What is the change in angular momentum of an electron in a hydrogen atom when it transitions from the $4^{th}$ orbit to the $5^{th}$ orbit? $(h = 6.6 \times 10^{-34} \, J \cdot s)$

In a hydrogen atom,the binding energy of the electron in the $n^{th}$ state is $E_n$. Then,the frequency of revolution of the electron in the $n^{th}$ orbit is:

What is the moment of inertia of the electron moving in the second Bohr orbit of a hydrogen atom? ($h=$ Planck's constant,$m=$ mass of electron,$\varepsilon_0=$ permittivity of free space,$e=$ charge on electron)

$A$ particle of mass $m$ moves in circular orbits with potential energy $V(r) = Fr$,where $F$ is a positive constant and $r$ is its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the particle's orbit is denoted by $R$ and its speed and energy are denoted by $v$ and $E$,respectively,then for the $n^{\text{th}}$ orbit (here $h$ is the Planck's constant)-
$(A)$ $R \propto n^{2/3}$ and $v \propto n^{1/3}$
$(B)$ $R \propto n^{2/3}$ and $v \propto n^{1/3}$
$(C)$ $E = \frac{3}{2} \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$
$(D)$ $E = 2 \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$

Positronium is just like a $H$-atom with the proton replaced by the positively charged antiparticle of the electron (called the positron,which is as massive as the electron). What would be the ground state energy of positronium?

The radius of the $2^{nd}$ orbit of $He^{+}$ in Bohr's model is $r_1$ and that of the fourth orbit of $Be^{3+}$ is represented as $r_2$. If the ratio $\frac{r_2}{r_1}$ is $x : 1$,then the value of $x$ is .........