Find the maximum wavelength of the Brackett series for a hydrogen atom in $\mathring A$.

In hydrogen atoms,the transition from the state $n=6$ to $n=1$ results in ultraviolet radiation. Infrared radiation will be obtained in the transition:

Match List $I$ with List $II$.

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

Choose the correct answer from the options given below:

The wavelengths corresponding to the first four spectral lines of the Lyman series of the $H$-atom are $\lambda = 1218 \, \mathring{A}, 1028 \, \mathring{A}, 974.3 \, \mathring{A}$,and $951.4 \, \mathring{A}$. Now,consider a deuterium atom instead of a hydrogen atom. Given the mass of the hydrogen atom is $1.6725 \times 10^{-27} \, kg$,the mass of the deuterium atom is $3.3374 \times 10^{-27} \, kg$,and the mass of the electron is $9.109 \times 10^{-31} \, kg$,calculate the percentage change in the wavelength of the first spectral line of the Lyman series for the deuterium atom relative to the hydrogen atom.

The minimum wavelength of Lyman series lines is $P$. What is the maximum wavelength of these lines?

When an electron in a hydrogen atom jumps to the innermost orbit,the radiation emitted belongs to which one of the following series?