Neils Bohr (1913) was the first to explain quantitatively the general features of the structure of hydrogen atom and its spectrum.
He used Planck's concept of quantisation of energy.
The main postulates are:
In an atom electron revolve around the nucleus in certain definite circular paths known as orbit or energy shells.
Each orbit is associated with definite energy hence known as energy levels or shells. These are numbered as 1, 2, 3,…. or K, L, M ….These integral numbers are known as Principal Quantum numbers
Only these orbits are permitted for electron which angular momentum (L = m vr) is whole no. multiple of $ \frac {h}{2 \pi}$ (where n = 1, 2, 3, …)
$L=n \frac {h}{2 \pi}$
$m v r = n\frac {h}{2 \pi}$
where, m = Mass of e = 9.1 × 10^{-31}Kg
v =revolving electron velocity
r = radius of orbit
As long as electron present in a particular orbit, it neither absorbs nor looses energy. Therefore energy remains constant and these orbits are called stationart states.It does not means the electrons are stationary but just their energy is fixed
When energy is supplied to electron it absorbs the energy & jumps to higher level & when energy is released it jumps back to the lower level. In doing so it emits energy
The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy
by $\Delta E$, is given by:
$\nu = \frac {\Delta E}{h}= \frac {E_2 - E_1}{h}$
Where E1 and E2 are the energies of the lower and higher allowed energy states respectively. This expression is commonly known as Bohr's frequency rule
The radii of the stationary state orbit is given by
R_{n} =n^{2} a_{0} Where n is the orbit number and a_{0} =52.9 pm
The energy of the stationary state is given by
$E_{n}=-2.18 \times 10^{-18}(\frac {1}{n^2})$ Joule
Here negative sign states that energy of the electron in a atom is less that energy of a free electron.A free electron at rest is infinitely away from the nuclues and is assigned a energy level 0. We can get this value by putting infinity in the above equation
Bohr’s theory can also be applied to the ions containing only one electron, similar to that present in hydrogen atom. For example, He+ ,Li2+, Be3+ and so on. The energies of the stationary states associated with these kinds of ions (also known as hydrogen like species) are given by the expression.
$E_{n}=-2.18 X 10^{-18}(\frac {Z^2}{n^2}) Joule $
And the radius of the orbit is given by
$r_n= \frac {52.9 n^2}{Z^2} pm $
The emission or absorption of energy can occur only when electrons jump from one energy level to another. The energy released or absorbed is quantised and can be calculated as follows
$\Delta E = E_f - E_i= -2.18 \times 10^{-18}(\frac {1}{n_f^2}) - [-2.18 \times 10^{-18}(\frac {1}{n_i^2})]$
$=2.18 \times 10^{-18} (\frac {1}{n_i^2} - \frac {1}{n_f^2})$
In terms of frequency of absorpa=tion or emission
$h \nu = 2.18 \times 10^{-18} (\frac {1}{n_i^2} - \frac {1}{n_f^2})$
$\nu = 3.29 \times 10^{15} (\frac {1}{n_i^2} - \frac {1}{n_f^2})$
In terms of wave number
$\bar{\nu} = \frac {\nu}{c} = 1.09677 \times 10^{7} (\frac {1}{n_i^2} - \frac {1}{n_f^2}) \; m^{-1}$
Here is calculated energies with energy level for hydrogen atom
In case of absorption spectrum, $n_f > n_i$ and energy is absorbed. On the other hand in case of emission spectrum $n_i > n_f$ , $\Delta E$ is negative and
energy is released.
Energy levels are not equally spaced. The energy difference is decreasing as we increase the n. The energy of the electron increases with increase in its distance from nucleus
Achievements of Bohr Theory:-
He explained atomic spectrum of hydrogen atom.
He explained stability of atom.
Bohr theory helped in calculating energy of electron in hydrogen atom & hydrogen like atoms. (i.e. species)
$E =\frac {-2r^{2} me^{4}Z^{2}}{n^2r^2}$
$E =\frac {-1312 Z^2}{n^2}$ unit – Kila joule/ mole
$E_{n}=-2.18 X 10^{-18}(\frac {1}{n^2}) $ unit – Joule/ atom
Limitations of Bohr Theory
It could not explain the spectrum of atoms containing more than 1 electron or multi electrons
Bohr Theory failed to explain fine structure of spectral lines.
It could not explain Zeeman effect & stark effect.
This theory failed to explain ability of atoms to form molecule by chemical bonds.
It was not in accordance with Heisenberg uncertainty principle.
Important Energy Conversion Units
$1 \; eV= 1.6020 \times 10^{-19} \; J$
$ 1 \; ergs = 10^{-7} \; j$
So, Energy state can also written as
$E=- \frac {13.6}{n^2}$ eV
Solved Example
Question 1
Calculate the energy associated with with first orbit in He^{+} ion.Also what is the radius of the orbit? Solution
$E_{n}=-2.18 \times 10^{-18}(\frac {Z^2}{n^2}) Joule $
Here n=1 and Z=2
E=8.72 X 10^{-18} Joule
And the radius of the orbit is given by
$r_n= \frac {52.9 n^2}{Z^2} pm $
So
=13.225 pm
Question 2
What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom? Solution
$\Delta E=2.18 \times 10^{-18} (\frac {1}{n_i^2} - \frac {1}{n_f^2})$
Here $n_i=5$ and $n_f=5$
Therefore
$\Delta E = 2.18 \times 10^{-18} (\frac {1}{5^2} - \frac {1}{2^2})$
$=-4.58 \times 10^{-19}$ J
It is an emission energy
The frequency of the photon (taking energy in terms of magnitude) is given by
$\nu = \frac {\Delta E}{h}$
$\nu = \frac {4.58 \times 10^{-19}}{6.626 \times 10^{-34}}= 6.91 \times 10^{14}$ Hz
Wavelength will be given by
$ \lambda = \frac {c}{\nu} = 434$ nm
Question 3
The energy of an electron in the first Bohr orbit of Hydrogen atom is -13.6eV. The possible energy value(s) of the excited state(s) for electrons in Bohr orbits to hydrogen is(are)
(i) -3.4eV
(ii) -4.2eV
(iii) -6.8eV
(iv) +6.8eV Solution
Values of energy can be
$E= \frac {-13.6}{n^2}$ eV where n=1,2,3
Subsituting n=2, we get
E=-3.4eV
Hence the answer is (i)
Question 4
Which of the following electron transition in a hydrogen atom will require the largest amount of energy ?
(i) From n=1 to n=2
(ii) From n=2 to n=3
(iii) From $n=\infty$ to n=1
(iv) From n=3 to n=5 Solution
Now Energy level is given by
$E_n=- \frac {13.6}{n^2}$ eV
For option 1:
$\Delta E=\frac{-13.6}{4}-\frac{-13.6}{1}=10.2 $ eV
For option 2:
$\Delta E=\frac{-13.6}{9}-\frac{-13.6}{4}=1.89 $ eV
For option 3:
$\Delta E=\frac{-13.6}{1}-\frac{-13.6}{\infty }=-13.6 $ eV
For option 4:
$\Delta E=\frac{-13.6}{25 }-\frac{-13.6}{9}=0.967 $ eV