For a hydrogen atom,when an electron jumps from $n = 2$ to $n = 1$,the wavelength of the radiation emitted is found to be $\lambda_0$. For which transition of an electron in a $He^+$ ion will the wavelength of the radiation emitted be equal to $\lambda_0$?

An electron in the ground state of a hydrogen atom is revolving in a circular orbit of radius $R$. The orbital magnetic moment of the electron is ($m =$ mass of electron,$h =$ Planck's constant,$e =$ electronic charge).

The radius of the orbit of an electron in a Hydrogen-like atom is $4.5 a_0$,where $a_0$ is the Bohr radius. Its orbital angular momentum is $\frac{3h}{2\pi}$. It is given that $h$ is Planck constant and $R$ is Rydberg constant. The possible wavelength$(s)$,when the atom de-excites,is (are) :
$(A)$ $\frac{9}{32R}$ $(B)$ $\frac{9}{16R}$ $(C)$ $\frac{9}{5R}$ $(D)$ $\frac{4}{3R}$

In the Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at $10.2 \, V$. The wavelength of light emitted by the hydrogen atom when excited to the first excitation level is $\qquad$ $nm$.
(Given $hc = 1245 \, eV \cdot nm$, $e = 1.6 \times 10^{-19} \, C$).

The ratio of radii of $3^{\text{rd}}$ and $6^{\text{th}}$ Bohr's orbit in a hydrogen atom is

The potential energy of an electron in an orbit of a hydrogen atom is $-6.8 \text{ eV}$. The de Broglie wavelength of the electron in this orbit is (where $r_0$ is the Bohr radius).