Question

The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to the stationary state, would be:[Rydberg constant $= 1.097 \times {10^7}{m^{ - 1}}$ ]
(A) $91nm$
(B) $192nm$
(C) $406nm$
(D) $9.1 \times {10^{ - 8}}$

Answer 1

Rydberg got the improvised method of finding the wavelength in the hydrogen emission spectrum. the wavelength by the Rydberg can be given by:
$\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$
By putting the appropriate values, wavelength can be determined.

Complete step by step solution:
Firstly it is important to note here that after the balmer series Rydberg get the improvised method of finding the wavelength in the hydrogen emission spectrum.we can say that the formula given by him directly generalises the equations used to calculate the wavelength of the hydrogen spectral lines.
As we know that the wavelength by the Rydberg can be given by:
$\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$
Where $\lambda$ is the wavelength to be calculated , $n_1^2$ and $n_2^2$ be the state where ${n_1}$ is the state where the value is one as it is the stationary state. ${n_2}$ is infinity as stated in the question.
Further R is the rydberg constant whose value is also given.
Now putting the values in the above equation we get:
$\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right] \\\ \dfrac{1}{\lambda } = 1.97 \times {10^7}\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right] \\\ \dfrac{1}{\lambda } = 1.97 \times {10^7}{m^{ - 1}} \\\ \lambda = 9.1 \times {10^8}m \\\$
So we found the wavelength. As this is the wavelength of the radiation emitted when an electron moves between the infinity and the stationary energy state.
We can here conclude that the correct wavelength for the above given question turns out to be the option D.

Note:
Also it is interesting to note here that there was an integer relationship between the wavenumbers of the successive lines which helps in finding the wavelength and on this basis the formula is made.