The series corresponding to minimum wavelength transition in H-atom is
A. Balmer series
B. Lyman series
C. Paschen series
D. Brackett series

Hint: Various spectral series of hydrogen are obtained based on the level between which transition of electron occurs. The lower level to which the electron jumps is used to identify the specific series for H-atom transitions.

Complete step by step answer:
A Hydrogen atom has a single electron in the ground state which can be excited to higher levels. To ionize the H-atom completely, energy of 13.6 eV is required. By ionization we mean that we take the electron from the ground state of the H-atom and completely pull it out of the atom so that the electron lies somewhere at infinity with respect to the atom.
The binding energies of electron to the nucleus in various orbits can be known with the help of formula:
$E = - \dfrac{13.6}{n^2}$.
This tells us that as the number of orbits increases the energy decreases.
We know that minimum wavelength implies maximum energy and maximum frequency. And we can see with the help of the given formula here that for the lowest level i.e., n = 1 we can find maximum energy. Therefore, any electron jumping to the lowest level can have maximum energy and hence minimum wavelength.
Now, we must know that n = 1 transitions correspond to Lyman series. Therefore, Lyman series transitions should give us minimum wavelength.

Hence, the correct answer is option (A).

Note:
Since the upper level can be any bound state of the hydrogen atom we need not consider that and besides series names are given based on the lower levels only. Here it can be a little confusing to understand what minimum wavelength relation is. There is also a Rydberg formula that gives direct wavelengths for hydrogen atom spectra.