Question

A proton has a mass $1.67 \times {10^{ - 27}}\,\,Kg$ and charges $+1.6 \times {10^{ - 19}}\,\,C$. If the proton is being accelerated through a potential difference of one million volts then its K.E. is:
A) $1.6 \times {10^{ - 25}}\,\,J$
B) $3.2 \times {10^{ - 13}}\,\,J$
C) $1.6 \times {10^{ - 15}}\,\,J$
D) $1.6 \times {10^{ - 13}}\,\,J$

Answer 1

Hint:- The above problem can be solved using the formula that is derived from the kinetic energy of the proton, that is with the respect to the mass of the proton, acceleration of the proton due to the potential difference of a voltage and the charge of the proton.

Useful formula:
The Kinetic Energy of the accelerated proton is given;
$K.E. = qV$
Where, $q$ denotes the charge on the proton, $v$ denotes the voltage acts on the accelerated proton.

Complete step by step solution:
The data given in the problem are;
Mass of the proton, $m = 1.67 \times {10^{ - 27}}$.
Charge of the proton, $q = 1.6 \times {10^{ - 19}}\,\,C$.
Potential difference of voltage, \[V = {10^6}\,\,V\]

The Kinetic Energy of the accelerated proton is given;
$K.E. = qV$
Substitute the values of charge of the proton and the potential difference in the above Kinetic energy formula;
$K.E. = 1.6 \times {10^{ - 19}}\,\,C\, \times {10^6}\,\,V$
On equating the above equation, we get;
$K.E. = 1.6 \times {10^{ - 13}}\,\,J$

Therefore, the kinetic energy of the mass of a proton that is accelerated is given as $K.E. = 1.6 \times {10^{ - 13}}\,\,J$.

Hence the option (D), $K.E. = 1.6 \times {10^{ - 13}}\,\,J$ is the correct answer.

Note: In the above given problem in case proton, if the mass of the proton increases due the addition of two or more protons, then the potential differences to move the proton increases and thus the acceleration acting on the proton increases.