Answer
(a) 6
(b) increase
(c) $-0.555eV$
Work Step by Step
(a) We know that
$U_n=-\frac{KZe^2}{n^2r_1}$
We plug in the known values to obtain:
$-1.20\times 10^{-19}=-\frac{(9\times 10^9)(1)(1.6\times 10^{-19})^2}{n^2(5.29\times 10^{-11})}$
This simplifies to:
$n=6$
(b) We know that $U_n\propto \frac{-1}{n^2}$. This relation shows that when $n$ increases then $U_n$ becomes a smaller negative number and thus $U_n$ increases.
(c) As $U_7=-\frac{ke^2}{n^2r_7}$
We plug in the known values to obtain:
$U_7=-\frac{(-9\times 10^{9})(1.6\times 10^{-19})^2}{(7)^2(5.29\times 10^{-11})}$
$\implies U_7=-8.88\times 10^{-20}J=-0.555eV$
