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 $\text{Å}$)

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

The largest wavelength in the ultraviolet region of the hydrogen spectrum is $122 \ nm$. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is (in $nm$)

Which series of the hydrogen atom lie in the infrared region?

The wavelength of the first spectral line in the Balmer series of a hydrogen atom is $6561 \mathring A$. The wavelength of the second spectral line in the Balmer series of a singly-ionized helium atom is

Taking Rydberg's constant $R_H = 1.097 \times 10^7 \ m^{-1}$,the first and second wavelengths of the Balmer series in the hydrogen spectrum are: