In accordance with the Bohr's model, find the quantum number that characterises the earth's revolution around the Sun in an orbit of radius $1.5\times {10}^{11} \mathrm{m}$ with orbital speed $3\times {10}^4 {\mathrm{ms}}^{-1}$.(Mass of the earth$=6\times {10}^{24} \mathrm{kg}$)

According to Bohr hypothesis the angular momentum of the electron in any stationary orbit or radius $\mathrm{r}$ is proportional to :

The ratio of energies of electron in second excited state to the first excited state in $\mathrm{H}-$atom is:

$13.6 \mathrm{eV}$ energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius of the corresponding electron. $\left(\frac{1}{4\pi {\epsilon }_o}=9\times {10}^9 \mathrm{N} {\mathrm{m}}^2/{\mathrm{C}}^2\right)$