What is the ratio of the wavelengths of radiations emitted when an electron in a hydrogen atom jumps from the fourth orbit to the second orbit and from the third orbit to the second orbit?

The wavelength of the first line of the Lyman series for a hydrogen atom is equal to that of the second line of the Balmer series for a hydrogen-like ion. The atomic number $Z$ of the hydrogen-like ion is:

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:

For a hydrogen atom,the ratio of the largest wavelength of the Lyman series to that of the Balmer series is:

The transition from the state $n = 4$ to $n = 3$ in a hydrogen-like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from:

$A$ doubly ionised $Li$ atom is excited from its ground state $(n = 1)$ to $n = 3$ state. The wavelengths of the spectral lines are given by $\lambda_{32}, \lambda_{31}$ and $\lambda_{21}$. The ratios $\lambda_{32}/\lambda_{31}$ and $\lambda_{21}/\lambda_{31}$ are,respectively: