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Showing posts with label STRUCTURE OF ATOM. Show all posts
Showing posts with label STRUCTURE OF ATOM. Show all posts

Nov 6, 2018

Rutherford Experiment

Rutherford Experiment:

Rutherford and his students describes some alpha particle scattering experiments. The alpha particles were already established by the group as He⁺² from their behavior in electric and magnetic fields. A beam of alpha particles obtained from spontaneously disintegrating Polonium, were directed on to a very thin foil of Platinum or Gold.
With the help of fluorescent zinc sulphide screen around the Platinum or Gold foil, any deflection of alpha particle was observed. 
Rutherford's Experiment on finding the Nucleus of an Atom
Scattering of Alpha Particles from a Metal Foil
The vast majority of the alpha particles passed straight line through the foil. But a very limited few were found to be deflected back from the foil, some even appearing on the side of incidence.
Rutherford concluded that since most of the alpha particles passed straight through there must be a very large volume of empty space in the atom of the platinum or gold.
A very small part of the platinum or gold atom must be responsible for the large scattering of the few alpha particles, central part was called atomic nucleus. 

Conclusions of Rutherford experiment:

The following Conclusions and an atomic model emerged from the Rutherford's experiment.
1. All the positively charged and almost entirely mass of the atom was concentrated in very small part of the atom and these central core called atomic nucleus.
2.The large deflection of alpha particle from its original path was due to Coulombic repulsion between the Alpha particle and Positive Nucleus of an atom. Simple impact between the two such massive particles can lead to a Scattering of the order of only .
3. An alpha particle suffers little deflection while passing by an electron. 
4.The Radius of atomic nucleus is ∼ 10⁻¹³ being the same as that of an electron. Since the radius of an atom is ∼ 10⁻⁸ it is obviously that an atom must have a very empty structure. From the above Conclusions an atomic structure is proposed by Rutherford.

Rutherford's Atomic Model:

According to Rutherford's Model the entire mass of the atom was concentrated in a tiny, positively charged nucleus around which the extra nuclear electrons were moving in a circular orbits.   
Rutherford divided the atom into two part,
(i) Centre of Atom or The Nucleus and 
(ii) Extra nuclear part of atom or The extra nuclear Electrons.
Rutherford's Experiment on finding the Nucleus of an Atom
Rutherford Atomic Model

(i) Centre of Atom or The Nucleus:

Almost the entire mass of the atom is concentrated in a very small, central core called the atomic Nucleus.
Since the extra nuclear electrons contribute negligible to the total mass of the atom ans since the atom is electrically neutral it follows that the nucleus must carry particles which will account both for the mass and positive charge of the atom. 

(ii) The Extra Nuclear Electrons:

A very small positive nucleus was considered surrounded by electrons. Such a system cannot be stable if the election were in rest. Therefore it is proposed that the electron moving in circular orbits around the nucleus so that the Coulombic attraction between the nucleus and the electron was equal to the centrifugal force of attraction.
Rutherford's Experiment on finding the Nucleus of an Atom
Rutherford's Model of the Hydrogen Atom

Defects of Rutherford's Model:

(i) The Rutherford model is however, not in conformity with the classical model of electromagnetic radiation. A moving charged particle will emit radiation, will then loss kinetic energy and eventually will hit the nucleus.
(ii) If the electrons lose energy continuously, the observed atomic spectra should be continuous, consisting of broad bands merging one into the other.

Atomic Structure

Atomic Structure: 

Constituent Particles of an Atom: 

Dalton’s Model to Modern Structure Of Atom: 

1.All the matter is made of atoms are indivisible and indestructible. 
2. All the atoms of a given element are identical mass and properties. 
3. Compounds are formed by combination of two or more same or different kind of atoms. 
4. A chemical reaction is rearrangement of atoms. 
Rutherford has remarked that it is not in the nature of things for any one man to make sudden violet discovery. Science goes step by step and every man depends on the work of his predecessor. 
The journey from Dalton model of the atom to modern structure of atom was long and arduous one. At turn this century many valuable information were being compiled. 
This clearly indicates that Dalton’s atomic model no longer enjoyed the exalted position to grant it. 
Today an atom is considered to made up of a tiny nucleus carrying neutrons and protons. This tiny nucleus has around itself a certain number of negatively charged particle carrying negligible mass, called electrons, arranged in a definite order.

Discussion on the Fundamental Particles of Atom:

Gases at low pressures, when subjected to high potential, becomes conducting and various luminous effect were observed. When the pressure is quite low(0.01 mm), the tube remains dark (Crooks dark space) but a streak of rays, named cathode rays, emanate from the cathode. This is confirmed from the fact that a fluorescence is produced on the opposite wall where the rays impinge. The cathode rays have been very carefully studied for their many definite characteristics.
Atom Structure and Constituent Particles Protons Neutrons and Electrons
Deflection of Cathode Rays on Electric Field

Characteristics of Cathode Rays:

(i) Cathode rays excite fluorescence on the glass walls where they impinge.
(ii) The rays have travel in straight lines, confirmed by the shadows of objects placed in their path.
(iii) The rays have penetrating power and can pass through thin metal foils.
(iv) They also possess considerable momentum, small paddle wheels placed in their path rotate from the impact with the rays.
(v) The cathode rays are deflected from their path by the application of magnetic or electrostatic field. From the direction of deflection, the charge accompanying the rays is a negative one.
(vi) When the rays impinge on a metal target, called anti cathode, and placed on the path , a different type of radiation, the X-rays is produced. these new radiation not deflected in an electric or magnetic field. X-rays are really electromagnetic radiation of very short wave length. 
Thus the cathode rays is a stream of negatively charged particles, called electrons and this was first established by J.J. Thomson in 1897. 

Charge of an Electron:

The electron carrying negatively charged and the charge of an electron(e), 4.8 × 10 ⁻¹⁰ esu = 1.60 × 10⁻¹⁹ coulomb.

Mass of an Electron:

Let the mass of an electron = m and charge = e, then e/m = 1.76 × 10⁸ Coulomb/gram.
Mass of an electron = (1.60 × 10⁻¹⁹)/(1.76 × 10⁸) gram
= 9.11 × 10⁻²⁸ gram.

Coulometric Determination of the Electronic Charge:

Electrode position of silver from an aqueous solution of a silver salt is a suitable experiment for the determination of the electronic charge.
Faraday's Low are readily interpreted by reference to the electrolysis of Silver Nitrate. The change at the cathode requires one electron for every Silver ion reduced. 
That is, Ag⁺ + e Ag 
If the electrons consumed at this electrode is equal to Avogadro Number (6.023 × 10²³ mol⁻¹), 1 mole of Silver metal (107.9 gm) is produced. At the same time, 1 mole of electrons is removed from the anode and 1 mole of nitrate ions is discharged.
Thus, 96500 coulomb of electricity which is necessary to produce 1 equivalent mass of a substance at the electrode will be total charge carried by 1 mole of electrons.
Hence Charge carried by each electron is given by,
e = (96500 C mol⁻¹)/(6.023 × 10²³ mol⁻¹)
= 1.60 × 10⁻¹⁹ C

Positive Ray - Discovery of Protons: 

Since the Electrons contribute negligible to the total mass of the atom and the atom is electrically neutral. That the nucleus must carry particles which will account both for the mass and positive charge of the atom.
We have so far deliberately restricted the discussion on the discharge phenomena at low gas pressure. The production of cathode rays in discharge tubes inspired physicists to look for the oppositely charged ions, namely positive ions.
Goldstein added a new feature to the discharge tubes by using holes in the cathode. With this modification it is observed that on operating such discharge tubes there appeared not only cathode rays travelling from the cathode to anode but also a beam of positively charged ions travelling from around anode to cathode. Some of the positively charged particles passed through the hole in the cathode and produces a spot on the far end of the discharge tube.
Atom Structure and Constituent Particles Protons Neutrons and Electrons
Goldstein Experiment (Discharge Tubes Using Holes In The Cathode)

The nature of these positive rays is extensively investigated by Thomson. It proved much more difficult to analyse the beam of the positive rays than to analyse a beam of electrons.
On deflection by a magnetic and electric field the positive ray beam produced a large defuse spot indicating that the e/m ratio of the constituents of the beam was not same and that the particles moved with different velocities.
Thomson further demonstrated that each different gas placed in the apparatus gave a different assortment of e/m.
Since a H⁺ is produced from a hydrogen atom by the loss of one electron, which has but a negligible mass, it follows that the mass of a H⁺ is same as that of the hydrogen atom(=1).
The particle represented by H⁺ is called a Proton and is considered a fundamental constituent that accounts for the positive charge of the nucleus.

Charge of a Proton:

The Proton carrying Positively charged and the charge of a proton(p), 4.8 × 10 ⁻¹⁰ esu = 1.60 × 10⁻¹⁹ coulomb.

Mass of an Proton:

Let the mass of an Proton = and charge = e, then e/m = 9.3 × 10⁴ Coulomb/gram.
∴ Mass of an Proton = 1.6725 × 10⁻²⁴ gm.

Discovery Of Neutron:

Attempts were now directed towards a correlation of atomic mass number ( = integer nearest to the atomic weight) and nuclear charge (= atomic number). If A stands for the mass number and Z for the nuclear charge of an element, then Z units of nuclear charge means Z number of proton inside the nucleus. But Z protons can contribute only Z mass units. 
The shortfall of the (A - Z) mass units bothered chemists and physicists for the quite same time. Rutherford then suggested this shortfall must be made up by another fundamental particle. This particle has electrically neutral, and mass equal to that of the proton, namely 1. He named the particle in advance as neutron.
The glory of discovering the neutron went to Chadwick, one of Rutherford Students.
An interpretation of the atomic nuclei on the basis of neutrons and protons is now a simple affair. Taking oxygen of mass number 16, for example and recalling that the atomic number of the element is 8, we have an atomic nucleus composed of 8 protons and 8 neutrons.
Since neutrons contribute only to the mass of the element but does nothing towards charge it follows that there may exist species with the same number of protons but varying numbers of neutrons inside the nucleus. 
Such species must belong to the same element and must vary only in their mass numbers. They are called isotopes.
Thus ₁H¹, ₁H², 1H³ are three isotopes of hydrogen, and ₈O¹⁶, ₈O¹⁷ and ₈O¹⁸ are three isotopes of oxygen.

Nov 5, 2018

Rydberg Equation

Rydberg Equation:

The emission spectrum of hydrogen atom was determined experimentally and it was possible to relate the frequencies of different spectral lines by simple equation,
ν = R[1/nI2 - 1/nII2]
Where nI and nII are integers and R is a constant, called Rydberg constant after the name of the discoverer.
The existence of discrete spectral emission could hardly be explained by any model at Rydberg's time.
The Energy associated with the permitted orbits is given by ,
En = -  1/2(e²/r) where r = n2h2/²m
Thus, En = - (2π2me4/n2h2)
Again E1 = - (2π2me4/h2)
∴ En  = E1/n2
Thus the values of E2E3E4E5 etc. in terms of E₁(- 13.6 eV) are given as,
E2 = -13.6/2² = - 3.4 eV
E3 = -13.6/3² = - 1.51 eV
E4 = -13.6/4² = - 0.85 eV
E5 = -13.6/5² = - 0.54 eV
As n increases the energy becomes less negative and hence the system becomes less stable.

Explanation of the Rydberg Equation:

The explanation of Rydberg equation is now simple. The most stable orbit must be the one with the lowest energy (ground state).
When n = 1 the energy of an electron is lowest and it the lowest energy state and the states with n 〉 1 are the higher energy exited states. Provided the right amount of energy is supplied the electron may jump from n = 1 to another higher level.
Conversely energy will be released in the form of light of definite frequency when the exited electron returns to its ground state. Since energy and frequency of the emitted light are connected by Plank Relation,
E = hν 
or, ν = E/h
The energy corresponding to a particular line in the emission spectrum of hydrogen atom is the energy difference between the initial state I and the final state II.
So that, EEII - EI
= - (2π2me4/nII2h2) - [- (2π2me4/nI2h2)]
= (2π2me4/h2)[1/nI2 - 1/nII2]
Then ν corresponding to the energy E is given by,
ν = E/h = (2π2me4/h3)[1/nI2 - 1/nII2]
= R[1/nI2 - 1/nII2]
Where R is the Rydberg Constant.
This relation tells us that as we go to higher and still higher orbits the energy gap between two immediate orbits continues to decreases.

Value of the Rydberg Constant:

Using the value of 9.108 × 10⁻²⁸ gm for the mass of an electron, the Bohr Theory Predicts the following value of Rydberg Constant.
R = (2π2me4/h3
= {2 × (3.1416)² × (9.108 × 10⁻²⁸ gm) × (4.8 × 10⁻¹⁰ esu)}/(6.627 × 10⁻²⁷ erg sec)³
= 3.2898 × 10¹⁵ cycles/sec
Evaluate R in terms of wave number () instead of frequency. Wave number and frequency are connected with,
λν = c and ν = c/λ = c
Thus ⊽ = ν/c
where c is the velocity of light. Recall the frequency (ν) indicates the number of waves passing a given point per second and is expresses as cycles per second. Wave number () on the other hand stands for the number of waves connected in unit length that is per centimeter (cm⁻¹) or per metre (m⁻¹).
Thus R = 3.2898 × 10¹⁵ sec⁻¹/2.9979 × 10¹⁰ cm sec⁻¹ = 109737 cm⁻¹ = 10973700 m⁻¹
The experimental values of R is 109677 cm⁻¹ (10967700 m⁻¹) showing a remarkable agreement between experiment and Bohr's Theory.

Electronic Transition and Origin of the Spectral Lines of Hydrogen Atom:

Putting n = 1, n = 2, n = 3, etc in Rydberg Equation we get the energies of the different stationary states for the hydrogen electron. The transition energies can be calculated from the Rydberg Equation.

The experimental spectra of hydrogen atom also exhibited several series of lines .
Rydberg Equation Derivation and Hydrogen Spectrum
Energy Level Diagram for the Hydrogen Spectrum

Hydrogen atom contains only one electron but its spectrum gives a large number of lines - Explain.
This is because any given sample of hydrogen contains almost infinite number of atoms. Under the normal conditions the electron of the each hydrogen atom remains in the ground state near the nucleus that is n = 1 (K - Shell). When energy is supplied to this sample of hydrogen gas by passing an electron discharge through the gas in the discharge tube, individual atom absorbed different amounts of energy. Some of atoms absorbed such energy to shift their electron to third energy level (n = 3) while some others may absorbed large amount of energy to shift their electron to the Fourth (n = 4), Fifth (n = 5), Sixth (n = 6) and Seventh (n = 7) energy level. The electrons in the higher energy level are relatively unstable and hence drop back to the lower energy level with the emission of energy in the form of line spectrum containing various lines in of particular frequency and wave length.

Lyman Series:

Transition to the ground state (n = 1) from n = 2, n = 3, n = 4, etc exited states constitute the Lyman Series in the ultraviolet region.
The wavelength of this spectral lines can be calculated by the following equation,
⊽ = 1/λ = R[1/nI2 - 1/nII2]
For Lyman Series, we have nI = 1 and nII = 2, 3, 4 .... 
Thus, (i) When nII = 2, 1/λ = 109677{1 - (1/4)} = {(109677 × 3)/4} cm⁻¹
∴ λ = {4/(109677 × 3)} = 1215 × 10⁻⁸ cm = 1215 Å
(ii) When nıı = ∞, 1/λ = 109677{1 - (1/∞²)} = 109677 cm⁻¹
∴ λ = (1/(109677) = 912 × 10⁻⁸ cm = 912 Å
The wave number of the experimental spectra of hydrogen atom in Lyman Series is 82200 cm⁻¹. Show that the transition occur to the Ground state (n =1) from n = 2. (R = 109600 cm⁻¹)
We know that, Wave number (⊽) = R[1/nI2 - 1/nII2]
For Lyman series n = 1 and Wave Number (⊽) = 82200 cm⁻¹ 
Thus, 82200 = 109600(1/1² - 1/nıı²) 
 or, 1/nıı² = 1 - 822/1096 = 274/1096 =1/4 
∴ nII = 2 
Thus the transition occurs to the ground state (n = 1) from n =2.

Balmer Series:

Transition to the n = 2 level from n = 3, n = 4, n = 5, etc exited states constitute the Balmer Series in the visible region.
The wavelength of this spectral lines can be calculated by the following equation,
⊽ = 1/λ = 109677[1/22 - 1/nII2]
Which of the Balmer lines fall in the visible region of the spectrum wave length 4000 to 7000 Å ? (Given  R = 109737 cm⁻¹

The Balmer Series of the hydrogen spectrum comprise the transition from n 〉 2 levels to the n = 2 level. 
4000 Å = 4000 × 10⁻⁸ cm = 4 × 10⁻⁵ cm and 7000 Å = 7 × 10⁻⁵ cm 
Therefore wave number (⊽) = 1/λ = (1/4) × 10⁵ to (1/7) × 10⁵ cm⁻¹ 
Transition energy of the Balmer lines, 
 ⊽ = 1/λ = 109937 {(1/2²) - (1/nıı²)} 
Taking (⊽) = (1/4) × 10⁵ cm⁻¹ we have: 
(1/4) × 10⁵ cm⁻¹ = 109737{(1/4) - (1/nıı²)} = 1.1 × 10 ⁵{(1/4) - (1/nıı²)} cm⁻¹ 
∴ 1/4 ⋍ 1.1 {(1/4) - (1/nıı²)} 
or, nII2 ⋍ 44 
So nII = 7 ( nearest whole number). 
Taking (⊽) = (1/7) × 10⁵ cm⁻¹ we have: 
(1/7) × 10⁵ cm⁻¹ = 109737{(1/4) - (1/nıı²)} = 1.1 × 10 ⁵{(1/4) - (1/nıı²)} cm⁻¹ 
∴ 1/7 ⋍ 1.1 {(1/4) - (1/nII2 )} 
or, nII ⋍ 3 
So nII = 3 ( nearest whole number). 
So all transition from n =7 to n = 3 while falling to the n = 2 will generate Balmer series.
Calculate the wave length of Hɑ and Hβ of the Balmer Series.
We know that, for the Balmer Series 1/λ = R {(1/2²) - (1/nıı²)}
Again for Hɑ line nıı = 3 and for Hβ line nıı = 4
Thus the wave length for Hɑ - line,
1/λ = 109677 {(1/2²) - (1/3²)} = 109677 × (5/36)
∴ λ = 36/(5 × 109677) = 6.564 × 10⁻⁵ cm = 6564 Å
Again the wave length for Hβ - line,
1/λ = 109677 {(1/2²) - (1/4²)} = 109677 × (3/16)
∴ λ = 16/(3 × 109677) = 4.863 × 10⁻⁵ cm = 4863 Å

Pschen Series:

Transition to the n = 3 level from n = 4, n = 5, n = 6, etc exited states constitute the Balmer Series in the Near Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
⊽ = 1/λ = 109677[1/32 - 1/nII2]

Brackett Series:

Transition to the n = 4 level from n = 5, n = 6, n = 7, etc exited states constitute the Balmer Series in the Far Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
⊽ = 1/λ = 109677[1/42 - 1/nII2]

Pfund Series:

Transition to the n = 5 level from n = 6, n = 7, n = 8, etc exited states constitute the Balmer Series in the Infrared region.
The wavelength of this spectral lines can be calculated by the following equation,
⊽ = 1/λ = 109677[1/52 - 1/nII2]
Rydberg Equation Derivation and Hydrogen Spectrum
Electronic Transitions and values of n₁ and n₂
for various spectral lines of Hydrogen Atom
A General of an Electron Moving about a Nucleus of Charge +Ze:
For this case, Centrifugal force = Electronic Force of Attraction
∴ mv²/r = Z e²/r²
Thus the Radius of the Bohr's Orbit (r) = n²h²/4π²mZe²
And the Energy of an Electron,
E = - (1/2) Ze²/r = - (2π²mZ²e⁴)/(n²h²)

Nov 2, 2018

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:

(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,
(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.
Bohr's model of Hydrogen atom and its Limitations.
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.
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 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:

Bohr's model of Hydrogen atom and its Limitations.
Bohr's Model of the Hydrogen Atom
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,
(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/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 × a where a₀ = /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}
= 0.529 × 10⁻⁸ cm = 0.529 Å = a₀
Since r = n² × a (n being 1,2,3, ......) the radius of first orbit r₁ = a, second orbit r₂ = 4 a and third orbit r₃ = 9 a and so on.

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.

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 = 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)
= 2.44 × 10¹⁴ sec⁻¹

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 ... En, these will be in the order,
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.
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 ?