[amp_mcq option1=”Ve > Vp” option2=”Ve < Vp" option3="Ve = Vp" option4="Cannot be determined" correct="option1"]
The correct answer is: A. Ve > Vp
The kinetic energy of a charged particle is given by the equation:
$KE = \frac{1}{2}mv^2$
where $m$ is the mass of the particle and $v$ is its speed.
The potential energy of a charged particle in an electric field is given by the equation:
$PE = qV$
where $q$ is the charge of the particle and $V$ is the potential difference.
When a charged particle is accelerated through a potential difference, its kinetic energy increases by an amount equal to its potential energy.
The mass of an electron is much less than the mass of a proton. Therefore, the kinetic energy of an electron that has been accelerated through a potential difference of 100 kV will be much greater than the kinetic energy of a proton that has been accelerated through the same potential difference.
Therefore, the speed of the electron will be much greater than the speed of the proton.
Option A is correct.