23.THE ATOMIC STRUCTURE -The Bohr Model(1).

In this post we shall start talking about another great physicist whose contribution to the atomic theory is unparalleled  – Niels Bohr.

220px-Niels_Bohr
Niels Bohr

 

Niels Bohr was born in Copenhagen on 7th October 1885.His father was a professor of physiology.It was his father who was responsible for kindling his interest in Physics.In 1911 , he spent 6 months with sir J.J.Thompson in Cavendish Lab and later in 1912 he spent 3 months working under Ernest Rutherford in Manchester.It was in Manchester where he proposed a quantitative model supporting Rutherford’s structure of atom  – Bohr model ! He was a theoretical physicist and so he devised some postulates to explain this atomic model.He devised his theory for one electron system in gas phase.A one electron system is an atom having only one electron e.g.H,He+ , Li2+ etc. It doesn’t matter how many protons are present in the nucleus of the atom, there should be only a single electron revolving around the nucleus. Like for example ,Li2+ has three protons but has only one electron (when Li donates two electrons ,Li2+ is formed).

Postulates/Assumptions of the Bohr model

1] Electrons move in fixed orbits and energy of the electron depends on size of the orbit – Rutherford model of atom is CORRECT !The protons occupy the centrally situated nucleus and the electrons revolve around it in fixed,concentric circular paths/orbits. Every orbit  is associated with a certain amount of energy and so these are also called as energy levels.These energy levels were given the numbers 1,2,3,4 .. or K,L,M,N .. (I shall explain why these alphabets were used in some later post).

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Bohr’s electron orbits / Energy levels.

 

2]The energy levels are quantized – The electrons can travel in certain discrete set of orbits only, at set of distances from the nucleus with specific energies.Thus, energy levels in an atom are quantized i.e they can only take certain discrete values of energy.The electron can occupy only these specific energy levels and cannot appear anywhere between these levels.

To understand quantisation , we could think of a ladder.

2391

Or we could think of currency in any country. For example , in India , we have rupee notes which are quantised. We have a 5,10 , 20 , 50 , 100 ,500 or 1000 rupees note. So, the notes can only take certain discrete values. One cannot have a 14.7 or 800.5 rupee note ! Similarly, electrons can occupy only certain specific orbits which satisfy the quantum condition.

THE QUANTUM CONDITION

The energy of an electron in an orbit is quantized through its angular momentum i.e the electrons can occupy  only those orbits , whose angular momentum is an integral multiple of the quantity h/2π .

234.jpg

 

Bohr  took Planck’s theory and proposed that the stationary orbits are only those in which angular momentum of an electron (revolving in that orbit) , is an integral multiple of  the quantity h/2π .

So, Angular Momentum , L = mvr = n (h/2π).

where,
m = mass of the electron
v = velocity of the revolving electron
n= integer (integral multiple) and n=1,2,3….etc.
h=Planck’s constant =6.626 × 
10-34 Js
π = 3.14(constant)
Thus, the angular momentum of an electron can have only certain discrete values.

3]Electromagnetic radiation occurs only when electron moves from one orbit to other – As long as the electron stays in a given orbit it does not radiate energy thus, its energy remains constant in a particular orbit.The energy of the electron changes only when it changes its energy levels i.e either it goes from a high energy level to a low energy level or vice versa.

 

 

4]Classical electromagnetic(EM) theory is NOT applicable to orbiting electron.

5]Newtonian mechanics is applicable to orbiting electron.

6] Energy of the orbiting electron , Eelectron = Ekinetic + Epotential.
Only the energy of the electron is considered as it is the electron which is moving and not the nucleus.The nucleus is stationary.In Rutherford’s language –

“When we have flees on an elephant,it’s the flees that do all the jumping”

7]Planck – Einstein relation applies to electronic transitions – 

ΔE = Efinal – Einitial= hν = hc / λ

ν (neu)= frequency = c/λ   [NOTE ⇒ v= velocity and ν is frequency]
c= velocity of light
λ = wavelength (length between successive crests or troughs of a wave).

8]The distance of the electron from the nucleus is maintained due to force balance –  The centrifugal force (mv2/r) – the force acting outward – and the centripetal electrostatic force (q1 q2 / 4π∈0r 2)– acting inwardly between the positive  nucleus and the negative electron – balance out each other.Thus, the drawback in Rutherford’s model – THE NUCLEAR COLLAPSE – is avoided.

 

Using these postulates he derived his atomic Model.The derivation is interesting as it explains Bohr’s postulate in better light. The derivation is as follows –

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One electron atom in gas phase.
  • Consider, an atom in one electron system and in gas phase as shown in the adjacent figure.The nucleus has a charge Ƶ+.Ƶ is the atomic number of the element in consideration.Suppose we consider,

He+ then Ƶ= 2
Li2+then  Ƶ= 3.

 

 

 

 

 

  • So ,

          Total nuclear charge = Atomic number × elementary charge .
q1 = (No.of protons) × (Charge on a single proton).
q1 = Ƶ × e … (1)
Similarly,

Charge on the electron = -(Charge on a single electron)

As electron is negatively charged, the sign is negative and as this is a one electron system , we consider only charge on that electron.

q2 = -e ….(2)

  • According to postulate no 5 , Newtonian Mechanics applies to the electron.

∴ Energy of the orbiting electron , Eelectron = Ekinetic + Epotential.

Ekinetic  = (1/2)mv2 ⇒ Energy on account of motion of the electron. The kinetic energy unit here is Joules (mass unit is kg and velocity is m/s2).This is a mechanical unit.

Epotential = (q1 q2) / 4π∈0r ⇒ Energy stored in the atom with opposite charges q1 q2 .In this formula charges q1 q2  have Coulomb(unit of charge) as their unit.This is an electrostatic unit.So, to convert the overall unit of potential energy to Joules(mechanical unit) we introduce a factor 4π∈0r.This factor thus helps us to put the electrostatic energy on the same plane as mechanical energies.Now the units for both kinetic and potential energy are the same and thus we can simply add them to find the energy of the electron.

(The above formulas used for kinetic and potential energies are the standard formulas for the respective energies).

∴ Eelectron = [(1/2)mv2 ]+[(q1 q2) / 4π∈0r ].
=[(1/2)mv2 ]+[(Ƶe)(-e) / 4π∈0r ]… substituting for q1 q2 … from (1) and (2)
=[(1/2)mv2 ]-[(Ƶe2/ 4π∈0r ].

∴ Eelectron = [(1/2)mv2 ]-[(Ƶe2/ 4π∈0r ]…………. 

According to postulate no 8, the centripetal and centrifugal forces acting on the electron revolving in an orbit balance each other, thus keeping the electron from moving from its orbit.

Centrifugal force = mv2/r

Centripetal force = Electrostatic force =  electrostatic or potential energy / distance (As energy = Force × Distance, ∴Force = Energy/Distance)
So, Centripetal force  = Potential energy / radius = (q1 q2 / 4π∈0r) × 1/ r = q1 q2 / 4π∈02

  ∑ Force =FCentrifugal + Fcentripetal = 0
= Fdynamic + Felectrostatic  = 0
=  mv2/r +q1 q2 / 4π∈0r 2 = 0

∴mv2/r = – (q1 q2 / 4π∈02 ).
mv2/r = – ( Ƶe2/ 4π∈02 )… [ as q1 q2 = Ƶe2.] ………②

  • And according to postulate no 3 ,

               L = mvr = n (h/2π),  n = 1,2,3.. ……

When we solve for equations ① ,② and ③ , we get three important parameters –

1.Radius of the orbit
2. Energy of the electron in a particular orbit.
3.Velocity of the electron in the given orbit. 

  1. Radius of the orbit (r) – 

r= (∈0h2/4πme2) × (n2 / Z)
= (Constant)× (n2 / Z).

Since,h = 6.62 x 10-27 erg.sec
π = 3.142
m = 9.109 x 10-28gm
e = 4.803 x 10-10esu

∴ r= 0.529 × (n2 / Ƶ), n= 1,2,3..

Thus, r can have values depending on n as r α n2. So, the radius of the electron orbit takes multiple values and they are quantized and are non-linear( as radius is proportional to the square of n).

e.g. H – atom , Ƶ= 1 and n= 1 (ground state) then r1 = 0.529 Å ≈ 1/2 Å – THIS IS THE BOHR RADIUS(a0).

Thus, radius of any 1 electron atom ⇒  r(n)  =a0n2 / Ƶ = 0.529×n2 / Ƶ.So,  r(n)  α n2 . Radius is directly proportional to the square of the orbit number , which means that as we go farther away from the nucleus, the radius of the orbits increase.

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2. Energy of the electron in the given orbit (E)-

Energy of the electron, E = -(me4/ 8∈0 2h2) (Ƶ2/ n2)

Now, the Rydberg constant ,RH = me4/ 8∈0 2h2.

Thus, E= RH .2/ n2).E(n)  α (1/ n2). Radius is inversely proportional to the square of  orbit number , which means that as we go farther away from the nucleus, the energy of the orbits decrease.

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Thus, energy is quantized as well.According to Bohr atomic model, the maximum energy value of electron at infinite is zero because of negligible attraction force between electron and nucleus at infinite distance.

Hence, as electron comes closer to nucleus, the energy becomes negative.

3.Velocity of an electron(v)

                              2355.jpg

 

The derivations may not be learnt.The formula for energy and radius of the orbits are important as they lead us to the understanding of many more phenomenon. We shall continue talking about the Bohr model in the next post as well.Till then,

Be a perpetual student of life and keep learning…

Good Day!

 

References and Further Reading –

1)http://chemistry.tutorvista.com/inorganic-chemistry/bohr-s-model-of-the-atom.html

2)http://guide.ceit.metu.edu.tr/thinkquest/apndx3.htm

 

 

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