Question

Two protons are place $1\stackrel{\circ }{A}$ apart. If they are released what will be the kinetic energy of each proton when they are at large separation?

• A. $2.56\times {10}^{-19}J$
• B. $11.52\times {10}^{-19}J$
• C. $23.04\times {10}^{-19}J$
• D. $2.56\times {10}^{-28}J$

Answer 1

The correct option is B $11.52\times {10}^{-19}J$

At large separation $KE=PE$
Potential energy $=KE$ at large separation
$\frac{1}{2}\times \frac{\left(e\right)\left(e\right)}{4\pi {ϵ}_{0}r}=\frac{\text{KE at large separation}}{2}$
$=$ KE of each proton
KE of each proton
$=\frac{1}{2}=\frac{9\times {10}^9\times (1.6\times {10}^{-19}{)}^2}{{10}^{-10}}$
$=11.52\times {10}^{-19}J$