Explain the following according to Bohr's model of hydrogen:
$(i)$ Principal quantum number
$(ii)$ Radius of stationary orbit $(r)$
$(iii)$ Energy of stationary state
$(iv)$ Isoelectronic ion of $H$
$(v)$ Velocity of electron

(N/A) $(i)$ Principal quantum number: The stationary states for the electron are numbered $n = 1, 2, 3, \dots$. These integral numbers are known as Principal quantum numbers.
$(ii)$ Radius of stationary orbit $(r)$: The radii of the stationary states are expressed as $r_n = n^2 a_0$,where $a_0 = 52.9 \ pm$. The radius of the first stationary state $(n = 1)$,called the Bohr orbit,is $52.9 \ pm$. As $n$ increases,the value of $r$ increases,meaning the electron is further from the nucleus.
$(iii)$ Energy of stationary state: The energy of the stationary state is given by $E_n = -R_H (1/n^2)$,where $R_H = 2.18 \times 10^{-18} \ J$. For the ground state $(n = 1)$,$E_1 = -2.18 \times 10^{-18} \ J$. For $n = 2$,$E_2 = -0.545 \times 10^{-18} \ J$. When the electron is free from the nucleus $(n = \infty)$,the energy is $0 \ J$,representing an ionized hydrogen atom $(H^+)$.
$(iv)$ Isoelectronic ion of $H$: Hydrogen has $1$ electron. An isoelectronic ion must also have $1$ electron,such as $He^+$,$Li^{2+}$,or $Be^{3+}$.
$(v)$ Velocity of electron: The velocity of an electron in a stationary orbit is given by $v_n = v_0 (Z/n)$,where $v_0 = 2.188 \times 10^6 \ m/s$ for hydrogen.