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?

$A$ free hydrogen atom after absorbing a photon of wavelength $\lambda_{a}$ gets excited from the state $n=1$ to the state $n=4$. Immediately after that,the electron jumps to $n=m$ state by emitting a photon of wavelength $\lambda_{e}$. Let the change in momentum of the atom due to the absorption and the emission be $\Delta p_{a}$ and $\Delta p_{e}$,respectively. If $\lambda_{a} / \lambda_{e} = 1/5$,which of the following options is/are correct?
[Use $hc = 1242 \text{ eV nm}$; $1 \text{ nm} = 10^{-9} \text{ m}$,$h$ and $c$ are Planck's constant and speed of light,respectively]
$(1)$ $\lambda_{e} = 418 \text{ nm}$
$(2)$ The ratio of kinetic energy of the electron in the state $n=m$ to the state $n=1$ is $1/4$
$(3)$ $m=2$
$(4)$ $\Delta p_{a} / \Delta p_{e} = 1/2$

The energy of $He^{+}$ ion in its first excited state is $......eV$ (The ground state energy for the Hydrogen atom is $-13.6\,eV$).

The following parameter is the same for all hydrogen-like atoms and ions in their ground state.

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

Let $R_1$ be the radius of the second stationary orbit and $R_2$ be the radius of the fourth stationary orbit of an electron in Bohr's model. The ratio $\frac{R_1}{R_2}$ is