Question

A proton has a mass $1.67\times {10}^{-27}kg$ and charge $+1.6\times {10}^{-19}C$. 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

Answer: (D)

$1.6\times {10}^{-13}J$

$\text{Step 1: Change in Kinetic Energy of a charge accelerated through a Potential Difference}$

Whenever a charge $q$ is accelerated through a Potential Difference $\Delta V$, then its change in Kinetic Energy equals $q\Delta V$.

i.e.

$\Delta K=q\Delta V$

$\Rightarrow K.E-0=1.6\times {10}^{-19}C\times {10}^{6}V$

$\therefore K.E=1.6\times {10}^{-13}J$
Option $D$ is correct.