Explanation

(a) $r_{n} \propto n^{2}$ in Bohr atom model
Potential energy $= \frac{- 1}{4\pi\varepsilon_{0}}.\frac{e^{2}}{r_{n}}$
For the $10^{th}$ orbit, it is $= \frac{- 1}{4\pi\varepsilon_{0}}.\frac{e^{2}}{r_{1}(100)}$
$\frac{1}{100}$ times less than the potential energy in the first orbit
Potential energy will decrease but with negative sign. Therefore when an electron in hydrogen atom jumps from $n = 1$ to $n = 10$ orbit, the potential energy of the system will increase