Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is :-

$A$ hydrogen atom emits a photon corresponding to an electron transition from $n=5$ to $n=1$. If $R$ is the Rydberg constant, then the wavelength of the emitted photon is

Taking the wavelength of the first Balmer line in the hydrogen spectrum ($n = 3$ to $n = 2$) as $660\,nm$,the wavelength of the $2^{nd}$ Balmer line ($n = 4$ to $n = 2$) will be....$nm$.

Ratio of longest wavelength corresponding to Lyman and Balmer series in hydrogen spectrum 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

In the line spectra of the hydrogen atom,the difference between the largest and the shortest wavelengths of the Lyman series is $304\,\mathring{A}$. The corresponding difference for the Paschen series in $\mathring{A}$ is: