Explain the hydrogen spectrum using Bohr's model.

The line spectrum observed in the case of a hydrogen atom can be explained quantitatively using Bohr's model.
Emitted Energy in the Spectrum $(\Delta E)$ : According to Bohr's postulate,radiation is emitted or absorbed when an electron moves between orbits.
The energy gap between the two orbits is given by the equation:
$\Delta E = E_{f} - E_{i}$
Since $E_{n} = -R_{H} \left( \frac{1}{n^{2}} \right)$,where $n = 1, 2, 3, \dots$
Substituting the values of $E_{n}$:
$\Delta E = -\left( \frac{R_{H}}{n_{f}^{2}} \right) - \left( -\frac{R_{H}}{n_{i}^{2}} \right)$
$\therefore \Delta E = R_{H} \left( \frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}} \right)$
$\therefore \Delta E = 2.18 \times 10^{-18} \ J \left( \frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}} \right)$
Frequency of the line spectrum $(\nu)$ :
Since $\Delta E = h\nu$,we have $\nu = \frac{\Delta E}{h}$.
Substituting the values:
$\nu = \frac{2.18 \times 10^{-18} \ J}{6.626 \times 10^{-34} \ J \cdot s} \left( \frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}} \right)$
$\therefore \nu = 3.29 \times 10^{15} \left( \frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}} \right) \ Hz$
Using this equation,the frequency of the emitted spectral line can be calculated.