The wavelength of the energy emitted when an electron transitions from the fourth orbit to the second orbit in a hydrogen atom is $20.397 \, cm$. The wavelength of the energy for the same transition in $He^+$ is .......... $cm$.

Using the formula for the radius of the $n^{th}$ orbit $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$,derive an expression for the total energy of an electron in the $n^{th}$ Bohr orbit.

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) :
$(A)$ $\frac{9}{32R}$ $(B)$ $\frac{9}{16R}$ $(C)$ $\frac{9}{5R}$ $(D)$ $\frac{4}{3R}$

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

The first four spectral lines in the Lyman series of a $H$ atom are $\lambda = 1218 \, \mathring{A}, 1028 \, \mathring{A}, 974.3 \, \mathring{A}, 951.4 \, \mathring{A}$. If instead of Hydrogen,we consider Deuterium,calculate the shift in the wavelength of these lines.

The electron of a hydrogen atom makes a transition from the $(n + 1)^{th}$ orbit to the $n^{th}$ orbit. For large $n$,the wavelength of the emitted radiation is proportional to: