The radius of an electron in the ground state of hydrogen is $1.06 \, \mathring{A}$. What will be the radius of the electron in the $10^{th}$ orbit in $\mathring{A}$?

The ratio of angular momentum of an electron in $n^{\text{th}}$ orbit of a hydrogen atom to the velocity of the electron in the $n^{\text{th}}$ orbit is proportional to

Hydrogen $(H)$,deuterium $(D)$,singly ionized helium $(He^+)$ and doubly ionized lithium $(Li^{2+})$ all have one electron around the nucleus. Consider the $n = 2$ to $n = 1$ transition. The wavelengths of emitted radiations are $\lambda_1, \lambda_2, \lambda_3$ and $\lambda_4$ respectively. Then approximately:

If in a hydrogen atom,the radius of the ${n^{th}}$ Bohr orbit is ${r_n}$ and the frequency of revolution of the electron in the ${n^{th}}$ orbit is ${f_n}$,choose the correct option.

In a hydrogen atom,the radius of the $n^{th}$ Bohr orbit is $r_n$. The graph between $\log \left( \frac{r_n}{r_1} \right)$ and $\log n$ will be:

The de Broglie wavelength of an electron in the first orbit of Bohr's model is . . . . . . .