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__Bohr's Model:__

__Bohr's Model:__

__Rutherford's planet__- like model of the atom is contested by

__Bohr__in

**1913**on two grounds,

(i) 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__.

(ii) 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:__

(i)

__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 emission or absorption of energy__.

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 E₁, E₂, E₃ ...... denotes the energy levels numbered as 1 (K - Shell), 2 (L - Shell), 3 (M - Shell) ......., these are in the order,

**E₁ㄑE₂ㄑE₃ㄑ......**

__(ii) 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 |

Thus,

**ΔE = hγ**
Where γ is the frequency of the energy (radiation) emitted or absorbed and h is the Plank Constant.

(iii) 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

####

The**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 of the Hydrogen Atom |

__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__,

(i)

__Centrifugal force__which is due to the motion of the electron and tends to take the electron away from the orbit. It is equal to +(

**mv²/r**) and acts

__outwards from the nucleus__.

(ii) 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**/

**r² = - e²**/

**r²**

__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, mv²/r = e²/r²

∴ mv² = e²/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, ......∞)

Again, v = n × (h/2𝜋) × (1/mr)

Then e²/r = mv² = m × n² × (h/2𝜋)² × (1/mr)² = n²h²/4𝜋²mr²

∴

**r = n²h²/4𝜋²me² = n² ×**where*a*₀*a*₀ = h²/4𝜋²me²__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**r₁**.
∴ r₁ = 1 × h²/4𝜋²me²

= {1 × (6.627 × 10⁻²⁷ erg sec)²}/{4 × (3.1416)² × (9.108 × 10⁻²⁸ gm) × (4.8 × 10⁻¹⁰ esu)²}

= {1 × (6.627 × 10⁻²⁷ erg sec)²}/{4 × (3.1416)² × (9.108 × 10⁻²⁸ gm) × (4.8 × 10⁻¹⁰ esu)²}

__= 0.529 × 10⁻⁸ cm = 0.529 Å =__*a₀*__Since__.

**r = n² ×**(*a*₀**n**being**1,2,3, ......**) the radius of first orbit**r₁ =**, second orbit*a*₀**r₂ = 4**and third orbit*a*₀**r₃ = 9**and so on*a*₀####
__Velocity of the Electron in Bohr Orbits:__

__Velocity of the Electron in Bohr Orbits:__

We have the following relations, mvr = nh/2𝜋 and r = n²h²/4𝜋²me².

Then, v = (nh/2𝜋m) × (1/r) = (nh/2𝜋m) × (4𝜋²me²/n²h²)

__∴ v = 2𝜋e²/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.

**Problem:**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.

**Answer:**Velocity of an electron in the Bohr orbit = 2𝜋e²/nh where n = 1, 2, 3, ........

Thus, the velocity of an electron in the first orbit (v₁) = 2𝜋e²/1 × h² = 2𝜋e²/h (when n = 1)

∴ v₁ = {2 × (3.14) × (4.8 × 10⁻¹⁰)²}/(6.626 × 10⁻²⁷ erg sec) =

**2.188 × 10⁸ cm sec⁻¹**
Again the velocity of the third orbit (v₃) = (1/3) × v₁ (when n = 3)

∴ v₃ = (2.188 × 10⁸ cm sec⁻¹)/3 = 7.30 × 10⁷ cm sec⁻¹

Radius of the third orbit = 3² × 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 × 10⁷ cm sec⁻¹)/(2 × 3.14 × 9 × 0.529 × 10⁻⁸ cm)

####

Radius of the third orbit = 3² × 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 × 10⁷ cm sec⁻¹)/(2 × 3.14 × 9 × 0.529 × 10⁻⁸ cm)

__= 2.44 × 10¹⁴ 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)mv² and V is the energy due to electric attraction and is given by,

V =

**∫**(e²/r²)dr = - (e²/r)
Thus the total energy E = (1/2)mv² - (e²/r)

= (1/2)mv² - mv² (where mv² = e²/r)

**= - (1/2)mv² = - (1/2)(e²/r)**

The energy associated with the permitted orbits is given by,

E = - (1/2)(e²/r) = - (2𝜋² me⁴/n²h²) = E₁/n² [where E₁ = - (2𝜋²me⁴/h²)]

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 E₁, E₂, E₃ ... E__

*n,*these will be in the order,**E₁ㄑE₂ㄑE₃ㄑ........ㄑEn**

Energy (E₁) of the moving in the 1st Bohr orbit is obtained by putting n=1 in the energy expression of E.

Thus, E₁ = - {2 × (3.14)² × (9.109 × 10⁻²⁸ gm) × (4.8 × 10⁻¹⁰ esu)⁴}/{1² × (6.6256 × 10⁻²⁷ esu sec)²} = - 21.79 × 10⁻¹² erg = - 13.6 eV = - 21.79 × 10⁻¹⁹ Joule = - 313.6 Kcal.

Problem:

Calculate the kinetic energy of the electron in the first orbit of He⁺². What will be the value if the electron is in the second orbit ?

Answer:

Problem:

Calculate the kinetic energy of the electron in the first orbit of He⁺². What will be the value if the electron is in the second orbit ?

Answer: