The third line of the Balmer series of an ion equivalent to a hydrogen atom has a wavelength of $108.5 \ nm$. The ground state energy of an electron of this ion will be.....$eV$

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

From a carbon nanotube of $1 \,\mu m$ length and $1 \,nm$ radius,$10$ electrons have been removed. Assume the resulting positive charge to be distributed uniformly over the surface of the tube. The energy of an electron moving in a stable circular orbit around the axis along the length of the tube is calculated by applying the Bohr model. Accordingly,the frequency of radiation required to excite an electron from its ground state to the next level is in the range of (charge of the electron,$e = 1.60 \times 10^{-19} \,C$; mass of the electron,$m_e = 9.11 \times 10^{-31} \,kg$; Planck's constant,$h = 6.63 \times 10^{-34} \,Js$; Permittivity of free space,$\varepsilon_0 = 8.85 \times 10^{-12} \,F/m$)

An electron revolving in the $n^{\text{th}}$ Bohr orbit has a magnetic moment $\mu_n$. If $\mu_n \propto n^x$,the value of $x$ is:

An excited hydrogen atom emits a photon of wavelength $\lambda$ in returning to the ground state. The quantum number $n$ of the excited state is ($R=$ Rydberg's constant).

If the difference between $(n + 1)^{th}$ Bohr radius and $n^{th}$ Bohr radius is equal to the $(n - 1)^{th}$ Bohr radius, then the value of $n$ is: