Consider the $3^{rd}$ orbit of $He^{+}$ (Helium) using a non-relativistic approach. The speed of the electron in this orbit will be (given $K = 9 \times 10^9 \; N \cdot m^2/C^2$,$Z = 2$,and $h = 6.6 \times 10^{-34} \; J \cdot s$).

$A$ hydrogen atom in an excited state transitions to the ground state by emitting a photon of wavelength $\lambda$. The value of the principal quantum number $n$ of the excited state is:
($R$: Rydberg constant)

In the Bohr model of the hydrogen atom,the linear momentum of the electron depends on the principal quantum number $n$ as:

For a certain hypothetical one-electron atom,the wavelength (in $\mathring{A}$) for the spectral lines for transition from $n = p$ to $n = 1$ is given by $\lambda = \frac{1500p^2}{p^2 - 1}$ (where $p > 1$). The ionization potential of this element must be .....$V$ (Take $hc = 12420\ eV\cdot\mathring{A}$).

$A$ 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 $v \propto n^{1/3}$
$(C)$ $E = \frac{3}{2} \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$
$(D)$ $E = 2 \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$

The total energy of an electron in the $n^{th}$ orbit of a hydrogen atom is $E_n$. The total energy of an electron in the $n^{th}$ orbit of a $He^+$ ion is .........