The lines in the Balmer series have their wavelengths lying between

Taking the wavelength of the first Balmer line in the hydrogen spectrum ($n = 3$ to $n = 2$) as $660\,nm$,the wavelength of the $2^{nd}$ Balmer line ($n = 4$ to $n = 2$) will be....$nm$.

If the wavelengths of the first Lyman line for Hydrogen,$He^+$,and $Li^{2+}$ ions are $\lambda_1$,$\lambda_2$,and $\lambda_3$ respectively,then the ratio of these wavelengths is:

Which of the following statement$(s)$ is(are) correct about the spectrum of hydrogen atom?
$(A)$ The ratio of the longest wavelength to the shortest wavelength in Balmer series is $9/5$.
$(B)$ There is an overlap between the wavelength ranges of Balmer and Paschen series.
$(C)$ The wavelengths of Lyman series are given by $\lambda = \frac{\lambda_0}{1 - 1/m^2}$,where $\lambda_0$ is the shortest wavelength of Lyman series and $m$ is an integer.
$(D)$ The wavelength ranges of Lyman and Balmer series do not overlap.

$A$ hydrogen atom in the ground state is excited by monochromatic radiation of $\lambda = 975 \; \mathring{A}$. The number of spectral lines in the resulting emission spectrum will be:

The ratio of the wavelengths of radiation emitted when an electron in the hydrogen atom jumps from the $4^{th}$ orbit to the $2^{nd}$ orbit and from the $3^{rd}$ orbit to the $2^{nd}$ orbit is: