$A$ hydrogen atom absorbs radiation of wavelength $975 \, \mathring{A}$ and transitions from the ground state to an excited state. How many spectral lines are possible in the emission spectrum?

If an electron in a hydrogen atom jumps from the $3^{rd}$ orbit to the $2^{nd}$ orbit,it emits a photon of wavelength $\lambda$. When it jumps from the $4^{th}$ orbit to the $3^{rd}$ orbit,the corresponding wavelength of the photon will be

In the Balmer series,the wavelength of the $2^{\text{nd}}$ line is $\lambda_1$ and for the Paschen series,the wavelength of the $1^{\text{st}}$ line is $\lambda_2$. Then the ratio $\lambda_1 : \lambda_2$ is:

If $\Delta \lambda_L$ is the difference between the shortest and longest wavelengths of the Lyman series and $\Delta \lambda_B$ is the difference between the shortest and longest wavelengths of the Balmer series,then $\frac{\Delta \lambda_B}{\Delta \lambda_L} = $

The wavelength of the first spectral line of the Lyman series of the hydrogen spectrum is

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.