The ionization potential of a hydrogen-like atom is $122.4 \, V$. Find its atomic number $Z$.

$A$ 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 $1/4$
$(3)$ $m=2$
$(4)$ $\Delta p_{a} / \Delta p_{e} = 1/2$

Speed of electron in its $1^{st}$ Bohr's orbit is given by $2.18 \times 10^6 \ m/s$. If the time period of electron in $n^{th}$ orbit is measured as $4.10 \ fs$,the value of $n$ is

The orbital magnetic moment $(m_{orb})$ of a revolving electron around the nucleus varies with the principal quantum number $(n)$ as

Suppose in a hypothetical world, the angular momentum is quantized to be even integral multiples of $\frac{h}{2 \pi}$. According to Bohr's model, what will be the largest possible wavelength emitted by hydrogen atoms in the visible range in this world (in $\text{ nm}$)? (Consider $hc = 1242 \text{ eV-nm}$)

The angular momentum of the orbital electron is an integral multiple of: