Answer
a) $n=4$
b) decrease
c) $K_5=8.674\times 10^{-20}J$
Work Step by Step
(a) We know that
$K=\frac{1}{2}(\frac{2\pi ke^2}{m})^2$
We plug in the known values to obtain:
$1.35\times 10^{-19}J=(\frac{1}{2})(9.11\times 10^{-31})[\frac{2(3.14)(9\times 10^9)(1.6\times 10^{-4})^2}{n(6.63\times 10^{-34})}]^2$
This simplifies to:
$n^2=\frac{2168.691\times 10^{-21}}{1.35\times 10^{-19}}$
$\implies n=4$
(b) We know that $K_n\propto \frac{1}{n^2}$. This relation shows that kinetic energy decreases when the electron moves away from the nucleus to the next higher Bohr orbit.
(c) We can determine the required kinetic energy as follows:
$K_5=(\frac{1}{2})(9.11\times 10^{-31})[\frac{(2)(3.14)(9\times 10^9)(1.6\times 10^{-19})^2}{5(6.63\times 10^{-34})}]^2$
$K_5=8.674\times 10^{-20}J$
