Answer
(a) electric force on electron
$F=8.2020\times10^{-8}N$
(b) orbital speed of electron
$v=2.1856\times10^{6}m/s$
(c)centripetal acceleration of electron
$a=9.013\times10^{21}g$
Work Step by Step
charge on electron is $q_{e}= -e=-1.6\times10^{-19}C$
charge on proton is $q_{p}= +e=+1.6\times10^{-19}C$
electron and proton are separated by distance $r=5.3\times10^{-11}m$
(a) Electric force on electron
from coulombs law electric force of attraction
$F=k\frac{q_{p}\times q_{e}}{r_{}^2}$
putting values of $q_{e}= 1.6\times10^{-19}C$,$q_{p}=1.6\times10^{-19}C$
$r=5.3\times10^{-11}m$, $k=9.0\times10^{9}N.m^2/C^2$
we will get $F=9.0\times10^{9}N.m^2/C^2\frac{1.6\times10^{-19}C\times 1.6\times10^{-19}C}{(5.3\times10^{-11}m)^2}$
$F=8.2020\times10^{-8}N$
(b) While orbiting around the nucleus electrical force of attraction on electron is balanced by its orbital circular motion.
in this case centripetal force =Electrical force
$\frac{m_{e}v^2}{r}= F$, here $ m_{e}=9.1\times10^{-31}kg$ is mass of electron
$v=\sqrt \frac{F\times r}{m_{e}}=\sqrt \frac{8.2020\times10^{-8}N\times5.3\times10^{-11}m}{9.1\times10^{-31}kg}$
$v=\sqrt(4.776989\times10^{12}m/s)$
$v=2.1856\times10^{6}m/s$
(c) centripetal acceleration
$a= \frac{v^2}{r}=\frac{(2.1856\times10^{6}m/s)^2}{5.3\times10^{-11}m}$
$a=9.013\times10^{22}m/s^2$
since $\approx 10m/s^2=g$
$a=\frac{9.013\times10^{22}g}{10}$
$a=9.013\times10^{21}g$
