24.THE ATOMIC STRUCTURE – The Bohr Model(2).

We continue our discussion on the Bohr model in this post and try to see its applications and implications . Neils Bohr , with his assumptions/postulates, derived formulas for the energy of electrons in specific orbits, velocity of electrons and the radius of the orbits.241.jpg

So,how are these formulae useful? What are the applications of the  Bohr model?

1)Bohr’s model allows us to calculate the radii of the orbits, that are allowed for an electron, to travel in an atom.

e.g. – For a H -atom , n= 1 and Ƶ = 1.  ∴r1 = 0.529 ×(12 / 1) = 0.529 Å ≈ 1/2 an angstrom
For He+atom, n=1 and   Ƶ = 2. ∴r1 = 0.529 ×(12 / 2) = 0.2645 Å≈ 1/4th of an angstrom.

Thus, as Ƶ ↑ , r(n) ↓ .. This is true as more the charge in the nucleus , more is the electrostatic pull experienced by the electron in the orbit towards the nucleus and thus small radius.

2)It also allows us to calculate energy in these orbits.Thus, we can theoretically calculate the energy of an electron by  just  plugging in  values of ‘n’ and ‘Ƶ’ for any one electron atom  in gas phase. 

e.g. – The energy for ground state electron of hydrogen atom can be calculated just by plugging in the necessary values, n= 1 and Ƶ = 1 and R=2.18x 10-18 J,

∴ E(1) = – RH × (12 / 12 ) = -2.18x 10-18 J/atom ⇒Energy of ground state electron in H atom. 

To find energy per mole, we simply divide the quantity by Avogadro’s number → 6.023x 1023.

∴ E(1) =-2.18x 10-18 J/atom × 6.023x 1023= -1.312 MJ/mol ⇒Energy of ground state electrons in 1 mole of H atom.(Generally, we never study energy of a single atom separately.So the energy calculated is for one mole of hydrogen atoms).
Thus, energy of hydrogen atom(gas phase) in ground state is 1.312MJ/mol.

Thus, we can theoretically calculate the energy of an electron by  just  plugging in  values of ‘n’ and ‘Ƶ’ for any one electron atom  in gas phase. 

We can thus draw an energy level diagram (diagram which shows energies of different orbits) for chemical species using the Bohr Model.

e.g. – Hydrogen atom (gas phase) energy level diagram is as follows –


In the above diagram, the energy of the electron in the ground state is -13.6 eV  we have already calculated this quantity).We can see that energy (E) is a function of the integer/orbit number ‘n’ i.e as ‘n’ changes, E also changes. Thus , we can also represent energy as E(n) .
e.g.–  We calculated that for hydrogen, E(1) = -2.18x 10-18 J/atom.
1.6 × 10-19 J = 1eV
∴-2.18x 10-18 J = (-2.18x 10-18 / 1.6×10-19 ) =- 13.6 eV.

Thus, the energy  of an electron in the ground state  of hydrogen atom= -13.6eV .

3) The Bohr model helps us calculate ionisation energy for certain species and shows us how cations are formed.

We know, that the energy of the ground state electron in H atom (gas phase) is -13.6 eV.So, to remove the electron from the atom (i.e to ionise it /to get it to n=∞) , minimum energy required is +13.6 eV.This is the ionisation energy – energy required to remove the ground state H atom electron from the atom.

∴Ionisation energy of atomic hydrogen = +13.6 eV.


4) This model can satisfactorily and quantitatively explain electronic transitions that take place in an atom.


When an electron changes its orbit it either absorbs or emits radiation.We can study this absorbed/emitted radiation and get an absorption/emission spectra,respectively, for a species. Bohr Model helps us to understand this phenomenon. The transition of an electron can be represented by the following formula –

The energy of the absorbed /emitted radiation = Efinal – Einitial
                                                                                           = – RH Ƶ2 (1/n2final – 1/n2initial)
ΔE  = – RH Ƶ2 (1/nf 2– 1/ni2)…(1)

Here, ni= orbit number where electron was found initially
nf= orbit number where electron goes finally.

If an electron in ground state H atom absorbs energy and gets excited to the next level This electron in its excited state is unstable , so it will fall down to its ground state i.e will go from n=2 to n=1.When an electron drops from a higher level to a lower level it sheds the excess energy, a positive amount, by emitting radiation /a photon.

According to the law of conservation of energy ,

Energy of the emitted photon = ΔEtransition= E2 – E1 = – RH Ƶ2 [1/(2) 2– 1/(1)2)]

Note – Because an electron bound to an atom can only have certain energies the electron can only absorb photons of certain energies.


5) It helps us to calculate the velocity of an electron in an orbit.

e.g. – For atomic H ,  n= 1 and Ƶ = 1, v = (h/2πma0) × Ƶ/n = 2.18 x 106 m/s.

We know that the speed of light c = 3 ×108m/s.

[Why is ‘c’ used as the symbol for speed of light ? This symbol has its origin in  the latin word ‘celeritas’  which means speed.So, words with ‘cele’ in them have to do something with speed. e.g.acceleration,deceleration].

So, speed of the electron moving in circular orbits around the nucleus is around than 1% of the speed of light. To understand this speed better, let us convert the speed into km/sec.The speed of light is 2180 kilometres per sec!!  Just imagine we drive cars at the speed of 100 kms per hour ! And the electron zooms around the nucleus at a speed of 2180 kilometres per second! So an electron could get around the earth in just 18 seconds!

The Bohr’s  atomic model was just a theoretical model constructed on basis of some postulates or assumptions. How do we know if it is true? Any theory can only be validated by experimental data. So did any experimental data corroborate this theory?  The answer is YES !


Angstrom’s painting.







Long before this theory was put forth by Neils Bohr (1913) , a Swedish scientist named Ångström,in 1853, had studied the emissions from gas discharge tube containing hydrogen gas.


Apparatus –
1) Gas discharge tube – made of borosilicate glass.
2)Two electrodes.
3)Variable voltage source , whose voltage can be varied .
4)vacuum pump

Ångström sealed hydrogen gas in a gas discharge tube  at low pressure.Two electrodes were sealed into the tube and they were connected to a variable voltage source.The electrode connected to the negative terminal was the cathode and the one connected to the positive terminal was the cathode.He started varying the volatge between the electrodes. At a high voltage value, he saw a set of 4 distinct lines(through a prism) as follows –

Line color









410 nm

He studied these lines and found out that these lines were exclusive to the hydrogen atom.As at that time the structure of the atom was not known in greater detail, the appearance of these lines couldnot be explained correctly.Almost after 50 years ,in 1913,the Bohr model however, was able to satisfactorily explain the phenomenon.The concept of quantised orbits explained the occurence of the specific lines.Why do we get these characteristic lines?

As the voltage in the experiment is raised, at a particular high voltage value, the electrons from the cathode start to boil off the surface of the cathode and go zooming through the discharge tube towards the anode (as electrons get attracted to the positive charge).These electrons which are ejected from the cathode may be termed as ‘ballistic electrons'(shown in red color in the above figure). These electrons however , at certain occasions encounter an H – atom in their path and collide with it. These collisons imparts energy to the electrons present inside the H-atom and make them excited (as the ballistic electron transfers energy to the H atom electrons).So, the electron inside the H- atoms will go to higher energy levels. But as we know that the excited state is not a stable state, so the excited electrons fall back down to the ground state , thus emitting photons.These emitted photons  have different energies depending on the various electronic transitions taking place inside of the H atom.

So, conceptually the theory successfully explains the discrete lines of hydrogen atom but scientific data has to be quantitative as well. So, what was the quantitative proof of the theory being correct?The story goes as follows –

As we know Ångström,in 1853, saw the discrete lines at particular wavelengths namely(656nm,486nm,434nm,410 nm)from a hydrogen gas discharge tube as follows –


220px-balmerIn 1885, J.J.Balmer, a Swiss mathematician started studying   Ångström’s experimental data.He was trying to find a pattern in the numbers 656,486,434 and 410.he successfully came up with an equation to represent those lines. He took a reciprocal of the wavelengths i.e wavenumber, and derived an equation as follows –247And in the above equation the value of the constant R = 1.1 × 107 m -1.

We already know that,

The value of the Rydberg constant RH is 1.097 × 10 7 per metre and that that the value of Bohr’s constanst K =RH . So , we can conclude that ,

RBalmer=KBohr =RH

All the three constants are equal . So, now quantitatively, we have a proof  from Balmer’s interpretation of Ångström’s experimental data that the Bohr Model is correct!!

The following is the Rydberg Equation is an general equation for one electron atom, 248.jpg


1)He+ electron goes from n= 4 to n=2.What is the wavelength of the photon emitted ?


We know,rydberg-formula

∴1/λ = (1.1 × 107 m -1) (2)2[(1/(2)2– 1/(4)2]

          =(1.1 × 107 m -1) (4) [(1/4) – (1/16)]

          =(1.1 × 107 m -1)(4) (0.1875)

     1/λ =0.825× 107 .

∴ λ = 1/ (0.825× 107)
λ = 1.21 × 10-7 m.

We shall discuss more about hydrogen spectra with respect to the Bohr model in our next post.Till then,

Be a perpetual student of life and keep learning !

Good Day !

Further Reading and References-


2)Lecture 4 ,MIT, 3.091SC Introduction to Solid State Chemistry, Fall 2010 by Professor Donald Sadoway.



Image sources –




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