Answer
a) $V=27.14V$
b) $E_K=2.17\times10^{-18}J=13.6eV$
c) $\sum E=-13.5eV=-2.16\times10^{-18}J$
d) $E_i=13.5eV=2.16\times10^{-18}J$
Work Step by Step
a) $V=\frac{kQ}{r}=\frac{(8.99\times10^9\frac{N\cdot m^2}{C^2})(1.60\times10^{-19}C)}{0.53\times10^{-10}m}=27.14V$
b) To find velocity, we can use the fact that the electron's orbit is circular.
$\frac{mv^2}{r}=\frac{ke^2}{r^2}$
$v=\sqrt{\frac{ke^2}{rm}}=\sqrt{\frac{(8.99\times10^9\frac{N\cdot m^2}{C^2})(1.60\times10^{-19}C)^2}{(0.53\times10^{-10}m)(9.11\times10^{-31}kg)}}$
$=2.18\times10^6\frac{m}{s}$
$E_K=\frac{mv^2}{2}=\frac{(9.11\times10^{-31}kg)(2.18\times10^6\frac{m}{s})^2}{2}=2.17\times10^{-18}J$
$2.17\times10^{-18}J\times\frac{1eV}{1.60\times10^{-19}J}=13.6eV$
c) $E_P=k\frac{Q_1Q_2}{r}=(8.99\times10^9\frac{N\cdot m^2}{C^2})\frac{(1.60\times10^{-19}C)(-1.60\times10^{-19}C)}{0.53\times10^{-10}m}$
$=-4.34\times10^{-18}J=-27.1eV$
$\sum E=E_K+E_P=13.6eV-27.1eV=-13.5eV=-2.16\times10^{-18}J$
d) At a distance of infinity, the electron will have no potential energy and it will have no kinetic energy. This means the energy required is equal in magnitude to the total energy of the electron but negative because we re removing the energy.
$E_i=13.5eV=2.16\times10^{-18}J$
