An electron in the ground state of the hydrogen atom has an orbital radius of $5.3 \times 10^{-11} \ m$,while that for the electron in the third excited state is $8.48 \times 10^{-10} \ m$. The ratio of the de Broglie wavelengths of the electron in the ground state to that in the third excited state is:

$A$ hydrogen atom,a deuteron atom,a $He^+$ ion,and a $Li^{++}$ ion each contain a single electron orbiting their nucleus. The wavelengths of the electromagnetic radiation emitted during the transition of the electron from the $n = 2$ orbit to the $n = 1$ orbit are found to be $\lambda_1, \lambda_2, \lambda_3$,and $\lambda_4$ respectively. Then:

In the Bohr model of the $H$-atom,the kinetic energy of an electron in any orbit depends on the principal quantum number $n$ as:

$A$ monochromatic radiation of wavelength $\lambda$ is incident on a hydrogen sample in ground state. The sample subsequently emits radiation of six different wavelengths, then the value of $\lambda$ is. [Use $hc = 1242 \text{ eV-nm}$] (in $\text{ nm}$)

In the Bohr model,an electron moves in a circular orbit around the nucleus. Considering an orbiting electron to be a circular current loop,the magnetic moment of the hydrogen atom,when the electron is in the $n^{th}$ excited state,is ($e=$ electronic charge,$m_{e}=$ mass of the electron,$h=$ Planck's constant).

The de-Broglie wavelength of an electron moving in the $n^{\text{th}}$ Bohr orbit of radius $r$ is