The difference between the $n^{th}$ and $(n+1)^{th}$ Bohr radius of an $H$ atom is equal to its $(n-1)^{th}$ Bohr radius. The value of $n$ is:

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:

If the radius of the first Bohr orbit in a hydrogen atom is $0.53 \, \mathring A$,then the radius of the third Bohr orbit will be ....... $\mathring A$.

The potential energy of an electron in an orbit of a hydrogen atom is $-6.8 \text{ eV}$. The de Broglie wavelength of the electron in this orbit is (where $r_0$ is the Bohr radius).

The de-Broglie wavelength of an electron in the $n^{th}$ Bohr orbit is $\lambda_n$ and the angular momentum is $J_n$,then:

The difference between the radii of $n^{\text{th}}$ and $(n+1)^{\text{th}}$ orbits of a hydrogen atom is equal to the radius of the $(n-1)^{\text{th}}$ orbit of hydrogen. The angular momentum of the electron in the $n^{\text{th}}$ orbit is $.........$ ($h$ is Planck's constant).