Question

The acceleration for electron and proton due to electrical force of their mutual attraction when they are $1\,\mathring{A}$ apart is

• A. $3.1\times10^{22}\,m\,s^{-2}$, $1.3\times10^{19}\,m\,s^{-2}$
• B. $3.3\times10^{18}\,m\,s^{-2}$, $3.2\times10^{16}\,m\,s^{-2}$
• C. $2.5\times10^{22}\,m\,s^{-2}$, $1.4\times10^{19}\,m\,s^{-2}$
• D. $2.5\times10^{18}\,m\,s^{-2}$, $1.3\times10^{16}\,m\,s^{-2}$

Answer 1

Answer: (C)

Force of mutual attraction between an electron and a proton
$F=\frac{1}{4\pi\varepsilon_{0}} \frac{e^{2}}{r^{2}}=\frac{9\times10^{9}\left(1.6\times10^{-19}\right)^{2}}{\left(10^{-10}\right)^{2}}=2.3\times10^{-8}\,N$
Acceleration of electron $=\frac{F}{m_{e}}=\frac{2.3\times10^{-8}}{9\times10^{-31}}$
$=2.5\times10^{22}\,m\,s^{-2}$
Acceleration of proton $=\frac{F}{m_{p}}=\frac{2.3\times10^{-8}}{1.66\times10^{-27}}$
$=1.4\times 10^{19}\,m\,s^{-2}$