$A$ hydrogen atom is excited from the ground state to a state with principal quantum number $n = 4$. The number of spectral lines emitted in the emission spectrum is:

Let $F_1$ be the frequency of the second line of the Lyman series and $F_2$ be the frequency of the first line of the Balmer series. Then,the frequency of the first line of the Lyman series is given by:

The shortest wavelength in the Lyman series of the hydrogen spectrum is $912 \ \mathring{A}$,corresponding to a photon energy of $13.6 \ eV$. The shortest wavelength in the Balmer series is about..... $\mathring{A}$.

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:

If a hydrogen atom is excited from the ground state to another state with a principal quantum number $n = 4$,the number of spectral lines in the emission spectrum will be:

Using the Rydberg formula,calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.