Q.Brackett series is formed in Hydrogen atom spectrum, when the electrons from higher orbits move to ................. energy level

When an electron jumps from a higher energy level to a lower one inside a hydrogen atom, it does not do so gradually. It vanishes from the higher orbit and reappears in the lower one, and the energy difference is carried away by a single photon of light. That photon has a very specific wavelength, determined by the size of the energy gap.

If you collect all the light emitted by a sample of hot hydrogen gas and pass it through a prism, you do not see a continuous rainbow. Instead, you see a handful of sharp, coloured lines against a dark background. That pattern of lines is the hydrogen spectrum.

Why lines and not a smear?

The key is that the electron in a hydrogen atom can only occupy certain allowed orbits (energy levels), labelled by the principal quantum number $n = 1, 2, 3, \dots$. An electron in level $n$ has a fixed energy $E_n$, given by the Bohr formula:

The negative sign means the electron is bound to the nucleus; the deeper (more negative) the energy, the more tightly bound it is. The ground state ($n=1$) has $E_1 = -13.6$ eV, the first excited state ($n=2$) has $E_2 = -3.4$ eV, and so on.

When the electron falls from a higher level $n_i$ to a lower level $n_f$, the energy of the emitted photon is exactly the difference:

The photon's wavelength $\lambda$ is then given by $hc/\lambda = \Delta E$, or more commonly in spectroscopy:

where $R_H \approx 1.097 \times 10^7 \text{ m}^{-1}$ is the Rydberg constant for hydrogen.

Because $n_i$ and $n_f$ can only be integers, only certain wavelengths are possible. Each jump produces one spectral line.

The spectral series

Transitions that end on the same lower level form a series, named after their discoverers:

The Balmer series is the only one that falls in the visible range — that is why you see four bright lines (H$\alpha$, H$\beta$, H$\gamma$, H$\delta$) when you look at hydrogen through a spectroscope.

A concrete example: the first Balmer line (H$\alpha$)

Take the transition from $n_i = 3$ to $n_f = 2$:

$\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \cdot \frac{5}{36}$ …