When the electron is accelerated between 500keV, what is the percentage increase in mass?
A. 82.35%
B. 97.85%
C. 42.35%
D. 59.45%

HINT: before attempting the question, one should have prior knowledge of kinetic energy of electrons, working of it and basic formulas given in kinetic energy of the accelerated electron. Using the given information will help you to approach the solution to the question.

Complete step by step answer:
Given: kinetic energy of the electron=500KeV
We can find the percentage increases in mass
So, here K.E of the electron=500KeV=$500 \times \mathop {10}\nolimits^3 eV$
Now the kinetic energy of the accelerated electron is given by,
$K.E = \mathop {mc}\nolimits^2 - \mathop {m _{\circ} c}\nolimits^2 $(in this we assume that the electrons are the photoelectrons)
Where, m= relativistic mass of electron
$m _{\circ} $= rest mass of the electron
And C= velocity of light
So, $\dfrac{{KE}}{{m _{\circ} \mathop c\nolimits^2 }} = \dfrac{{m\mathop c\nolimits^2 - m _{\circ} \mathop c\nolimits^2 }}{{m _{\circ} \mathop c\nolimits^2 }} = \dfrac{{m - m _{\circ} }}{{m _{\circ} }}$
Thus, percentage increase in mass will is equal to
$\dfrac{{m - m _{\circ} }}{{m _{\circ} }} \times 100 = \dfrac{{KE}}{{m _{\circ} \mathop c\nolimits^2 }} \times 100$
$ \Rightarrow \dfrac{{500 \times \mathop {10}\nolimits^3 }}{{0.511 \times \mathop {10}\nolimits^6 }} \times 100 = 97.85\% $
Where $m _{\circ} \mathop c\nolimits^2 $=0.511MeV
So, the correct answer is option (B).

NOTE: SI unit of energy is joule. There are many conversions of energy in other units, that is we can express energy in many units. Rest mass is a relatively different concept. Rest mass energy is dependent on rest mass and velocity of light in vacuum only. Relativistic mass is the mass which is assigned to the body in the motion.