Answer
a) $ K_3=1.51eV$
b) $ U_3=-3.024eV$
c) $E_3=-1.51eV$
Work Step by Step
(a) We know that
$K_n=\frac{1}{2}m(\frac{2\pi ke^2}{nh})^2$
For $n=3$ and plugging in other values, we obtain:
$K_3=\frac{1}{2(9.11\times 10^{-31})[\frac{(2)(3.14)(9\times 10^9)(1.6\times 10^{-19})^2}{(3)(6.63\times 10^{-34})}]^2}$
$\implies K_3=241\times 10^{-21}J$
$\implies K_3=1.51eV$
(b) We know that
$U_n=\frac{-KZe^2}{r_n}$
We plug in the known values to obtain:
$U_3=-\frac{9\times 10^9(1)(1.6\times 10^{-19})^2}{(9) (5.29\times 10^{-11})}$
$U_3=-0.4839\times 10^{-18}J$
$\implies U_3=-3.024eV$
(c) We can find the total energy for state $n=3$ as follows:
$E_3=K_3+U_3$
We plug in the known values to obtain:
$E_3=1.50625+(-3.024)$
$E_3=-1.51eV$
