The shortest wavelength in the Balmer series of the hydrogen atom spectrum is approximately equal to (use $R_{H} = 1.097 \times 10^7 \ \text{m}^{-1}$) (in $\text{Å}$)

$A$ hydrogen atom is excited from the ground state to the energy level $n = 3$. According to Bohr's model,the number of spectral lines emitted is:

The wavelength of radiation emitted is $\lambda_0$ when an electron jumps from the second excited state to the first excited state of a hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom,the wavelength of the radiation emitted will be $\frac{20}{x} \lambda_0$. The value of $x$ is

Find the ratio of the wavelength of the first line of the Lyman series for a doubly ionized lithium atom $(Li^{2+})$ to the wavelength of the first line of the Lyman series for deuterium $(D)$.

In the hydrogen spectrum,the shortest wavelengths of the Lyman and Balmer series are $\lambda_1$ and $\lambda_2$ respectively. The Rydberg constant of hydrogen is

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