### Bohr’s Model

Rutherford's planet - like model of the atom is contested by

**on two grounds, According to classical mechanics, whenever a charged particle is subjected to acceleration it emits radiation and loses energy. an electron revolving round the nucleus would therefore be continually accelerated towards the centre of the orbit and consequently emitting radiation. The result of this would be that the radius of curvature of its path would go on decreasing and due to spiral motion, the electrons will finally fall on the nucleus when all its rotational energy spent on the electromagnetic radiation and the atom would collapse.**__Bohr__in 1913
If the electrons lose energy continuously, the observed

__atomic spectra__should be continuous, consisting of broad bands merging one into the other. The observed atomic, however, consists of well defined lines of definite frequencies. To resolve the anomalous position__,__**Bohr**proposed several novel postulates###
__Postulates of Bohr's Theory:__

__Postulates of Bohr's Theory:__

- An atom possesses several stable circular orbits in which an electron can stay. So long as an electron can stays in a particular orbit there is no
or__emission__of energy.__absorption__These orbits are called Energy Levels or Main Energy Shells.These shells are numbered as**1**,**2**,**3**, ........ starting from the nucleus and are designated as capital letters,**K**,**L**,**M**, ........ respectively.The energy associated with a certain energy level increases with the increase of its distance from the nucleus. Thus, if**E**,_{1}**E**,_{2}**E**...... denotes the energy levels numbered as_{3}**1**(**K**- Shell),**2**(**L**- Shell),**3**(**M**- Shell) ......., these are in the order,**E**_{1}ㄑE_{2}ㄑE_{3}ㄑ...... - An electron can jump from one orbit to another higher energy on absorption of energy and one orbit to another lower energy orbit with the emission of energy.
__Emission and Absorption of Energy of an Electron.__The amount of energy (**ΔE**) emitted or absorption in this type of jump of the electron is given by.__Plank's Equation__**ΔE = hν**Where**ν**is the frequency of the energy (radiation) emitted or absorbed and**h**is the Plank Constant. - The angular momentum of an electron moving in an orbit is an integral multiple of
**h/2π**.This is known as principle of quantisation of angular momentum according to which,**mvr = n × (h/2π)**Where**m**= mass of the electron, v = tangential velocity of electron in its orbit,**r**= distance between the electron and nucleus and**n**= a whole number which has been called__principle quantum number__by Bohr.

###
__Radii of Bohr Orbit:__

__Radii of Bohr Orbit:__

Bohr's Model |

The nucleus has a mass

**m'**and the electron has mass**m**. The radius of the circular orbit is**r**and the linear velocity of the electron is**v**.
Evidently on the revolving electron two types of forces are acting,

- Centrifugal force which is due to the motion of the electron and tends to take the electron away from the orbit.
__Centrifugal force__**(mv**^{2}/r)Acts outwards from the nucleus. - The attractive force between the nucleus and the electron. Two attractive forces are in the operation, one being the
__electric force of attraction__between nucleus (Proton) and the electron, the other being the__Gravitational force__is comparatively weak and can be neglected. It is given by__Coulomb's inverse squire low__and is therefore equal to,**e × (e/r**^{2}) = e^{2}/r^{2}It acts towards the nucleus.

In order that the electron may keep on revolving in its orbit, these two forces, which acts in opposite direction must balanced each other.

That is, mv

^{2}/r = e^{2}/r^{2}**∴ mv**

^{2}= e^{2}/r
Now

**Bohr**made a remarkable suggestion that the__angular momentum__of the system, equal to**mvr**, can assume certain definite values or quanta. Thus all possible**r**values only certain definite**r**values are permitted. Thus only certain, definite orbits are available to the revolving electron.
According to

**Bohr's theory**the quantum unit of angular momentum is**h/2π**(**h**being**Plank's Constant**).Thus, mvr = nh/2π(where n have values 1,2,3, ......∞)or, v = n × (h/2π) × (1/mr) Then, e ^{2}/r = mv^{2}= m × n ^{2} × (h/2π)^{2} × (1/mr)^{2}= n ^{2}h^{2}/4π^{2}mr^{2} |

∴ r = n^{2}h^{2}/4π^{2}me^{2} = n^{2} × a_{0} where a_{0} = h^{2}/4𝜋^{2}me^{2} |

We thus have a solution for the radius of the permitted electron orbits in terms of quantum number

**n**. Taking**n = 1**, the radius of the first orbit is**r**._{1}
∴

**r**= 1 × h_{1}^{2}/4π^{2}me^{2}
= {1×(6.627×10

^{-27})^{2}}/{4×(3.1416)^{2}×(9.108×10^{-28}×(4.8 × 10^{-10})^{2}}
= 0.529 × 10

^{-8}cm**= 0.529 Å = a**

_{0}
∴

**r = n × a**(_{0}**n**being**1,2,3, ......**)
Thus the radius of first orbit

**r**=_{1}**a**,_{0}
second orbit

**r**= 4_{2}**a**and_{0}
third orbit

**r**= 9_{3}**a**and so on._{0}

__Velocity of the Electron in Bohr Orbits:__
We have the following relations,

mvr = nh/2π and r = n

^{2}h^{2}/4π^{2}me^{2}
Then,

**v**= (nh/2πm) × (1/r)
= (nh/2πm) × (4π

^{2}me^{2}/n^{2}h^{2})∴ v = 2𝜋e^{2}/nh |

Putting the values of

**n**(**1, 2, 3, .....**) we see the velocity in the second orbit will be one half of the first orbit and that in the third orbit will be one third of that in the first orbit and so on.
Calculate the

**velocity of the hydrogen**electron in the**first**and**third**orbit. Also calculate the**number of rotation of an electron per second in third orbit**.
Velocity of an electron in the Bohr orbit,

= 2πe

^{2}/nh
where n = 1, 2, 3, ........

Thus, the velocity of an electron in the first orbit,

v

_{1}= 2πe^{2}/1 × h
= 2πe

^{2}/h (when n = 1)
∴

**v**= {2×(3.14)×(4.8×10_{1}^{-10})^{2}}/(6.626×10^{-27})**= 2.188 × 10**

^{8}cm sec^{-1}
Again the velocity of the third orbit,

**v**= (1/3) ×

_{3}**v**(when

_{1}**n = 3**)

∴

**v**= (2.188 × 10_{3}^{8}cm sec^{-1})/3**= 7.30 × 10**

^{7}cm sec^{-1}
Radius of the

**,**__third orbit__
= 3

^{2}× 0.529 × 10^{-8}cm
Thus the circumference of the third orbit,

=

**2πr**= 2 × 3.14 × 0.529 × 10^{-8}cm
∴ Rotation of an electron per second in third orbit,

= (7.30 × 10

^{7})/(2 × 3.14 × 9 × 0.529 × 10^{-8})**= 2.44 × 10**

^{14}sec^{-1}###
__Energy of an Electron in Bohr Orbits:__

__Energy of an Electron in Bohr Orbits:__

The energy of an electron moving in one such Bohr orbit be calculated remembering that the total energy is the sum of the kinetic energy (T) and the Potential energy (V).

Thus,

**T**=**(1/2)mv**and^{2}**V**is the energy due to electric attraction and is given by,

V

**=**∫**(e**^{2}/**r**^{2}**)dr****= - (e**

^{2}/r)
Thus the total energy,

**E = (1/2)mv**

^{2}- (e^{2}/r)
= (1/2)mv

^{2}- mv^{2}
(where mv

^{2}= e^{2}/r)**= - (1/2)mv**

^{2}**= - (1/2)(e**

^{2}/r)
The energy associated with the permitted orbits is given by,

E = - (1/2)(e^{2}/r)= - (2π² me⁴/n^{2}h^{2}) = E_{n=1}/n^{2}
[where E_{n=1} = - (2π^{2}me^{4}/h^{2})] |

The energy being governed by the value of quantum number

**n**. As**n**increases the energy becomes less negative and hence the system becomes less stable. Also note that with increasing n, r also increases. Thus increasing r also makes the orbit less stable.
Thus if the energies associated with

**1st**,**2nd**,**3rd**, .... ,**nth**orbits are**E**,_{1}**E**,_{2}**E**..._{3}**E**, these will be in the order,_{n}Eㄑ_{1}Eㄑ_{2}Eㄑ........ㄑ_{3}E_{n} |

Energy (

**E**) of the moving in the 1st Bohr orbit is obtained by putting_{1}**n=1**in the energy expression of**E**.
Thus,

**E**= - {2 × (3.14)_{1}^{2}× (9.109 × 10^{-28})×(4.8 × 10^{-10})^{4}}/{1^{2}× (6.6256 × 10^{-27})^{2}}
= - 21.79 × 10

^{-12}erg
=

**- 13.6 eV**
= - 21.79 × 10

^{-19}Joule**= - 313.6 Kcal.**

Calculate the kinetic energy of the electron in the first orbit of

**He**. What will be the value if the electron is in the second orbit ?^{+2}**Kinetic Energy**= 1/2 mv

^{2}

= 1/2 m (2πZe

^{2}/nh)^{2}
= 2π

^{2}mZ^{2}e^{4}/n^{2}h^{2}
Where, e = 4.8 × 10

^{-10}esu,
Plank's Constant(h) = 6.626 × 10

^{-27 }erg sec and
m = 9.1 × 10

^{-28}g**Kinetic Energy = 871 × 10**

^{-13}erg