April 2019

    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 give the birth of an equation of state for Ideal gas.

Ideal Gas Equation its derivation

Boyle's law,
V ∝ 1/P
when n and T are constant for a gas

Charl's law,
V ∝ T
when n and P are constant for a gas

Avogadro's Law,
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
Ideal gas equation its derivation
Ideal gas equation
    Where R is the universal gas constant. This is called the ideal gas equation of state for an 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

    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 value 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

Significance of universal gas constant

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 gas when its temperature is raised by one kelvin.

Molar mass from the 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

Gas molecule present in an ideal gas

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, P = (N/V) × (R/N₀) × T


∴ P = N′ kT
where N′ = number of molecules present per unit volume and
k = Boltzmann constant = R/N₀ = 1.38 × 10-16 erg molecule⁻¹ K⁻¹

Questions and answers

Problem
    Determine the value of gas constant R when pressure is expressed in Torr and Volume in dm³.
Answer
    61.54 Torr dm³ mol⁻¹ K⁻¹
Problem
    Derive the value of R when, (a) pressure is expressed in the atom, and volume in cm³and (b) Pressure in dyne m⁻² and volume mm³.
Answer
    (a) 82.05 atm cm³ mol⁻¹K⁻¹
    (b) 8.314 × 10¹⁴ dyne m⁻² mm³ mol⁻¹ K⁻¹
Problem
    Find the Molar mass of ammonia at 5 atm pressure and 30⁰C temperature (Density of ammonia = 3.42 gm lit⁻¹).
Answer
    17 gm mol⁻¹
Problem
    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?
Answer
We have, PV = nRT
or, PV = (g/M)RT (molecular wt. of N₂ = 28)
∴ P = (7/28) × (0.082 × 300)/3
= 2.05 atm.
Here R = 0.082 lit atm mol⁻¹ K⁻¹
Problem
    Calculate the number of molecules present per ml of an ideal gas maintained at a pressure of 7.6 × 10⁻³ mm of Hg at 0°C.
Answer
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 × 10⁻²³ mol⁻¹)
= 2.68 × 10⁻¹¹

Problem
    Assuming ideal behavior finds out the total pressure exerted by 2 gm ethane and 3 gm CO₂ contained in a vessel of 5-liter capacity at 50°C.
Answer
No of molecules of ethane(n₁) = 2/30
= 0.0667

No of molecules of CO₂(n₂) = 3/44
= 0.0682

So the total moles (n₁ + n₂) = (0.0667 + 0.0682)
= 0.1349

Thus the total pressure P = (n₁ + n₂)RT/V
or, P = {0.1349 × 0.082 × (273+50)}/5

= 0.715 atm

Problem
    At 2 atm constant pressure slope of a one-mole ideal gas in V vs T graph is x L mol⁻¹ K⁻¹. Find out the value of R by x.
Answer
The ideal gas equation for 1-mole gas is,
PV = nRT
or, V = (nR/P) × T.
Thus the slope of the V vs T graph = nR/P where P = 2 atm.
∴ (R/2 atm) = x L mol⁻¹ K⁻¹
or, R = 2x L atm mol⁻¹ K⁻¹

Formulation of kinetic theory of gas
Kinetic gas equation
    After knowing the experimental gas lows, it is of interest to develop a theoretical model based on the structure and properties of gases, which can be correlated to the experimental facts.
    Fortunately, such theory has been developed for the formulation of the kinetic theory of gases based upon the certain postulates which are supposed to be applicable to an Ideal gas.

Assumptions of the kinetic theory of gases

  1. The gas is composed of very small discrete particles, now called molecules. For gas, the mass and size of the molecules are the same and different for different gases.
  2. The molecules are moving in all directions with the verity of speeds. Some are very fast while others are slow.
  3. Due to random motion, the molecules are executing collision with the walls of the container (wall collision) and with themselves (intermolecular collision). These collisions are perfectly elastic and so there occurs no loss of kinetic energy or momentum of the molecules by this collision.
  4. The gas molecules are assumed to be point masses, that is their size is very small in comparison to the distance they travel.
  5. There exist no intermolecular attraction especially at low pressure, that is one molecule that can exert pressure independent of the influence of other molecules.
  6. The pressure exerted by a gas is due to the uniform wall collisions. Higher the frequency of the wall collision greater will be the pressure of the gas.
  7. This explains Boyle's law since when the volume is reduced, wall collision becomes more frequent and pressure is increased.
  8. Through the molecular velocity is constantly changing due to the intermolecular collision, the average kinetic energy(є) of the molecules remains fixed at a given temperature. This explains the Charl's law that when T is increased, velocity is increased, wall collision becomes more frequent and pressure(P) is increased when T kept constant or Volume(V) is increased when P kept constant.

Root mean square velocity

    Root mean square(RMS) speed is defined as the square root of the average of the squares of speeds.
CRMS² = (N₁C₁² + N₂C₂² + ...)/N

Formulation of Kinetic gas equation

    Let us take a cube of edge length l containing N molecules of gas of molecular mass m and RMS speed is CRMS at temperature T and pressure P.
Formulation of kinetic theory of gas
Kinetic theory of gas

    Let in gas molecules, N₁ have velocity C₁, N₂ have velocity C₂, N₃ have velocity C₃, and so on.
    Let us concentrate our discussion to a single molecule among N₁ that has resultant velocity C₁ and the component velocities are Cx, Cy, and Cz.
C₁² = Cx² + Cy² + Cz²
    The molecule will collide walls A and B with the component velocity Cx and other opposite faces by Cy and Cz.
Change of momentum along X-direction for a single collision,
= m Cx - (- m Cx) = 2 m Cx

Rate of change of momentum of the above type of collision,
= 2 mCx × (Cx/l)

= 2 mCx²/l
    Similarly, along Y and Z directions, the rate of change of momentum of the molecule is 2 mCy²/l and 2 mCz²/l respectively.
Total rate change of momentum for the molecule,
= 2 mCx²/l + 2 mCx²/l +2 mCz²/l
= 2 (m/l) (Cx² + Cy² + Cz²)

= 2 mC₁²/l

For similar N₁ molecules, it is 2 mN₁C₁²/l

Taking all the molecules of the gas, the total rate of change of momentum,
= (2 mN₁C₁²/l)+(2 mN₂C₂²/l)+(2 mN₃C₃²/l)+ ..
= 2 mN {(N₁C₁² + N₂C₂² + N₃C₃² ..)/N}

= 2 mN CRMS²

where CRMS² = root means square velocity of the gases.
    According to Newton's 2nd low of motion, the rate of change of momentum due to wall collision is equal to force developed within the gas molecules.
That is, P × 6l² = 2 mN CRMS²/l
or, P × l³ = 1/3 m N CRMS²

∴ PV = 1/3 m N CRMS²

Here, l³ = volume of the cube contains gas molecules.

The other form of the equation is,
P = 1/3 × (mN/V) × CRMS²

∴ P = 1/3 d CRMS²
where (mN/V) is the density(d) of the gas molecules.
This equation is also valid for any shape of the gas container.

Root mean square velocity

Let us apply the kinetic equation for a 1-mole ideal gas.
In that case mN = mN₀ = M
Ideal gas equation,
PV = RT.
Hence from the kinetic gas equation, PV = 1/3 m N CRMS²
or, RT = 1/3 m N CRMS²

or, CRMS² = 3RT/M

∴ CRMS = √3RT/M
    Thus root means square velocity depends on the molar mass(M) and temperature(T) of the gas.

Average kinetic energy

The average kinetic energy(Ē) is defined as,
Ē = 1/₂ m CRMS²

Again from the kinetic gas equation,
PV = 1/3 m N CRMS²
or, PV = ²/3 N × 1/3 m CRMS²
or, PV = ²/3 N Ē

For 1 mole ideal gas,PV = RT and N = N₀
Thus RT = ²/3 N₀ Ē

or, Ē = ³/₂R/N₀T
∴ Ē = ³/₂ kT
Where k = R/N0 and is known as the Boltzmann constant.
Its value is 1.38 × 10-23 JK⁻¹

The total kinetic energy for 1 mole of the gas is,

ETotal = N₀ (Ē) = ³/₂RT
    Thus average kinetic energy is dependent on T only and it is not dependent on the nature of the gas.

Problems solutions

Problem
    Calculate the pressure exerted by 10²³ gas particles each of the mass 10⁻²² gm in a container of volume 1 dm³. The root means square speed is 10⁵ cm sec⁻¹.
Solution
    We have, N = 10²³, m = 10⁻²² gm = 10⁻²⁵ Kg, V = 1 dm³ = 10⁻³ m³ and CRMS² = 10⁵ cm sec⁻¹ = 10³ m sec⁻¹.
Therefore, from the kinetic gas equation,
PV = 1/3 mN CRMS²
or, P = 1/3 × (m N/V) × CRMS²
Putting the value we have,
P = (1/3)(10⁻²⁵ Kg×10²³/10⁻³ m³) × (10³ m sec⁻¹)²
∴ P = 0.333 × 10⁷ atm
Problem
    Calculate the root mean square speed of oxygen gas at 27⁰ C.
Solution
We know that,
CRMS² = (3RT/M)

Here, M = 32 gm mol⁻¹, and T = 270 C = (273+27)K = 300 K.

Thus,CRMS² = (3 × 8.314 × 107 erg mol⁻¹ K⁻¹ × 300 K)/(32 gm mol⁻¹)

∴ CRMS = 48356 cm sec⁻¹
Problem
    Calculate the RMS speed of NH3 at N.T.P.
Answer
    At N.T.P, V = 22.4 dm³ mol⁻¹ = 22.4 × 10⁻³ m³ mol⁻¹, P = 1 atm = 101325 Pa and M = 17 × 10⁻³ Kg mol⁻¹.
Thus, CRMS² = 3RT/M
= (3 × 101325 × 22.4 × 10⁻³) /(17 × 10⁻³)
∴ CRMS = 632 m sec⁻¹
Problem
    How the root means square velocity for oxygen compares with that of the hydrogen?
Answer
We know that, CRMS² = 3RT/M
Hence at a given temperature,
(CRMS of H₂)²/(CRMS of O₂)² = MO₂/MH₂
= 32/2 = 16
That is CRMS of O₂ = 4 × CRMS of H₂.
Problem
    Calculate the kinetic energy of translation of 8.5 gm NH₃ at 27⁰ C.
Solution
We know that total kinetic energy for 1 mole of the gas is,
ETotal = (3/2)RT

= (3/2)(2 cal mol⁻¹ K⁻¹ × 300 K)
= 900 cal mol⁻¹
Again 8.5 gm NH3 = (8.5/17) mol = 0.5 mol
Thus the kinetic energy of 8.5 gm NH3 at 27⁰ C is, (0.5 × 900) cal

= 450 cal
Problem
    Calculate the RMS velocity of oxygen molecules having a kinetic energy of 2 K.cal mol⁻¹. At what temperature the molecules have this value of KE?
Solution
T = 673.9K or 400.9⁰ C

A different application of dipole moment is taken one after another.

Determination of partial ionic character

    Let us consider a molecule AB having the dipole moment μobs and the bond length l cm. If the shared electron pair lies at the midpoint of the atoms, the bond will be purely covalent and the present of ionic character is zero.
    But if the bond is ionic, and B is more electronegative then A, B will carry a unit negative charge and a uni-positive charged respectively. In that case, the dipole moment of the molecule would be,
μionic = e × ℓ
= (4.8 × 10⁻¹⁰) ℓ esu cm
    But the dipole moment of AB is neither zero or nor μionic. Thus we can calculate partial ionic character by the application of dipole moment.
Ionic character from dipole moment
Ionic character

Induced polarization

The induced polarization (Pi)
= (4/3) π N₀ αi

But when the substance is in the gaseous state,
Pi = {(D₀ - 1)/(D₀ + 2)} M/ρ
(for the covalent substance)

The value of D൦ is close to unity under this condition,
Hence, {(D0 - 1)/3} × 22400 = ( 4/3 ) π N₀ r³

At NTP, M/ρ = molar volume = 22400cc/mole and
αi= r³ taking the spherical shape of the molecule

or, r³ = (22400/4πN₀ ) (D₀ - 1)
= 2.94 ×10⁻²¹ (D₀ - 1)
    Hence, the radius of the molecule r can be determined by measuring D൦ of the substance at NTP.

Molecular structure form dipole moment

Mono-atomic molecules

    The mono-atomic inert gases are non-polar, and it indicates the symmetrical charge distribution in the molecule.

Di-atomic molecules

    The homonuclear diatomic molecules are largely non-polar such as N₂, O₂, Cl₂ indicating that there are symmetrical charge distributions. The bonding electron pair is equally shared by the two bonding atoms.
    However, Br₂, I₂ have non zero values of dipole moment and this indicates the unsymmetrical charge distribution,
H⁺ㄧI⁻
    Heteronuclear diatomic molecules are always polar due to the difference in electronegativity of the constituent atoms. The example is, HCl, HBr, HF, etc. this indicates that the electron pair is not equally shared and shifted to the more electronegative atom.
Application of Dipole Moment
Dipole moment of HCl, HBr, HI, HF

Tri-atomic molecules

    The molecules, like CO₂, BeCl₂, BeF₂, SnCl₂, etc have zero dipole moment indicating that the molecules have a symmetrical linear structure.
    For example, CO₂ has structure, The electric moment of one C - O bond ( known as bond moment ) cancels the electric moment of the other CO bond.
Application of Dipole Moment
Dipole moment of CO₂ and BeCl₂
    The electric moment associated with the bond arising from the difference of electronegativity is called the bond moment(m). In molecules, the vectorial addition of the bond moments (m) gives the resultant dipole moment(μ) of the molecule.
μ2 = m12 + m22 + 2m1m2Cosθ
Where m₁ and m₂ are the bond moments.
    These help to calculate the bond angle (θ) in the molecule. Thus in the carbon dioxide molecule, μ = 0 and m₁ = m₂
Hence, 0 = 2m²(1 + cosθ) or, θ = 180° that is the molecule is linear.

Non-linear structure

    Another type of molecules such as H₂O, H₂S, SO₂᠌᠌, etc. have μ ≠ 0 indicating that they have non-linear structure. The bond angle can be calculated from the bond moments of the molecules.

Tetra atomic molecules

    The molecules like BCl₃, BF₃, etc have dipole moment zero indicating that they have regular planar structure. Their halogen atoms are on a plane at the corner of the equilateral triangle and B atom is at the intersection of the molecules.
Dipole moment of BF3 in application of dipole moment
Application of dipole moment of BF₃
    While other types of the molecule such as NH₃, PH₃, AsH₃ are polar (μ≠0) indicating that that the molecule has a pyramidal structure in which the three H-atoms are on the plane and N-atom at the apex of the pyramid in NH₃. ;
    But NF₃ has a very small dipole moment though there is a great difference of electronegativity between N and F atoms and similar structure of NH₃.
    A low value of μ of NF₃ is explained by the fact that the resultant bond moment of the three N - F bonds are acting in the opposite direction to that of the lone pair placed at the N-atom. But in NH₃, the resultant bond moment is acting in the same direction as that of the lone pair electrons.
Application of dipole moment of NH₃ and BF₃
Dipole Moment of NH₃ and BF₃

Penta-atomic molecules

    The molecules of this type CH₄, CCl₄, PtCl₄ are the examples of having zero dipole moment. This suggests that they are either regular tetrahedral or square planer. But polar molecules of this type have pyramidal structures.
    Let us discuss the structure of CH₄ that has regular tetrahedral structure and the angle of each H-C-H is 109°28ˊ.
Dipole moment of CH₄ molecule
CH₄ molecule
    The electric moment associated with a group is called group moment and it depends on the arrangement of the bonds in the group and the difference of the electronegativity of the constituent atoms forming the bonds in the group.
    It can be shown that the group moment of CH₃ (mCH₃) is identical to the bond moment of C-H (mC-H) and so the two moments cancel each other resulting in zero dipole moment of the molecule in CH₄.
Thus , mCH₃ = 3 mCH Cos(180° -109°28՛)
= 3 mCH Cos 70°32՛
= 3 mCH × (1/3)
= mCH

Dipole moment of CH₄

Thus the resultant dipole moment of CH₄
= mCH (1 + 3 Cos 109°28՛)
= mCH {1 - (3 ×(1/3)}
= 0

Dipole moment of CCl₄, CHCl₃, and CH₃Cl

Application of Dipole Moment
CCl₄, CHCl₃, and CH₃Cl molecules
Dipole moment of CH₃Cl molecule

μ² = m₁² + m₂² + 2 m₁m₂ Cosθ
But here θ = 0° hence Cosθ = 1

∴ μ² = m₁² + m₂² + 2 m₁m₂= (m₁ + m₂)²
or, μ = (m₁ + m₂)
= (mCCl + mCH)
= (1.5 D + 0.4 D)
= 1.9 D

Dipole moment of CHCl₃ molecules

μ = (m₁ + m₂)= (mCH + mCCl)
= (0.4 D + 1.5 D)
= 1.9 D
    A similar calculation can be done for the group moment of C₂H₄, C₃H₇, C₄H₉, etc. have the same value and equal to the bond moment of C-H.
    The identical value of the dipole moment of the homologous alcohols and saturated hydrocarbons confirms the above statement.

Hexa atomic molecules

    The molecules of this type are PCl₅, AsCl₅, PF₅, etc have μ =0 indicating that they have a pyramidal structure having the center of symmetry.

Hepta-atomic molecules

    Hepta-atomic molecules like SF₆, XeF₆, WF₆, etc have μ =0 indicating that these have an octahedral symmetrical structure.

Questions answers

Question
    The displacement of HCl is 1.03 Debye and its bond length is 1.27 A. Calculate the (a) charge on the constituent atom and (b) the % of the ionic Character of HCl.
Answer

(a) Charge on the constituent atom(q), = 0.8 × 10⁻¹⁰ esu
(b) Percentage of the ionic character of HCl, = 16.89%
Question
    The difference between the electronegativity of carbon and oxygen is large but the dipole moments of carbon monoxide are very low - why?

Answer

    However, in CO, there is a large difference in electronegativity between C and O but the molecule is the very low value of dipole moment. This suggested that the charge density in O-atom is somehow back-donated to C-atom.
    This explains by forming a coordinate covalent bond directing towards C-atom.
Question
    H₂O molecule has a dipole moment-explain. Does it invalidate a linear structure?
Answer
Bond Moment of H2O and H2S in application of Dipole Moment
Dipole Moment of H₂O and H₂S
For Water (H₂O), μ = 1.84 D and mOH = 1.60 D
Thus, μ² = 2 m² (1+ cosθ )
or, (1.84)² = 2 (1.60)² (1+ cosθ )
or, θ = 105°
    The contribution of non-bonding electrons towards the total dipole moment is included within the bond moment.
Question
    The bond angle in H₂S is 97° and dipole moment = 0.95 D. Find the bond moment of the S - H link. ( Given, Cos97° = -0.122)
Answer
We have, μ = 0.95 D and θ = 97°.
From the equation, μ² = 2 m² (1+ cosθ )
Putting the value we have, (0.95)² = 2 m² (1 + cos97° )
Here m = mS-Hor, 0.9025 = 2 m² (1-0.122)
or, m² = 0.9025/ (2 × 0.878)
or, m² = 0.5139
or, m = 0.72
Thus the bond moment of the S - H link is 0.72 D

Pairing of the nuclear spins

Since radioactivity is a nuclear phenomenon it must be connected with the instability of the nucleus. We know that the nucleus of an atom is composed of two fundamental particles, protons, and neutrons. Since all elements are not radioactive, the ratio of the neutron to the proton of the unstable, radioactive nucleus is the factor responsible for radioactivity.
    Nuclear scientists studied this problem and concluded that stability or instability is connected with the pairing of the nuclear spins.
    Just as electrons spin around their own axes and just as electron- spin pairing leads to be stable chemical bonds, so also the nuclear protons and neutrons spin around their own axes, and pairing of spins of neutrons among neutrons, and pairing of spins of protons among protons leads to nuclear stability. When all the non-radioactive, stable isotopes of elements are considered it is observed that,
  1. Nuclei with an even number of protons and even number of neutrons are most abundant and most stable. It is observed that even number leads to spin pairing, and odd number leads to unpaired spins.
  2. Nuclei with an even number of neutrons and odd number protons or an odd number of neutrons and even number of protons slightly less stable than even number of neutrons and protons.
  3. By far the least stable isotopes are those which have odd numbers of protons and an odd number of neutrons. The nuclear spin pairing is a maximum when odd numbers of both are present.

Protons and neutrons ratio

    Along with the odd or even number of protons and neutrons, nuclear stability also influenced by the relative numbers of protons and neutrons.
    The neutron/proton ratio (n/p) helps us to predict which way an unstable radioactive nucleus will decay.
Protons to neutrons is cause of radioactivity in radioactive elements
Protons to neutrons are the cause of radioactivity
    The above graph is obtained by plotting the number of neutrons in the nuclei of the stable isotopes against the respective number of protons.
    A study of this figure shows that the actual n/p plot of stable isotopes breaks from hypothetical 1:1 plot around an atomic number 20, and thereafter rises rather steeply, this indicates that as the number of protons increases inside the nucleus more and more neutrons are needed to minimize the proton-proton repulsion and thereby to add to nuclear stability.
    Neutrons, therefore, serve as binding martial inside the nucleus. The way an unstable nucleus disintegrates will be resided by its position with respect to the actual n/p plot of nuclei.
    When the isotope located above this actual n/p plot it is too high an n/p ratio, and when it is located below the plot it is too low in n/p ratio. In either case, the unstable nucleus should decay so as to approach the actual n/p plot. We now discuss the two cases of decay.

Neutron- to - proton ratio too high

    Isotopes with too many neutrons in the nucleus ( that is, with more neutrons than its needs for stability) can attain greater nuclear stability if one of the decays of the neutron to the proton. Such a disintegration leads to the ejection of an electron from inside the nucleus.
    0n1 1H1 + -1n0(electron)
    Thus beta ray emission will occur whenever the n/p ratio is higher than the n/p value expected for stability.
    This is almost always the case when the mass number of radioactive isotopes is greater than the average atomic weight of the element.

Neutron- to - Proton ratio too low

    A nucleus deficient in neutrons will tend to attain nuclear stability by converting one of its protons to a neutron and this will be achieved either by the emission of a position or by the capture of an electron.
    1H1 0n1 + +1e0(Positron)
    Such Decay occurs with a radioactive isotope whose mass number is less than the average atomic weight of the element. Positron emission occurs with light isotopes of the elements of low atomic number.
7N13   6C13 + +1e0
53I121   52Te121 + +1e0
    Orbital electron capture occurs with too light isotopes ( too low n/p) of the elements of relatively high atomic numbers. For such elements, the nucleus captures an electron from the nearest orbital ( K shell; n = 1) and thus changes one of its protons to a neutron.
37Rb82 + -1e0 36Kr82
79Au194 + -1e0 78Pt194
    Among the heaviest nuclei, the total proton-proton repulsion is so large that the binding effect of the neutron is not enough to lead to a stable nonradioactive isotope. For such nuclei, alpha particle emission is the common mode of decay.
90Th232 88Ra228 + 2He4
92U235 90Th231 + 2He4
92U238 90Th234 + 2He4

Units of radioactivity measurement

    In Practice, radioactivity is expressed in terms of the number of disintegration per second. One gram of Radium undergoes about 3.7 × 10¹⁰ disintegrations per second.
    The quantity of 3.7 × 10¹⁰ disintegrations per second is called curie, which is the older unit of radioactivity. Millicurie and microcurie respectively correspond to 3.7×10⁷ and 3.7×10⁴ disintegration per second.
  • Explanation:
    On this basis, the radioactivity of radium is 1 curie per gram. Phosphorus-32, a beta - emitter, has an activity of 50 millicuries per gram. This means that for every gram of phosphorus-32 in some material containing this species, there are 50 × 3.7 × 10⁷ disintegrations taking place per second.
    1 disintegration per second is called becquerel(Bq), it is the S.I. unit of radioactivity.
1 Curie = 3.7 × 10¹⁰ Bq
    Another practical unit of radioactivity is Rutherford(Rd).

Radioactive disintegration

    When an Alpha particle ejected from within the nucleus the mother element loss two units of atomic number and four units of mass number. Thus, if a radioactive element with mass number M and atomic number Z ejected an alpha particle the newborn element has mass number = (M - 4) and atomic number = (Z - 2).
    ₈₈Ra²²⁶ ₈₈₋₂Rn²²²⁻⁴ + ₂He⁴(∝) ₈₈Ra²²⁶ ₈₈Rn²²² + ₂He⁴(∝)
    When a beta particle is emitted from the nucleus, the daughter element nucleus has an atomic number one unit greater than that of the mother element nucleus. Thus, if a radioactive element with mass number M and atomic number Z ejected a beta particle the newborn element has mass number the same and atomic number = (Z + 1).
    ₉₀Th²³⁴ ₉₁Pa²³⁴ + ₋₁e⁰(β)

Disintegration series

    We have just seen that the radioactive elements continue to undergo successive disintegration till the daughter elements become stable, non- radioactive isotopes of lead. The mother element along with all the daughter elements down to the stable isotope of lead is called a radioactive disintegration series.

Uranium - 238 disintegration series

    Uranium - 238 decays ultimately to an isotope of lead. The entire route involves alpha emissions in eight stages and beta emission in six stages, the overall process being,
    ₉₂U²³⁸ ₈₂Pb²⁰⁶ + 8 ∝ + 6 β
    The mass number of all the above disintegration products are given by (4n +2) where n = 59 for Uranium - 238. This disintegration series is known as (4n + 2) series.
Uranium - 238 Radioactive Disintegration Series (4n + 2) Series
Uranium - 238 radioactive series in cause of radioactivity

Uranium - 235 disintegration series

    The (4n+3) series (n=an integer) starts with Uranium - 235(n=58) and end with the stable isotope Lead - 207(n=51).
    The entire route involves alpha emissions in seven stages and beta emission in four stages, the overall process being,
    ₉₂U²³⁵ ₈₂Pb²⁰⁷ + 7 ∝ + 4 β
Uranium - 235 Disintegration Series (4n + 3 Series)
Uranium - 235 radioactive series in the cause of radioactivity

Thorium- 232 disintegration series

    The (4n) series (n=an integer) starts with Thorium - 232(n=58) and end with the stable isotope Lead - 208(n=52). 
    The entire route involves alpha emissions in six stages and beta emission in four stages, the overall process being, 
    ₉₀Th²³² ₈₂Pb²⁰⁶ + 6 ∝ + 4 β
Thorium- 232 Disintegration Series (4n Series):
Thorium - 232 radioactive series in the cause of radioactivity

Group displacement law

    When an alpha particle is emitted in a radioactive disintegration step, the product is displaced two places to the left in the Periodic Table but the emission of a beta particle results in a displacement of the product to one place to the right.
Problem
    79Au197 is nonradioactive but 88Ra226 is Radioactive-Why?
Answer
    The number of the neutron in gold(Au) is 118 and proton is 79. So, n/p ratio = 118/79 =1.49 Which is less than 1.5 thus Au-197 is stable.
    But in Ra - 226 the number of Neutrons(n) = 138 and number of proton(p) = 88. So n/p ratio = 1.57. Thus, Ra-226 is radioactive.
Problem
    Which of the following elements are beta emitter and why? (i) 6C12 and 6C14 (ii) 53I127 and 53I133.
Answer
  1. The n/p ratio for stable carbon C-12 is 1.0(6n + 6p) but that for C - 14 is 1.3 (8n + 6p). It will predict that carbon -14 will be radioactive and will emit beta rays.
    • 6C14 7N14 + -1e0( beta ray)
  2. Similarly the n/p ratio of the stable Iodine-127 is 1.4 (74n + 53p). For Iodine-133 the n/p ratio equals 1.5 (80n + 53p) and it is again predicted to the beta emitter.
    53I127 54Xe137 + -1e0(beta ray)
Problem
    If a radioactive element x number disintegration per second. Express the radioactivity of this element in Curie.
Answer
    3.7 × 10¹⁰ disintegrations per second = 1 curie .
    Thus x number of disintegration per second = x/(3.7 × 10¹⁰) Curie. 

Oxidation and reduction are always found to go hand in hand during a redox reaction. Whenever an element or compound is oxidized, another element or another compound must be reduced.
    An oxidant is reduced and simultaneously the reductant is oxidized. It follows from our above discussion that we should be able to balance oxidation-reduction reactions on the basis of
  • Ion-electron method
  • Oxidation number method

Ion-Electron Method

  1. Ascertain the reactants and products, and their chemical formulas.
  2. Set up the partial equations representing the reduction of the oxidant, and the oxidation of the reductant. 
  3. If the reaction occurs in acid medium use a requisite number of H⁺ for balancing the number of atoms involved in the partial equation. for alkaline medium use OH⁻ ions.
  4. Balancing the charge in the partial equations by adding a suitable number of electrons. These electrons indicate the electrons involves in the oxidation or the reduction half-reactions.
  5. Multiply each partial equation by a suitable factor so each of the two partial equations involves the same number of electrons.
  6. Add the partial equations and cancel out species that appear on both sides of the equations.
    An aqueous acid medium, potassium permanganate oxidizes ferrous state to the ferric state
    In this reaction, permanganate ions are the oxidant and ferrous ion the reductant. The left-hand side of the ultimate equation will carry, MnO₄⁻ (KMnO₄), H⁺ (H₂SO₄) and Fe²⁺ (FeSO₄).
    The Right-hand side will have as products, Mn²⁺ (MnSO₄), H₂O and Fe³⁺ [Fe₂(SO₄)₃]. The partial equation representing the reduction of the oxidant MnO₄⁻ will involve,
MnO₄⁻ → Mn²⁺
    Since the reaction occurs in an acid medium we utilize eight H⁺ ions to balance the four oxygen atoms in MnO₄⁻.
Thus, MnO₄⁻+8H⁺ → Mn²⁺+4H₂O
    Since the above partial equation is still unbalanced from the viewpoint of charge, the equation is balanced by bringing in five electrons:
MnO₄⁻ + 8H⁺ + 5e ⇆ Mn²⁺ + 4H₂O
    If we now apply the above steps to obtain the partial equation representing the oxidation of reductant we have the following:
Fe⁺² ⇆ Fe⁺³ + e
    This equation on multiplication by 5 and then adding to the partial equation of the oxidant gives the final balanced equation:
MnO₄⁻ + 8H⁺ + 5e ⇆ Mn²⁺ + 4H₂O
5×( Fe⁺² ⇆ Fe⁺³ + e)

MnO₄⁻ + 8H⁺ + 5Fe⁺² → Mn²⁺ + 5Fe⁺³ + 4H₂O
    Oxidation of potassium iodide by potassium dichromate in dilute acid medium.
    In this reaction, dichromate is reduced to (+6) state to (+3) state and iodide ion is oxidized to elementary iodine. Taking the dichromate side, balancing the atoms and charges provides the partial equation
Cr₂O₇⁻² + 14H⁺ + 6e ⇆ 2Cr⁺³ + 7H₂O
(Reduction of Oxidant)
Considering the case of iodide ion we get
2I⁻ ⇆ I₂ + 2e
(Oxidation of Reductant)
    Multiplying the above equation by 3 and adding these two partial equations we have the following final equation
Cr₂O₇⁻² + 14H⁺ + 6e ⇆ 2Cr⁺³ + 7H₂O
3 (2I⁻ ⇆ I₂ + 2e)

Cr₂O₇⁻²+14H⁺+6I⁻→2Cr⁺³+3I₂+7H₂O 
    The reaction of permanganate ion with sodium stannite (Na₂SnO₂) in alkaline solution in which manganese dioxide (MnO₂) and sodium stannate(Na₂SnO₃) are formed.
    The Partial equation representing the reduction of oxidant is, Since the medium is alkaline we put the requisite number of OH⁻ ions to effect atom balance as
MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻
    To balance the charge on both sides three electrons are added on the left-hand side
MnO₄⁻ + 2H₂O + 3e ⇆ MnO₂ + 4OH⁻

For the reductant, the balance partial equation is
SnO₂⁻² + OH⁻ ⇆ SnO₃⁻² + H₂O + 2e
    To balance the number of electrons involved in the two partial equations the first equation is multiplied by 2 and the second by 3 and on the adding, we have the final form of the ionic equation
2MnO₄⁻ + 4H₂O + 6e ⇆ 2MnO₂ + 8OH⁻3
SnO₂⁻² + 3OH⁻ ⇆ 3SnO₃⁻² + 3H₂O + 6e

2MnO₄⁻+3SnO₂⁻²+H₂O→2MnO₂+3SnO₃⁻²+2OH⁻
    Oxidation of Mn⁺² to MnO₄⁻ by PbO₂ or NaBiO₃ in acid medium
Balancing oxidation reduction reactions
Balancing oxidation reaction

Oxidation number method

    Assuming that the oxidation number of oxygen remains unchanged manganese in MnO₄⁻ has an oxidation number +7. Calculation of the decrease and increase in oxidation number gives:
How Balancing Oxidation Reduction Reactions by oxidation number method?
Balancing oxidation-reduction reactions
    Putting the right factors the decreases and increases in oxidation numbers are balanced, giving,
MnO₄⁻ + 5Fe⁺² → Mn⁺² + 5Fe⁺³
    Since the reaction medium is acidic and the changes are not equal on the two sides of the above expression, H⁺ ions are added and the requisite number of H₂O written:
MnO₄⁻+8H⁺+5Fe⁺² → Mn⁺²+ 5Fe⁺³+4H₂O
    Oxidation of iodide ion by dichromate ion in acid medium
The reaction represented as Cr₂O₇⁻² + I⁻ → 2Cr⁺³ + I₂
    The oxidation number of Cr decreases by 2 and the oxidation number of Sn increases by 1. Equalizing the increase and decrease in oxidation number,
Cr₂O₇⁻² + 6I⁻ → 2Cr⁺³ + 3I₂
    The medium is acidic and because the ionic charges are not equal on the two sides, fourteen H⁺ ions are added to the left-hand side and the requisite number of H₂O added to the left-hand side.
Cr₂O₇⁻² + 6I⁻ + 8H⁺ → 2Cr⁺³ + 3I₂ +7H₂O
    Oxidation of sodium stannite to stannate in alkaline medium
This reaction represented as,
MnO₄⁻ + SnO₂⁻² → MnO₂ + SnO₃⁻²
    The oxidation number of Mn decreases by 3 and the oxidation number of Sn increases by 2. Equalizing the increase and decrease in oxidation number,
2MnO₄⁻ + 3SnO₂⁻² → 2MnO₂ + 3SnO₃⁻²
    Because the medium is alkaline and because the ionic charges are not equal on the two sides, two OH⁻ ions are added to the right-hand side.
2MnO₄⁻ + 3SnO₂⁻² → 2MnO₂ + 3SnO₃⁻² + 2OH⁻
    Since there are two hydrogen atoms on the right and none on the left, H₂O is added to make atom balance. The final equation is
2MnO₄⁻ + 3SnO₂⁻² + H₂O → 2MnO₂ + 3SnO₃⁻² + 2OH⁻
Problem
    Express by ion-electron method the reduction of permanganate to manganous stat by hydrogen peroxide in acid medium.
Answer
    This Reaction occurs in acid medium and the partial equation is
MnO₄⁻ + 8H⁺ + 5e ⇆ Mn⁺² + 4H₂O
(Reduction of Oxidant)
    In an acid medium, H₂O₂ will give O₂ and the partial equation being
H₂O₂ ⇆ 2H⁺ + O₂ +2e
(Oxidation of Reductant)
    The first equation is multiplying by 2 and the second is 5 to have electron balanced. We have,
2MnO₄⁻ + 16H⁺ + 10e ⇆ 2Mn⁺² + 8H₂O
5H₂O₂ ⇆ 10H⁺ + 5O₂ +10e

2MnO₄⁻+5H₂O₂+6H⁺⇆2Mn⁺²+8H₂O+5O₂
Problem
    Express by ion-electron method the reduction of nitrate ion to ammonia by aluminum in aqueous NaOH.
Answer
    In the alkaline medium, we use H₂O and OH⁻ ion according to convenience. The partial equation with charge balance is
NO₃⁻ + 6H₂O + 8e ⇆ NH₃ + 9OH⁻
    Metallic aluminum will go over to aluminate ion, the partial equation being
Al + 4OH⁻ ⇆ AlO₂⁻ + 2H₂O +3e
    Multiplying by right factors for electron balance we have the balanced equation
3NO₃⁻ + 18H₂O + 24e ⇆ 3NH₃ + 27OH⁻
8Al + 32OH⁻ ⇆ 8AlO₂⁻ + 16H₂O +24e

3NO₃⁻+8Al+2H₂O+5OH⁻→3NH₃+8AlO₂⁻
Problem
    Use the Oxidation Number method to balance the reaction of iodide ion and iodate ion in an acid medium to liberate iodine.
Answer
The reaction represented as
I⁻ + IO₃⁻ → I₂
    Representation of the above equation with the oxidation number of iodine,
I⁻(-1) + IO₃⁻(+5) → I₂(0)
    Since I⁻ (-1) oxidation number of iodide increases by 1 and IO₃⁻(+5) oxidation number decreases by 5. Putting the right factors the decrease and increase in oxidation number balanced
5I⁻ + IO₃⁻ → 3I₂
    Since the medium is acidic and the charges are unequal on two sides of the above expression we put Six H⁺ and write three H₂O to balanced the equation
I⁻ + IO₃⁻ + 6H⁺ → I₂ + 3H₂O
Problem
    Use the oxidation number method to balance the reaction between Sulfurous acid and dichromate in acidic medium.
Answer
The reaction represented as,
SO₃⁻² + Cr₂O₇⁻² → SO₄⁻² + 2Cr⁺³
    Oxidation of two Cr decreases by 2 × (+3) = 6 and the oxidation number of S increases by 2. Equalizing the above equation increase and decrees in oxidation number
3SO₃⁻² + Cr₂O₇⁻² → 3SO₄⁻² + 2Cr⁺³
    Because the medium is acidic and the ionic charges are not equal on the two sides, eight H⁺ ions are added to the left-hand side.
3SO₃⁻² + Cr₂O₇⁻² + 8H⁺ → 3SO₄⁻² + 2Cr⁺³
    Since there are eight hydrogen ion on the left and none on the right, four H₂O added to make atom balance. Thus the final equation is
3SO₃⁻²+Cr₂O₇⁻²+8H⁺→3SO₄⁻²+2Cr⁺³+4H₂O

The history of the atomic chronicle must also induce another great discovery, namely, the Phenomenon of Radioactivity.
    In 1896 the French scientist Becquerel, while investigating the nature of the mysterious X -rays discovered by Rontgen a few months earlier, found that a photographic plate wrapped in thick black paper was affected by a sample of potassium - uranyl - sulfate placed over it. In fact, any uranium compound would be effective the plate through covered by paper and kept away from light.
    The oblivious conclusion that some radiations emanating from the uranium compound could penetrate through the cover and attack the photographic plate.
    This penetrating radiation had its source in uranium itself and Becquerel christened this amazing behavior as Radioactivity. The properties of these radiations were very similar to those of X -rays.
  1. They were highly penetrating, they affected photographic plates, they would ionize gases and would also induce the fluorescence in some substances.
  2. The rays are not influenced by heat, light or chemical composition.

Discovery of radioactive elements

    Marie Curie found that the activity of mineral pitchblende was far greater than what was expected of its uranium content. In 1898 Pierre and Marie Curie actually isolated two new elements Polonium and Radium which were more radioactive compared to uranium, the heaviest atom known at the time.
    In 1900, Debierne and Giesel discovered actinium which was also radioactive. That the radioactive effects were essentially atomic was recognized early and this helps the isolation to a considerable extent.
    It was immaterial how uranium and radium chemically combined. The same number of radium atoms will always have the same activity independent of the physical state or environmental conditions.
    The phenomenon of radioactivity is associated with atoms that are heavier than lead or bismuth.
    The phenomenon of emission of radiation as a result of spontaneous disintegration in atomic nuclei was termed as radioactivity.

Radiation from radioactive elements

    The radiations emitted by naturally radioactive elements were shown to split by an electric or magnetic field into three distinct parts: Alpha(α), Beta(β) and Gamma(ɣ) Rays.

Alpha rays

    These consist of a stream of positively charged particle which carries +2 charge and has mass number is 4.
    These particles are shown by Rutherford to be identical with, the nuclei of the helium atom, that is, these are doubly charged helium ion He⁺²(atomic number 2, mass number 4).
    When an Alpha particle ejected from within the nucleus the mother element loss two units of atomic number and four units of mass number.
    92U238 92U234 + 2He4

Beta rays

    These are made up of a stream of negatively charged particles (beta particles). They have been shown to be identical with electrons from a study of their behavior in electric and magnetic fields and from the study of their e/m values (1.77 × 108 coulomb/gm). 
    The ejection of a beta particle (charge -1, mass 0) results from the transformation of a neutron (mass 1, charge 0) somewhere at the surface of the nucleus into a proton (mass 1, Charge +1). 
    0n11H1 + -1e0
    When a beta particle is emitted from the nucleus, the daughter element nucleus has an atomic number one unit greater than that of the mother element nucleus.
    90Th234  91Pa234 + -1e0
    Although beta particles and electrons are identical in their electrical nature and charge/mass ratio, there is a fundamental difference between them.
    The ejection of an electron from an atom converts a neutral atom into a positively charged ion but leaves the nucleus undisturbed. The ejection of a beta particle changes the very composition of the nucleus and produces an atom of the next higher atomic number.

Gamma rays

    These consist of electromagnetic radiation of very short wavelength (λ ∼ 0.005 - 1 Å). These are high energy photons.
    The emission of gamma rays accompanies all nuclear reactions. During all nuclear reactions there occurs a change in the energy of the nucleus due to the emission of alpha or beta particles. The unstable, excited nucleus resulting from the emission of an alpha or beta particle gives off a photon and drops a lower and more stable energy state.
    Gamma rays do not carry charge or mass, and hence emission of these rays cannot change the mass number or atomic of the mother nucleus.

Positrons

    Since the works of the Curies and Rutherford yet another mode of nuclear transformation has been discovered. This involves the ejection of a positron ₊₁e⁰ from within the nucleus.
    This ejection is made possible by the conversion of a proton into a neutron.
    ₁H¹ → ₀n¹ + ₊₁e⁰ 
    The ejection of positron lowers the atomic number one unit but leaves the mass number unchanged. 
    ₅₁Sb¹²⁰ → ₅₀Sn¹²⁰ + ₊₁e⁰

Neutrino

    Breaking down of a neutron into a proton and a beta particle creates a problem with the principle of conservation of angular momentum. Particles like Neutron, Proton, and Electron have the spin angular momentum of ±1/2 (h/2π) each. It is thus seen that the equation:
    ₀n¹ ₁H¹ + ₋₁e⁰
    This is not balanced in so far as angular momentum is concerned. If the angular momentum of the proton and the electron are +1/2 (h/2π) they exceed the angular momentum of the neutron.
    If they oppose each other then the momentum becomes zero in violation of that of the neutron. Pauli, therefore, postulated that along with the ejected beta particle another tiny neutral particle called neutrino is also ejected.
    This neutrino has also spin angular momentum of ±1/2 (h/2π). The sum of angular momentum of the particles ejected {say +1/2 (h/2π) for proton, -1/2 (h/2π) for the electron and +1/2 (h/2π) for the neutrino} may now be +1/2 (h/2π) being the same as that of the neutron.
    The mass of the neutrino is around 0.00002 with respect to the oxygen scale. Ejection of an electron from within the nucleus should be represented as:
NeutronProton+Electron+Neutrino
Radioactivity and artificial transmutation reactions
Radioactive reaction

Alpha, beta and gamma rays

 Radioactivity and properties of alpha beta and gamma rays
Alpha, beta and gamma rays

Nuclear and chemical reaction

Nuclear reactions are different from chemical reactions in many respects:
  1. Chemical reactions involve some loss, gain or overlap of outer orbital electrons of the reactant atoms. Such reactions cannot alter the composition of the nuclei so that the atomic number of the chemical reactions unchanged.
    CH4 + H2O CO + 3H2
    On the other hand cause of nuclear decay involves emission of alpha particles, beta particles or positrons from inside the nucleus, which leads to change in the atomic number of the nucleus.
    51Sb120  50Sn120 + +1e0
    In some artificially induced radioactive decay reactions, neutrons are absorbed by target nucleus producing isotopes. Nuclear reactions, therefore, leads either to the birth of another element or produce isotopes of the parent element.
  2. The nuclear reactions are accompanied by energy changes which far exceed the energy changes in chemical reactions.
    For example, the energy evolved in the radioactive transformation of one gram of radium is five hundred thousand times as large as the energy released when one gram of radium combine with chlorine to form RaCl2.

Rutherford's planet-like model of the atom is contested by Bohr's Model in 1913 on two grounds, According to classical mechanics, whenever a charged particle is subjected to the acceleration it emits radiation and loses energy.
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,
      E1ㄑE2ㄑE3ㄑ......
      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.
    Hydrogen energy levels and Bohr's model
    Hydrogen energy levels
      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 the principle of quantization of angular momentum according to which,
      mvr = n × (h/2π)
      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,
    Bohr's model radius, velocity, and energy of an electron
    Bohr's model energy of an 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 of an atom 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
      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/2π (where n have values 1,2,3, ......∞)
    or, v = n × (h/2π) × (1/mr)

    Then, e²/r = mv²

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

    or, e²/r = n²h²/4π²mr²

    ∴ r = n²h²/4π²me²

    Bohr's nth and first orbit

    ∴ rn = n²h²/4π²me²
    or, rn = n² × (h²/4π²me²)

    Thus, rn = n² × r₁

    Radius of first Bohr orbit

      Taking n = 1, the radius of the first orbit is r₁.
    ∴ r₁ = 1 × h²/4π²me²
    = {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
      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/2π, and r = n²h²/4π²me²
    Then, v = (nh/2πm) × (1/r)
    = (nh/2πm) × (4π²me²/n²h²)

    ∴ v = 2πe²/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)
      The energy associated with the permitted orbits is given by,
    E = - ½ (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,
    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)×(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

    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.
    Answer
    Velocity of an electron in the Bohr's model, = 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)
    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,
    2πr = 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
    Question
      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
    Kinetic Energy = ½ 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 × 10⁻¹³erg
    Question
      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?
    Answer

    The energy of an electron in the excited state of Li⁺²
    ELi⁺² = - 30.6 eV
    Question
      What is the radius of the second orbit of a hydrogen atom? Give radius of first orbit = 0.529 Å.
    Answer
      Thus the radius of the second orbit of a hydrogen atom = 2.12 Å
    Question
      H, H⁺, He⁺ and Li⁺² - for which of the species Bohr's theory is not applicable? Explain
    Answer
      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.

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