In a Thomson set-up for the determination of e/m, electrons accelerated by $2.5$ $kV$ enter the region of crossed electric and magnetic fields of strengths $3.6 \times {10^4}V{m^{ - 1}}$ and $1.2 \times {10^{ - 3}}T$ respectively and go through undeflected. The measured value of $e/m$ of the electron is equal to
$1.0 \times {10^{11}}C{\rm{ - }}k{g^{ - 1}}$
$1.76 \times {10^{11}}C{\rm{ - }}k{g^{ - 1}}$
$1.80 \times {10^{11}}C{\rm{ - }}k{g^{ - 1}}$
$1.85 \times {10^{11}}C{\rm{ - }}k{g^{ - 1}}$
In an electron gun, the electrons are accelerated by the potential $V$. If e is the charge and $m$ is the mass of an electron, then the maximum velocity of these electrons will be
From the following, what charges can be present on oil drops in Millikan's experiment
(Here e is the electronic charge)
In an electron gun the control grid is given a negative potential relative to cathode in order to
The fact that electric charges are integral multiples of the fundamental electronic charge was proved experimentally by
Positive rays consists of