The angular momentum of the electron in the third Bohr orbit of hydrogen atom is $l$. Its angular momentum in the fourth Bohr orbit is

The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum is

According to Bohr's theory,the expressions for the kinetic and potential energy of an electron revolving in an orbit are given respectively by:

The de Broglie wavelength of the electron in the ground state is $\lambda_1$ and that in the $n = 3$ level is $\lambda_3$. Then $\lambda_3$ is given by:

The de-Broglie wavelength $(\lambda_B)$ associated with the electron orbiting in the second excited state of a hydrogen atom is related to that in the ground state $(\lambda_G)$ by:

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,$\varepsilon_0 = 8.85 \times 10^{-12} \,F/m$)