The frequency of one of the lines in the Paschen series of a hydrogen atom is $2.340 \times 10^{11} \ Hz$. The quantum number $n_2$ which produces this transition is

If in a hydrogen atom,an electron jumps from the $3^{rd}$ Lyman line to the $1^{st}$ Lyman line,then the obtained wavelength is:

For the Balmer series in the spectrum of $H$ atom,$\bar{v}=R_{H}\left\{\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right\}$,the correct statements among $(I)$ to $(IV)$ are:
$(I)$ As wavelength decreases,the lines in the series converge.
$(II)$ The integer $n_{1}$ is equal to $2$.
$(III)$ The lines of longest wavelength corresponds to $n_{2}=3$.
$(IV)$ The ionization energy of hydrogen can be calculated from wave number of these lines.

The frequency of a line in the Pfund series of a hydrogen atom is $2.340 \times 10^{14} \ Hz$. The value of the quantum number $n_2$ for this transition is .....

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Which of the following is correct for any $H^-$-like species?