m}\alpha \dfrac{1}{\sqrt{V}}$$
D$$\dfrac{e}{m}\alpha V^{2}$$

Solution

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
$$eV=\dfrac {mv^2}{2}$$
or, $$\dfrac{e}{m}= \dfrac{v^2}{2V}$$.
So the specific charge of an electron (e/m) is related to potential difference (V) between the cathode and the anode as
$$\dfrac{e}{m}\propto\dfrac{1}{V}$$.

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