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Atoms — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsAtoms5 Marks
Question
Explanation hydrogen spectrum on the basis of Bohr's principle.
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Answer

Explanation of Hydrogen spectrum on the basis of Bohr's principle:
In general the energy of electron in nth orbit of the hydrogen atom is given by: $E _n=-\left(13.6 / n^2\right) eV$.
$E_1=-\left(13.6 / 1^2\right) eV =-13.6 eV$
$E _2=-\left(13.6 / 2^2\right) eV =-1.51 eV$
$E_3=-\left(13.6 / 3^2\right) e V=-0.85 e V$
$E _4=-\left(13.6 / 4^2\right) eV =-0.54 eV$
$E _5=-\left(13.6 / 5^2\right) eV =-0.37 eV$
...................................................................
$E_{\infty}=-\left(13.6 / \infty^2\right) e V=0$
From these values it is obvious that the electron in hydrogen atom can have only certain discrete values of energy while revolving in different orbits. It is also obvious that the first orbit corresponds to the lowest energy and is called ground state. As $n$ increases, the energy becomes less negative i.e, goes on increasing. When $n$ becomes very large $(n \rightarrow \infty)$ the energy becomes zero. This zero level represents the state of maximum energy. The minimum energy required to remove an electron from n = 1 to n = ∞ is called ionisation energy of H2 atom and it is equal to difference of E1 and $E_{\infty}$ infty i.e., it is 13.6 eV. Zero energy state is called ionisation state of the atom. It is also obvious that with increasing value of $n$ the difference between the energy of different states goes on decreasing.
Image
Taking all the above points into account if a diagram is drawn in which the energies of the different orbits (energy states) of H2 atom are represented by parallel horizontal lines with some suitable energy scale, this diagram is called energy level diagram of H2 atom. In this diagram the transitions corresponding to different spectral series of hydrogen atom are also shown.
formula :
$\frac{1}{\lambda}= R \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ ...(1)
On the basis of this formula production of all the five spectral series can be explained as follows :
(1) Lymann Series : When the electron of hydrogen atom transits from some higher energy level to the first energy level (lowest energy level), (i.e., n1 = 1 and n2 = n = 2 ,3,4,....) then the emitted spectral lines are obtained in the ultraviolet region of the spectrum and make the Lymann series. Hence the wavelength of these lines can be found by the following form of the above equation (1),
$\frac{ 1 }{\lambda}= R \left(\frac{ 1 }{ 1 ^2}-\frac{ 1 }{ n ^2}\right), \quad$ where $n=2,3,4, \ldots$
Where R is Rydberg's constant and its value is 1.097$\times 10^7 m^{-1}$. Somewhere it is also denoted by RH.
The series lies in the ultra-violet. Lymann, in 1916, photographed the lines of this series of hydrogen spectrum. Hence this series is named 'Lymann series'. The longest wavelength of this series (for n = 2 ) is 1216 Å and the shortest wavelength (for n = ∞ ) is 912 Å. The wavelength (912 Å) corresponding to n = ∞ is called the 'series limit'.
(2) Balmer Series : Its wavelengths are given by
$\frac{1}{\lambda}= R \left(\frac{1}{2^2}-\frac{1}{n^2}\right),(n=3,4,5, \ldots)$ ...(2)
The lines of this series were seen and studied for the first time by Balmer in 1885. Hence this series is named as Balmer series. The longest wavelength of this series (for n = 3 ) is 6563 Å and the shortest wavelength (for n = ∞) is 3646 Å.
For n = 3 4, 5, 6, ...the spectral lines of Balmer series were termed as $H _\alpha, H _\beta, H _\gamma, H _\delta, \ldots$ and so on.
(3) Paschen Series : Its wavelengths are given by
$\frac{1}{\lambda}= R \left(\frac{1}{3^2}-\frac{1}{n^2}\right),(n=4,5,6, \ldots$ ...(3)
The series lies in the near infrared region.
This series was discovered by Paschen.
The greatest wavelength of this series is 18751 Å and the shortest is 8107 Å.
(4) Brackett Series : Its wavelengths are given by
$\frac{1}{\lambda}= R \left(\frac{1}{4^2}-\frac{1}{n^2}\right),(n=5,6,7, \ldots$ ...(4)
The series lies in the far infrared region.
This series was discovered by Brackett in 1922. The greatest wavelength of this series is 40500 Å and the shortest is 14516 Å.
(5) Pfund Series : Its wavelengths are given by
$\frac{1}{\lambda}= R \left(\frac{1}{3^2}-\frac{1}{n^2}\right),(n=6,7,8, \ldots)$ ...(5)
The series lies in far infrared region.
This series was discovered by Pfund is 1924. The largest wavelength of this series is 74000 Å and shortest is 23673 Å.

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