As the electron in a Bohr orbit of a Hydrogen atom transitions from state $n = 2$ to $n = 1$,how do the kinetic energy $K$ and potential energy $U$ change?

In the Bohr's hydrogen atom model,the radius of the stationary orbit is directly proportional to ($n =$ principle quantum number)

In a mixture of $H-He^{+}$ gas ($He^{+}$ is a singly ionized $He$ atom),$H$ atoms and $He^{+}$ ions are excited to their respective first excited states. Subsequently,$H$ atoms transfer their total excitation energy to $He^{+}$ ions by collisions. Assume that the Bohr model of the atom is exactly valid.
$1.$ The quantum number $n$ of the state finally populated in $He^{+}$ ions is
$(A) 2$ $(B) 3$ $(C) 4$ $(D) 5$
$2.$ The wavelength of light emitted in the visible region by $He^{+}$ ions after collisions with $H$ atoms is
$(A) 6.5 \times 10^{-7} \ m$ $(B) 5.6 \times 10^{-7} \ m$ $(C) 4.8 \times 10^{-7} \ m$ $(D) 4.0 \times 10^{-7} \ m$
$3.$ The ratio of the kinetic energy of the $n=2$ electron for the $H$ atom to that of the $He^{+}$ ion is
$(A) 1/4$ $(B) 1/2$ $(C) 1$ $(D) 2$

In a hydrogen-like atom,the velocity of an electron in the second orbit is $v$. What will be the velocity of the electron in its fifth orbit?

The ground state energy of a hydrogen atom is $-13.6 \text{ eV}$. The potential energy of the electron in this state is: (in $\text{ eV}$)

The kinetic energy of an electron revolving around a nucleus will be