An electron in the hydrogen atom excites from the $2^{nd}$ orbit to the $4^{th}$ orbit. The change in angular momentum of the electron is (Planck's constant $h = 6.64 \times 10^{-34} \ J \cdot s$).

If in Bohr's atomic model,it is assumed that the force between the electron and the proton varies inversely as $r^4$,the energy of the system will be proportional to

In a hydrogen-like atom with atomic number $Z$,an electron is in an excited state with principal quantum number $2n$. The maximum energy of a photon that can be emitted from this state is $204 \ eV$. If the electron transitions from the $2n$ orbit to the $n$ orbit,a photon with energy $40.8 \ eV$ is emitted. The value of $n$ is:

If Bohr's quantisation postulate (angular momentum $= n h / 2 \pi$) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?

Assuming the atom is in the ground state,the expression for the magnetic field at the nucleus in a hydrogen atom due to the circular motion of the electron is: $[\mu_0 \rightarrow \text{permeability of free space, } m \rightarrow \text{mass of electron, } \varepsilon_0 \rightarrow \text{permittivity of free space, } h \rightarrow \text{Planck's constant}]$

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 $\mathring{A}$?