Given below are two statements

Statement I: When electric discharge is put on hydrogen, it emits discrete frequency in electromagnetic spectrum.

Statement II: Frequency of He⁺ ion of 2nd line of Balmer series is equal to first line of Lyman series.

The correct answer is: Both statement I and statement II are correct

Analysis of Statement I:
When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the resulting hydrogen atoms get excited to higher energy levels. As these excited electrons return to lower energy states, they emit electromagnetic radiation. Since the electronic energy levels of a hydrogen atom are quantized (discrete), the energy difference between any two levels is also discrete. Consequently, the emitted photons have specific, discrete frequencies, giving rise to a line spectrum rather than a continuous spectrum. Therefore, Statement I is correct.

Analysis of Statement II:
The frequency ($\nu$) of the spectral lines in hydrogen-like species is given by the Rydberg formula:
$\nu =Rc{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
where $R$ is the Rydberg constant, $c$ is the speed of light, $Z$ is the atomic number of the species, and $n_1$ and $n_2$ are the principal quantum numbers of the lower and higher energy orbits, respectively.

For the ${\text{He}}^{+}$ ion:
The atomic number is $Z=2$.
For the Balmer series, the lower energy level is $n_1=2$.
The 2nd line of the Balmer series corresponds to the transition from the $n_2=4$ state to the $n_1=2$ state.
Calculating the frequency for this transition:
${\nu }_{{\text{He}}^{+}}=Rc\cdot 2^2\cdot \left(\frac{1}{2^2}-\frac{1}{4^2}\right)$
${\nu }_{{\text{He}}^{+}}=4Rc\cdot \left(\frac{1}{4}-\frac{1}{16}\right)$
${\nu }_{{\text{He}}^{+}}=4Rc\cdot \frac{3}{16}=\frac{3}{4}Rc$

For the hydrogen atom (H):
The atomic number is $Z=1$.
For the Lyman series, the lower energy level is $n_1=1$.
The 1st line of the Lyman series corresponds to the transition from the $n_2=2$ state to the $n_1=1$ state.
Calculating the frequency for this transition:
${\nu }_{\text{H}}=Rc\cdot 1^2\cdot \left(\frac{1}{1^2}-\frac{1}{2^2}\right)$
${\nu }_{\text{H}}=Rc\cdot \left(1-\frac{1}{4}\right)=\frac{3}{4}Rc$

Comparing the two results:
${\nu }_{{\text{He}}^{+}}={\nu }_{\text{H}}=\frac{3}{4}Rc$
Since the two frequencies are identical, Statement II is correct.

Thus, both Statement I and Statement II are correct.