Question

A proton has a mass $1.67 \times 10^{-27} \,kg$ and charge $+1.6 \times 10^{-19} \, C$. If the proton is accelerated through a potential difference of million volts, then the kinetic energy is

• A. $1.6 \times 10^{-15} \,J$
• B. $1.6 \times 10^{-13} \,J$
• C. $1.6 \times 10^{-21} \,J$
• D. $3.2 \times 10^{-13} \, J$

Answer 1

Answer: (B)

$\left(1.6 \times 10^{-19}\right)\left(10^{6}\right)$
$=\frac{1}{2}\left(1.67 \times 10^{-27}\right) v^{2}$
$1.6 \times 10^{-13}=\frac{1.67}{2} \times 10^{-27} v^{2}= KE$
$KE =1.6 \times 10^{-13}\, J$