The speed of the electron in a hydrogen atom in the $n=3$ level is (Planck constant $= 6.6 \times 10^{-34} \ J \ s$):

The figure shows standing de-Broglie waves due to the revolution of an electron in a certain orbit of a hydrogen atom. Then,the expression for the orbit radius is (All notations have their usual meanings).

If one takes into account the finite mass of the proton,then the correction to the binding energy of the hydrogen atom is approximately (take,mass of proton $= 1.60 \times 10^{-27} \, kg$ and mass of electron $= 9.10 \times 10^{-31} \, kg$) (in $\%$)

In a hydrogen-like ion,the energy difference between the $2^{\text{nd}}$ excitation state and the ground state is $108.8 \ eV$. The atomic number of the ion is:

Using the formula for the radius of the $n^{th}$ orbit $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$,derive an expression for the total energy of an electron in the $n^{th}$ Bohr orbit.

In the hydrogen atom, the electron makes a transition from the higher orbit $(i)$ to a lower orbit $(f)$. The ratio of the radius of the orbits is given by $r_i : r_f = 16 : 4$. The wavelength of the photon emitted due to this transition is . . . . . . nm. (Given Rydberg constant $R = 1.0973 \times 10^7 \text{ m}^{-1}$)