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Van't Hoff Equation - Effect on temperature on Equilibrium Constant:

Equilibrium Constant KP of a reaction is constant at a given temperature but when temperature is changed, the value of equilibrium constant also changed. The quantitative relation, relation, Known as Van't Hoff equation is derived by using Gibbs - Helmholtz equation.
The Gibbs - Helmholtz equation is,
ΔG0 = ΔH0 + T[d(ΔG0)/dT]P
Zero superscript is indicating the stranded values.
or, - (ΔH0/T2 )= -(ΔG0/T2 )+(1/T)[d(ΔG0)/dT]P
or, - (ΔH0/T2 ) = [d/dT(ΔG0/T)]P
Again Van't Hoff isotherm is,
- RT lnKP = ΔG0
or, - R lnKP = ΔG0/T
Differentiating with respect to T at constant P,
- R [dlnKP/dT]P = [d/dT(ΔG0/T)]P
Comparing the above two equation we have,
dlnKP/dT = ΔH0/T2
This is the differential form of Van't Hoff reaction equation.
Greater the value of ΔH0, the faster the equilibrium constant (KP) changes with temperature (T).
Separating the variables and integrating,
dlnKP = (ΔH0/R)(dT/T2)
Assuming that ΔH0 is independent of temperature.
or, lnKP = - (ΔH0/R)(1/T)+C(integrating constant)
The integration constant can be evaluated and identified as ΔS0/R using the relation,
ΔG0 = ΔH0 - TΔS0
Thus the Van't hoff Equation is,
lnKP = - (ΔH0/R)(1/T) + ΔS0/R
For a reaction 2A + B ⇆ 2C, ΔG0(500 K) = 2 KJ mol-1 Find the KP at 500 K for the reaction A + ½B ⇆ C.
ΔG0(500K) for the reaction A + ½B C is,
2 KJ mol-1/2 = (1 KJ mol-1)/2. 
The relation is ΔG0 = - RT lnKP
or, 1  = - (8.31 × 10-3 ) × (500) lnKP 
 or, lnKP = 1/(8.31 × 0.5) 
or, lnKP = 0.2406
KP = 1.27
Plot of lnKP vs 1/T:
For exothermic reaction, ΔH0 = (-) ve.
Examples are the formation of ammonia from H2 and N2.
N2 + H2 2NH3 ΔH0 = (-) ve
For endothermic reaction, ΔH0 = (+) ve.
Examples are the dissociation of HI into H2 and I2.
HI H2 + I2 ΔH0 = (+) ve
For the reaction, ΔH0 = 0
lnKP is independent of T. Provided ΔS0 does not change much with T.
Van't Hoff Equation - Effect on temperature on Equilibrium Constant and Expression of Le -Chatelier Principle from Van't Hoff Equation
Plot of lnKP vs 1/T
Since for ideal system, H is not a function of P and ΔH0 = ΔH and Van't Hoff equation is,
dlnKP/dT = ΔH/RT2
and the integrated equation is,
lnKP = (ΔH/R)(1/T) + ΔS/R
However S of an ideal gas depends strongly on P, so ΔS and ΔG per mole of reaction in the mixture differ quite substantially from ΔS0 and ΔG0.
If we consider the integrated Van't Hoff equation at two temperature then, it becomes,
ln(KP2/KP1) = (ΔH/R){(T2 -T1)/T1T2)}
Where KP1 and KP2 are the equilibrium constants of the reaction at two different temperature T1 and T2 respectively.
Thus determination of KP1 and KP2 at two temperature helps to calculate the value of ΔH of the reaction.
The above relation is called Van't Hoff reaction isobar since P remains constant during the change of temperature.
Use Gibbs - Helmholtz equation to derive the Vant Hoff reaction isochore. Under What condition do you expect a linear relationship between logk and 1/T ?
Again KP = KC (RT)ΔƔ
or, lnKP = lnKC + ΔƔ lnR + ΔƔlnT
Differentiating with respect to T,
dlnKP/dT = dlnKC/dT + ΔƔ/T
But dlnKP/dT = ΔH0/RT2
Hence, dlnKC/dT = ΔH0/RT2 - ΔƔ/T
or, dlnKC/dT = (ΔH0 - ΔƔRT)/RT2
∴ dlnKC/dT = ΔU0/RT2
ΔU0 is standers heat of reaction at constant volume. This is really the Van't Hoff reaction isochore in differential form.
The integrated form of the equation at two temperature is,
ln(KC2/KC1) = (ΔU0/R){(T2 -T1)/T1T2)}
For ideal system ΔU0 = ΔU. Since the reaction occurs in an equilibrium box at constant Volume this is called Van't Hoff reaction isochore.
Two important assumptions are:
  1. The reacting system is assumed to behave ideally.
  2. ΔH is taken independent of temperature for small range of temperature change. 
Due to assumption involved ΔH and ΔU do not produce precise value of these reactions. As ΔG is independent of pressure for ideal system ΔH and ΔU also independent of pressure.
Hence, ΔG0 = - RT lnKP
or, [dlnKP/dT]T = 0
Show that the equilibrium condition for any chemical reaction is given by ΔG = 0.
Vant Hoff reaction isotherm is, 
ΔG = - RT lnKa + RT lnQa
But when the reaction attain equilibrium Qa = Ka
∴ ΔG = 0
Expression of Le -Chatelier Principle from Van't Hoff Equation:
Van't Hoff equations gives quantitative expression of Le-Chatelier Principle.
From the equation,
lnKP = -(ΔH/R)(1/T) + C
It is evident that for endothermic reaction (ΔH〉0), increase of T increases the value of a of the reaction.
But for exothermic reaction (ΔHㄑ 0), with rise in temperature, a is decreased.
This change of KP also provides the calculation of quantitative change of equilibrium yield of products.
This above statement is in accordance with Le - Chatelier Principle which states that whenever stress is placed on any system in a state of equilibrium, the system is always react in a direction to reduce the applied stress.
Thus when T is increased, the system in equilibrium will try to move in a direction in which heat is absorbed that is endothermic reaction is favors.

A gas at equilibrium has definite value of Pressure(P), Volume(V), Temperature(T) and Composition(n). These are called state Variables and are determined experimentally. The state of the gas can be defined by these variables. Boyle's(1662), Charles's(1787) and Avogadro laws gives the birth of an equation of state for Ideal Gas.
Boyle's law V ∝ 1/P
When n and T are constant for a gas.
Charl's LowVT
When n and P are constant for a gas.
Avogadro's Low,  V ∝ n
When P and T are constant for a gas.
When all the variables are taken into account, The variation rule states that,
V ∝ (1/P) × T × n
or, V = R × (1/P) × T × n 
PV = nRT
Ideal Gas Equation.
Where R is the Universal Gas Constant. This is called ideal gas equation of state for ideal gas.
This equation is found to hold most satisfactory when P tense to zero. At ordinary temperature and pressure, the equation is found to deviated about 5%.
Value Of Universal Gas constant (R) at NTP:
At NTP 1 mole gas at 1 atm Pressure occupied 22.4 lit of gas.
Thus, R = (PV)/(nT)
Putting the values above equation,
We have, R = ( 1 atm × 22.4 lit)/(1 mol × 273 K)
= 0.082 lit atm mol-1 K-1
Value of R in C.G.S. and S.I. system:
P = 1 atm = 76 cm Hg
= 76 cm × 13.6 gm cm-2 × 981 cm sec-2
= 76 × 13.6 × 981 dyne cm-2
Thus, R = (76 × 13.6× 981 dyne cm-2 ×22.4 × 103 cm3)/(1 mol × 273 K)
= 8.314 × 107 dyne cm2 mol-1 K-1
Again, Work (W) = Force(F) × Displacement(d),
So, erg = dyne cm2.
Thus, R = 8.314 × 107 erg mol-1 K-1
We Know That, 1 J = 107 erg,
Thus the vale of R in S.I. Unit,
= 8.314 J mol-1 K-1
Again, 4.18 J = 1 Cal,
hence, R = 8.314 / 4.18 Cal mol-1 K-1
= 1.987 Cal mol-1 K-1
≃ 2 Cal mol-1 K-1
Physical Significance of Gas Constant R:
The universal gas constant R = PV/nT
Thus, it has the units of (Pressure × Volume)/(amount of gas × temperature).
Now the dimension of pressure and volume are,
Pressure = (force/area)
= (force/ length2)
= force × length-2
and Volume = length3
R = (force×length-2×length3)/(amount of gas×Kelvin)
= (force × length)/(amount of gas × kelvin)
= (Work or Energy)/(amount of gas × kelvin)
Thus, the dimensions of R are energy per mole per kelvin and hence it represents the amount of work or energy) that can be obtained from one mole of a gas when its temperature is raised by one kelvin.
Determine the value of gas constant R when pressure is expressed in Torr and Volume in dm3.
61.54 Torr dm3 mol-1 K-1
For Solution See Problem 6 
Properties of Gases
Derive the value of R when, (a) pressure is expressed in atom, and volume in cm3and (b) Pressure in dyne m-2 and volume mm3.
(a) 82.05 atm cm3 mol-1K-1
(b) 8.314 × 1014 dyne m-2 mm3 mol-1 K-1
For Solution See Problem 7 
Properties of Gases
Determination of Molar mass from Ideal Gas Equation:
The Ideal Gas Equation is,
PV = nRT
or, PV= (g/M)RT
Where g = weight of the gas in gm and M = Molar mass of the gas.
Again, P = ( g/V) (RT/M)
We know that, Density (d) = Weight (g)/Volume (V).
P = dRT/M
Find the Molar mass of ammonia at 5 atm pressure and 300C temperature (Density of ammonia = 3.42 gm lit-1).
17 gm mol-1
For Solution See Problem 8 
Properties of Gases
What is the molecular weight of a gas, 12.8 gms of which occupy 10 liters at a pressure of 750 mm and at 27°C ?
31.91 gm mol-1
For Solution See Problem 9 
Properties of Gases
Determination of Number of Molecule Present in Ideal Gas From Ideal Gas Equation:
The Ideal gas equation for n mole gas is,
PV = nRT
Again, PV= (N/N0) RT
Where N = Number of molecules present in the gas and N0 = Avogadro Number.
Thus, = (N/V) × (R/N0) × T
P = N′ k T
Where N′ = number of molecules present per unit Volume and
k = Boltzmann Constant = R/N0
= 1.38 × 10-16 erg molecule-1 K-1
Calculate the number of molecules present per ml of an ideal gas maintained at pressure of 7.6 × 10-3 mm of Hg at 0°C.
We have given that, V = 1ml = 10-6 dm3
P = 7.6 × 10-3 mmHg
= (7.6 × 10-3 mmHg) (101.235 kPa/760 mmHg)
= 1.01235 × 10-3 kPa
Amount of the gas, n = PV/RT 
= (1.01235 × 10-3 kPa)(10-6 dm3)/(8.314 kPa dm3 mol-1 K-1)(273 K)
= 4.46 × 10-13 mol
Hence the number of molecules, N = n N0
= (4.46 × 10-13 mol)(6.023 × 1023 mol-1)
= 2.68 × 10-11

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:

Cathode rays - Discovery of Electron:

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.
Atomic Structure
Deflection of Cathode Rays on Electric Field
Characteristics of Cathode Rays:
  1. Cathode rays excite fluorescence on the glass walls where they impinge.
  2. The rays have travel in straight lines, confirmed by the shadows of objects placed in their path. 
  3. The rays have penetrating power and can pass through thin metal foils. 
  4. They also possess considerable momentum, small paddle wheels placed in their path rotate from the impact with the rays.
  5. 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.
  6. 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.
  • 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.
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.
Structure of Atom and Constituent Particles of Atom (Electron, Proton and Neutron)
Goldstein Experiment.
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 = m 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.

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.
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 10. 
  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,
  1. 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.
  2. 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.
Defects of Rutherford's Model:
  1. 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.
  2. If the electrons lose energy continuously, the observed atomic spectra should be continuous, consisting of broad bands merging one into the other.

Rutherford's planet - like model of the atom is contested by Bohr in 1913 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. 
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:

  1. 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,
  2. 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 for Hydrogen Atom
    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. 
  3. 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 for Hydrogen Atom
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,
  1. 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.
  2. 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/ (h being Plank's Constant).
Thus, mvr = nh/
(where n have values 1,2,3, ......∞)
or, v = n × (h/2π) × (1/mr)
Then, e2/r = mv2
= m × n2 × (h/2π)2 × (1/mr)2
= n2h2/2mr2
∴ r = n2h2/2me2 = n2 × a0 
where a0 = 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 × h2/2me2
= {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 Å = a0
∴ r = n × a0(n being 1,2,3, ......)
Thus the radius of first orbit r1 = a0,
second orbit r2 = 4 a0 and
third orbit r3 = 9 a0 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 × h
= 2πe2/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)
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)/(2 × 3.14 × 9 × 0.529 × 10-8 )
= 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) = En=1/n2 [where En=1 = - (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)×(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.
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 = 1/2 mv2
= 1/2 m (2πZe2/nh)2
= 2π2mZ2e4/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 × 10-13 erg

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