(A) The two important points of Bohr's model that can be used to derive the given formula are as follows:
$(i)$ Electrons revolve around the nucleus in a circular path of fixed radius and energy. These paths are called orbits,stationary states,or allowed energy states.
$(ii)$ Energy is emitted or absorbed when an electron moves from a higher stationary state to a lower stationary state or from a lower stationary state to a higher stationary state,respectively.
Derivation:
The energy of an electron in the $n^{\text{th}}$ stationary state is given by: $E_{n} = -R_{H} \left( \frac{1}{n^{2}} \right)$,where $R_{H}$ is the Rydberg constant $(2.18 \times 10^{-18} \ J)$.
The energy change $(\Delta E)$ when an electron transitions from an initial orbit $(n_{i})$ to a final orbit $(n_{f})$ is:
$\Delta E = E_{f} - E_{i} = -R_{H} \left( \frac{1}{n_{f}^{2}} \right) - \left( -R_{H} \frac{1}{n_{i}^{2}} \right) = R_{H} \left[ \frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}} \right]$
Since $\Delta E = h\nu$,the frequency $\nu$ is:
$\nu = \frac{\Delta E}{h} = \frac{R_{H}}{h} \left[ \frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}} \right]$
The wavenumber $\bar{\nu}$ is defined as $\bar{\nu} = \frac{\nu}{c} = \frac{\Delta E}{hc}$. Substituting the values:
$\bar{\nu} = \frac{R_{H}}{hc} \left[ \frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}} \right]$
Using $R_{H} = 2.18 \times 10^{-18} \ J$,$h = 6.626 \times 10^{-34} \ J \ s$,and $c = 3 \times 10^{10} \ cm \ s^{-1}$,the constant $\frac{R_{H}}{hc} \approx 109677 \ cm^{-1}$,which is the Rydberg constant for wavenumber $(R)$.
Thus,$\bar{\nu} = 109677 \left[ \frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}} \right] \ cm^{-1}$.