Answer
$\frac{1}{16}$
Work Step by Step
(a) $E_{n}=-(13.6\,eV)\frac{Z^{2}}{n^{2}}$
For hydrogen atom, $Z=1$. Given that $n=4$.
$E_{4}=-(13.6\,eV)\frac{1^{2}}{4^{2}}=-0.85\,eV$
(b) $\frac{E_{4}}{E_{1}}=\frac{-(13.6\,eV)\frac{1^{2}}{4^{2}}}{-(13.6\,eV)\frac{1^{2}}{1^{2}}}=\frac{1}{16}$
