If a proton and an electron are accelerated through the same potential difference:
A) both the proton and electron have same K.E
B) both the proton and electron have same momentum
C) both the proton and electron have same velocity
D) both the proton and electron have same temperature

Hint: In these types of questions, if a proton and an electron are accelerated through the same potential difference, we will answer by checking for each physical quantity for which both electron and proton possess the same value.
When a point charge q is accelerated between two points having potential difference V, then the K.E. of the point charge will be qV.

Step-by-step solution:
Let a proton and an electron are accelerated through the same potential difference V then
Kinetic energy possessed by proton is given by-
$K.{E_P} = {q_p}V$
We know that the Charge on proton,${q_p} = 1.6 \times {10^{ - 19}}C$
⇒$K.{E_P} = 1.6 \times {10^{ - 19}} \times V$
Kinetic energy possessed by electron is given by-
$K.{E_e} = {q_e}V$
We know that the Charge on electron,${q_e} = 1.6 \times {10^{ - 19}}C$
⇒$K.{E_e} = 1.6 \times {10^{ - 19}} \times V$
Hence, we can see that $K.{E_p} = K.{E_e}$
Therefore, both the proton and electron have the same K.E.

Option A is correct answer.

Note: It may be noted that both electron and proton have different masses so they both do not have the same momentum and velocity.
As the formula of momentum of proton and electron is given by-
$p = \sqrt {2mK.E.} $
We can see that momentum depends on mass and K.E. As both electron and proton possess the same K.E. but different masses so they will both have different values of momentum.
Now, the formula of velocity of proton and electron is given by-
$v = \sqrt {\dfrac{{2K.E.}}{m}} $
As we can see that velocity is also mass dependent and as both electron and proton have different values of masses so they both will possess different velocity.
Also option D is also incorrect because we can’t determine whether both will have the same temperature or not because in this there is no relation with temperature.