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

An electron revolving around the nucleus would, therefore, be continually accelerated towards the center of the orbit and consequently emitting radiation.

**in 1913 on two grounds, According to classical mechanics, whenever a charged particle is subjected to the acceleration it emits radiation and loses energy.***Bohr's Model*An electron revolving around the nucleus would, therefore, be continually accelerated towards the center 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.

- 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's model proposed several novel postulates,

### Postulates of Bohr's model

- An atom possesses several stable circular orbits in which an electron can stay. So long as an electron can stay 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,

- E1ã„‘E2ã„‘E3ã„‘......

- An electron can jump from one orbit to another higher energy on the absorption of energy and one orbit to another lower energy orbit with the emission of energy.

Emission and absorption energy |

- The amount of energy (Î”E) emitted or absorption in this type of jump of the electron is given by Plank's Equation.

- Î”E = hÎ½

- Where Î½ 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/2Ï€. This is known as the principle of quantization of angular momentum according to which,

- mvr = n × (h/2Ï€)

- Where m = mass of the electron, v = tangential velocity of the electron in its orbit, r = distance between the electron and nucleus and n = a whole number which has been given the principal quantum number and atomic orbitals by Bohr.

### Radius of the nth orbit in 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,

Bohrs model radius velocity of the electron |

- The centrifugal force which is due to the motion of the electron and tends to take the electron away from the orbit.

- Centrifugal force = mv²/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 the 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 balance each other.

That is, mv²/r = e²/r²

∴ mv² = e²/r

∴ 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.

Thus, mvr = nh/2Ï€ (where n have values 1,2,3, ......∞)

or, v = n × (h/2Ï€) × (1/mr)

Then, e²/r = mv²

or, e²/r = m × n² × (h/2Ï€)² × (1/mr)²

or, e²/r = n²h²/4Ï€²mr²

∴ r = n²h²/4Ï€²me²

or, v = n × (h/2Ï€) × (1/mr)

Then, e²/r = mv²

or, e²/r = m × n² × (h/2Ï€)² × (1/mr)²

or, e²/r = n²h²/4Ï€²mr²

∴ r = n²h²/4Ï€²me²

#### Bohr's nth and first orbit

∴ r

or, r

Thus, r

_{n}= n²h²/4Ï€²me²or, r

_{n}= n² × (h²/4Ï€²me²)Thus, r

_{n}= n² × r₁#### Radius of first Bohr orbit

- Taking n = 1, the radius of the first orbit is r₁.

∴ r₁ = 1 × h²/4Ï€²me²

= {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 â„« = a₀

= {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 â„« = a₀

- Thus the radius of first orbit r₁ = a₀, second orbit r₂ = 4 a₀ and third orbit r₃ = 9 a₀ and so on.

### Velocity of an electron in nth orbit in Bohr's model

- We have the following relations, mvr = nh/2Ï€, and r = n²h²/4Ï€²me²

Then, v = (nh/2Ï€m) × (1/r)

= (nh/2Ï€m) × (4Ï€²me²/n²h²)

∴ v = 2Ï€e²/nh

= (nh/2Ï€m) × (4Ï€²me²/n²h²)

∴ v = 2Ï€e²/nh

- This is the velocity of the electron in the nth orbit in Bohr's model of the hydrogen atom.

- 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.

### Energy of electron in nth orbit in Bohr's model

- 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).

Kinetic energy(T) = (1/2)mv²

- Potential energy(V) is the energy due to electric attraction and is given by,

V = ∫(e²/r²)dr = - (e²/r)

Thus the total energy, E = ½ mv² - (e²/r)

= ½ mv² - mv² (where mv² = e²/r)

= - ½ mv²

= - ½ (e²/r)

Thus the total energy, E = ½ mv² - (e²/r)

= ½ mv² - mv² (where mv² = e²/r)

= - ½ mv²

= - ½ (e²/r)

- The energy associated with the permitted orbits is given by,

E = - ½ (e²/r)

= - (2Ï€² me⁴/n²h²)

= (En =1)/n²

[where (En =1) = - (2Ï€²me⁴/h²)]

= - (2Ï€² me⁴/n²h²)

= (En =1)/n²

[where (En =1) = - (2Ï€²me⁴/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 ... En, these will be in the order,

E₁ã„‘E₂ã„‘E₃ã„‘........ã„‘En

- 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)×(4.8 × 10⁻¹⁰)⁴}/{12 × (6.6256 × 10⁻²⁷)²}

= - 21.79 × 10⁻¹² erg

= - 13.6 eV

= - 21.79 × 10⁻¹⁹ Joule

= - 313.6 Kcal

= - 21.79 × 10⁻¹² erg

= - 13.6 eV

= - 21.79 × 10⁻¹⁹ Joule

= - 313.6 Kcal

### Bohr's model questions answers

Question- 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 the third orbit.

Velocity of an electron in the Bohr's model, = 2Ï€e²/nh

where n = 1, 2, 3, ........

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

v₁= 2Ï€e²/1 × h

= 2Ï€e²/h (when n = 1)

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

= 2.188 × 10⁸ cm sec⁻¹

where n = 1, 2, 3, ........

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

v₁= 2Ï€e²/1 × h

= 2Ï€e²/h (when n = 1)

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

= 2.188 × 10⁸ cm sec⁻¹

Again the velocity of the third orbit,

v₃= (1/3) × v₁ (when n = 3)

Thus 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

v₃= (1/3) × v₁ (when n = 3)

Thus 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,

2Ï€r = 2 × 3.14 × 0.529 × 10⁻⁸ cm

∴ Rotation of an electron per second in the third orbit,

= (7.30 × 10⁷)/(2 × 3.14 × 9 × 0.529 × 10⁻⁸ )

= 2.44 × 10¹⁴ sec⁻¹

Question2Ï€r = 2 × 3.14 × 0.529 × 10⁻⁸ cm

∴ Rotation of an electron per second in the third orbit,

= (7.30 × 10⁷)/(2 × 3.14 × 9 × 0.529 × 10⁻⁸ )

= 2.44 × 10¹⁴ sec⁻¹

- 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?

Kinetic Energy = ½ mv²

= 1/2 m (2Ï€Ze²/nh)²

= 2Ï€²mZ²e⁴/n²h²

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 × 10⁻¹³erg

Question= 1/2 m (2Ï€Ze²/nh)²

= 2Ï€²mZ²e⁴/n²h²

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 × 10⁻¹³erg

- The energy of an electron in the first Bohr orbit of the hydrogen atom is -13.6 eV. What is the energy value of the electron in the excited state of Li⁺²?

The energy of an electron in the excited state of Li⁺²

E

QuestionE

_{Li⁺²}= - 30.6 eV- What is the radius of the second orbit of a hydrogen atom? Give radius of first orbit = 0.529 Ã….

- Thus the radius of the second orbit of a hydrogen atom = 2.12 Ã…

- H, H⁺, He⁺ and Li⁺² - for which of the species Bohr's theory is not applicable? Explain

- From the above species H, He⁺ and Li⁺² contain one electron but H⁺-ion has no electron. We know that Bohr's theory is applicable for one electronic system thus for H⁺-ion Bohr's theory is not applicable.