For the wavelength of visible radiation of the Hydrogen spectrum, Balmer gave an equation as $\lambda = \frac{xm^2}{m^2 - 4}$, where $m$ is an integer value. The value of $x$ in terms of Rydberg's constant $R$ is

In the hydrogen spectrum,the wavelengths of light emitted in a series of spectral lines is given by the equation,$\frac{1}{\lambda}=R\left(\frac{1}{4^{2}}-\frac{1}{n^{2}}\right)$,where $n=5, 6, 7, \ldots$ and $R$ is Rydberg's constant. Identify the series and wavelength region.

Ionisation potential of hydrogen atom is $13.6 eV$. When hydrogen atoms in ground state are excited by a supply of $12.1 eV$,then the number of spectral lines emitted by hydrogen atoms according to Bohr's theory is

The wavelength of the first line of Balmer series of hydrogen atom is $\lambda$. What will be the wavelength of the same line in doubly ionized lithium?

Which of the following statements is true?

The difference in the wavelength between the maximum and minimum of Balmer series [Use $R_{H} = 1 \times 10^7 \ m^{-1}$]. (in $Å$)