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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.

Ideal Gas Equation its derivation:

    Boyle's law, V ∝ 1/P
    When n and T are constant for a gas.
    Charl's Low, V ∝ T
    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 its derivation and universal gas constant
Ideal gas equation its derivation
    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:

    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 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 a gas when its temperature is raised by one kelvin.

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

Number of Gas Molecule Present in 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′ 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

Ideal Gas Low problems and solutions:

  • Problem 1:
    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 2:
    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.
  • Answer:
    (a) 82.05 atm cm3 mol-1K-1
    (b) 8.314 × 1014 dyne m-2 mm3 mol-1 K-1
  • Problem 3:
    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
    For the details solution see
  • Problem 4:
    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 5:
    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:
    PV = nRT
    or, PV = (g/M)RT
    or, M = (g/PV) × RT
    = (g × R × T)/(P × V)
    = (12.8 × 0.082 × 300)/{(750/760) × 10}
    = 31.91
  • Problem 6:
    Calculate the pressure excreted on the walls by ideal gas of a 3 litre of flask when 7 gms of nitrogen are introduced into the same 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 7:
    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.
  • Solution:
    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

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 correlated to the experimental facts. Fortunately such theory has been developed for Formulation of Kinetic Theory of Gases based upon the certain postulates which are supposed to be applicable to an Ideal Gas.

Assumptions of Kinetic Theory of Gases:

  1. The gas is composed very small discrete particles, now called molecules. For a gas the mass and size of the molecules are same and different for different gases.
  2. The molecules are moving in all directions with verity of speeds. Some are very fast while other are slow.
  3. Due to random motion, the molecules are executing collision with the walls of the container (wall collision) and with themselves (inter molecular collision). These collision 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 inter molecular attraction specially at law pressure, that is one molecule 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 volume is reduced, wall collision becomes more frequent and pressure is increased.
  8. Through the molecular velocity are constantly changing due to inter molecular collision, average kinetic energy(є) of the molecules remains fixed at a given temperature. This explain the Charl's law that when T is increased , velocity are increased, wall collision become more frequent and pressure(P) is increased when T kept constant or Volume(V) is increases when P kept constant.

Root Mean Square Speed:

    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 a gas of molecular mass m and RMS speed is CRMS at temperature T and Pressure P.
Molecular Velocity and its Components in Formulation of Kinetic Theory of Gas
Formulation of Kinetic Theory of Gas
    Let in a 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 have resultant velocity C1 and the component velocities are Cx, Cy and Cz.
    So that, 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,
    = mCx - (- mCx)
    = 2 mCx
    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 are 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 m N CRMS2
    Where CRMS2 = Root Mean Square Velocity of the Gases.
    According to the Newton's 2nd Low of Motion, rate of change of momentum due to wall collision is equal to force developed within the gas molecules.
    That is, P × 6l² = 2 m N CRMS2/l
    or, P × l³ = 1/3 m N CRMS2
∴ PV = 1/3 m N CRMS2
    Here, l³ = Volume of the Cube Contain 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.
Formulation of Kinetic Theory of Gas and Kinetic gas equation
Kinetic Gas Equation
    This equation are also valid for any shape of the gas container.

Derivation of root mean square velocity:

    Let us apply the kinetic equation for 1 mole Ideal Gas.In that case mN = mN₀ = M and Ideal Gas Equation, PV = RT.
    Hence from Kinetic Gas Equation,
    PV = 1/3 m N CRMS2
    or, RT = 1/3 m N CRMS2
or, CRMS2 = 3RT/M
∴ CRMS2 = √3RT/M
    Thus root mean square velocity depends on the molar mass(M) and temperature(T) of the gas.

Expression of Average Kinetic Energy:

    The average kinetic energy(Ē) is defined as,
    Ē = 1/2 m CRMS2.
    Again from 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 of T only and it is not dependent on the nature of the gas.
  • Problem 1:
    Calculate the pressure exerted by 1023 gas particles each of mass 10-22 gm in a container of volume 1 dm3. The root mean 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 Kinetic Gas Equation,
    PV = 1/3 mNCRMS2
    or, P = 1/3 × m NV × 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 P
  • Problem 2:
    Calculate the root mean square speed of oxygen gas at 270C.
  • 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-1K-1 × 300 K)/(32 gm mol-1)
    ∴ CRMS = 48356 cm sec-1
  • Problem 3:
    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 4:
    How the root mean 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 5:
    Calculate the kinetic energy of translation of 8.5 gm NH3 at 270C.
  • 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 Kinetic Energy of 8.5 gm NH3 at 270C is,
    (0.5 × 900) cal
    = 450 cal
  • Problem 6:
    Calculate the RMS velocity of oxygen molecules having a kinetic energy of 2 mol-1. At what temperature the molecules have this value of KE ?
  • Solution:
    T = 673.9K or 400.90C

Different application of dipole moment are taken one after another.
  • Determination of Partial Ionic Character and Residual Charge on the Constituent Atoms in a Molecule:
    Let us consider a molecule A - B 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 100% ionic, and B is more electronegative then A, B will carry 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.
Application of dipole moment for calculation of partial ionic character
Partial Ionic Character

Induced polarization (Pi):

    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, radius of the molecule r can be determined by measuring D൦ of the substance at NTP.

Determination of molecular structure :

  • Mono-atomic Molecules (A):
    The mono-atomic inert gases are non-polar and it indicates the symmetrical charge distribution in molecule.
  • Di-atomic Molecules (AX):
    The homonucler 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 vale of dipole moment and this indicates the unsymmetrical charge distribution,
    Heteronucler diatomic molecules are always polar due to difference of electronegativity of the constituent atoms. The example are , HCl, HBr, HF etc. this indicates that 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 (AX₂):
    The molecules, like CO₂, BeCl₂, BeF₂, SnCl₂ etc have zero dipole moment indicating that the molecules have symmetrical linear structure.
    For example, CO₂ has structure, The electric moment of one C - O bond ( known as bond moment ) cancel the electric moment of the other C - O bond.
Application of Dipole moment for Tri-atomic Molecules
Dipole moment of CO₂ and BeCl₂
    The electric moment associated with the bond arising from 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 helps 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.
    For 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 (AX₃):
    The molecules like BCl₃, BF₃ etc have dipole moment zero indicating that they have regular planar structure. There 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
  • Problem 1:
    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%
    For Solution Follow the link
Questions and Answers
  • Problem 2:
    The Difference between electronegativity of Carbon and Oxygen Is large but the dipole moments of carbon monoxide is very low - Why?
  • Answer:
    However, in CO, there are large difference of electronegativity between C and O but the molecule is very low value of dipole moment. This suggested that the charge density in O-atom is somehow back-donated to C-atom.
    This explain by forming a coordinate covalent bond directing towards C-atom.
  • Problem 3:
    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.
  • Problem 4:
    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-H
    or, 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
  • Problem 5:
    Show that the bond moment vectors of NF₃ molecule adds up to zero.
  • Answer:
    While other type of the molecule such as NH₃, PH₃, AsH₃ are polar (μ≠0) indicating that that the molecule have 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 very small dipole moment though there is great difference of electronegativity between N and F atoms and similar structure of NH₃.
    Low value of μ of NF₃ is explain by the fact that resultant bond moment of the three N - F bonds is acting in 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

The rate at which a radioactive sample disintegrates can be determined by counting the number of particles emitted in a given time.
    Radioactive decay is one important natural phenomenon obeying the first order rate process. Rate of these reactions depends only on the single power of the conc. of the reactant. The rate can be expressed as,
(dN/dt) = - k N
    Where N is the number of the atoms of the disintegrating radio-element present at the any time, dt is the time over which the disintegration is measured and k is the radioactive decay rate constant.
    Again, k = - (dN/dt)/N
    Thus, the rate constant(k) is defined as the fraction decomposing in unit time interval provided the conc. of the reactant in kept constant by adding from outside during this time interval.
    The negative sign shows that N decreases with time.
    Let N₀ = number of the atoms present at the time t = 0 and N = Number of atom present after t time interval.
    Rearranging and integrating over the limits N₀ and N and time, 0 and t.
Rate of radioactive decay half life for radioactive elements
Rate of radioactive decay half life

Half life period of radioactive elements:

    After a certain period of time the value of (N₀/N ) becomes one half, that is, half of the radioactive elements have undergone disintegration.
    This period is called half life of radioactive element and is a characteristic property of a radioactive element.
Half life of the radioactive elements
Half life and rate of radioactive decay
    If a radioactive element is 100% radioactive and the half life period of this element 4 hour.
    Thus after four hour it decompose 50% and remaining 50%.
    After 8 hour it decompose 75% and reaming 25% and the process is continued. The half life is given by,
    2.303 log(N₀/N ) = kt
    When t = t½, N = N₀/2 
    Putting in the above equation we have,
    2.303 log{N₀/(N₀/2)} = k t½
    or, t½ = 0.693/k
k = 0.693/t½
    The above relation shows that both the half-life and radioactive decay rate constant are independent of the amount of the radio-element present at a given time.
    ₈₄Po²¹³ has t½ = 4.2 × 10⁻⁶ sec, whereas ₈₃Bi²⁰⁹ is 3 × 10⁷ years.

Calculation of the Age Of Radioactive Elements:

  • Age of Organic Material:
    A method of determining the age of organic martial based on the accurate determination of the ratio of carbon-14 and carbon-12. carbon 14 is produced in the atmosphere by the interaction of neutron with ordinary nitrogen.
    ₇N¹⁴ + ₁n⁰ ₆C¹⁴ + ₁H¹
    ( Cosmic Reaction)
    The carbon-14 ultimately goes over to carbon-14 dioxide. A steady state concentration of one ¹⁴C to ¹²C is reached in the atmospheric CO₂ .
    This carbon dioxide is taken in or given out by plants and plant eating animals or human beings so they all bear this ratio. 
    When a plant or animals died the steady state is disturbed since there is no fresh intake of stratospheric CO₂ the dead matter is out of equilibrium with the atmosphere.
    The ¹⁴C continues to decay so that there after a number of years only a fraction of it is left on the died matter. Therefore the ratio of the ¹⁴C/¹²C drops from the steady state ratio in the living matter.
    ₆C¹⁴ ₇N¹⁴ + ₋₁e⁰(t½ = 5760 years)
    By measuring this ratio and comparing it with the ratio in living plants one can estimate when the plant died.
  • Determination of the Age of Rock Deposits:
    A knowledge of the rate of decay of certain radioactive isotopes helps to determine the age of various rock deposits. Let us consider a uranium containing rock formed many years ago.
    The uranium started to decay giving rise to the uranium - 238 to lead -206 series. The half lives of the intermediate members being small compared to that of uranium -238(4.5 × 10⁹ years), it is reasonable to assume that those uranium atoms that started decaying many many years ago must have been completely converted to the stable lead-206 during this extra long period.
    The uranium-238 remaining and the lead-206 formed must together account for the uranium 238 present at zero time , that is, when the rock solidified.
    Thus both N₀ and N are known k is known from a knowledge of the half life of uranium -238. Therefore the age of the rock can be calculated.

Determination of the Avogadro Number:

    If 1 gm of a radioactive element contain N number of atoms.
    Then, N = N₀/m
    Where N₀ = Avogadro Number and m =
    Mass Number
    Therefore, kN = (k × N₀)/m
    Where k = 0.693/t½
    Thus, kN = (0.693 × N₀)/(t½ × m)
∴ N₀ = (kN × t½ × m)/0.693
    If each atom of the radio-element expels one particle then kN, the rate of decay, is also the rate at which such particles are ejected.
    One gram of radium undergoes 3.7 × 10¹⁰ disintegration per second, mass number of Ra-226 is 226 and t½ = 1590 year, so that the Avogadro Number can be calculated by the above equation,
    Avogadro Number =(3.7 × 10¹⁰ sec⁻¹ × 1590 × 365 × 24 × 60 sec × 226)/0.693
    = 6.0 × 10²³

Average life period of radioactive elements:

    Besides half life of a radio-element another aspect of the life of a radio-element must also be known. It is now possible to determine the average life period of a radioactive atom present in an aggregate of a large number of atoms.
    Average life period of an atom of the radio-element tells us the average span of time after which the atom will disintegrate.
    The length of time a radio-element atom can live before it disintegrates may have values from zero to infinity.
    This explains the gradual decay of the radio-element instead of decay of all the atoms at the same time. The average life period may be calculated as follows- 
    tav = Total Life Period/Total Number of Atoms
    Let N₀ atoms of a radioactive element are present in an aggregate of large no of atoms at time zero. Now in the small time interval t to (t+dt), dN atoms are found to disintegrate.
    Since dt is a small time period, we can take dN as the number of atoms disintegrating at the time t. So the total life time of all the dN atoms is t dN.
    Again the total number of atoms N₀ is composed of many such small number of atoms dN₁, dN₂, dN₃ etc, each with its own life span t₁, t₂, t₃ etc.
Average life of the radioactive elements
Average life and rate of radioactive decay
    The average life of Radio-element is thus the reciprocal of its radioactive disintegration constant. This result can also be derived in a very simple alternative way.
    Since the radioactive atoms may be regarded as having an average life, then the product of the fractions of atoms disintegrating in unit time (that is k) and average life must be unity. 
    tav k = 1
    So that, tav = 1/k
    Thus the relation between average life and half life is,
t½ = 0.693/k = 0.693 tav
  • Problem 1:
    A piece of wood was found to have ¹⁴C/¹²C ratio 0.7 times that in the living plant. Calculate the approximate period when the plant died(t₁/₂ = 5760 years).
  • Answer:
    We know that radioactive decay constant,
    k = 0.693/(t½)
    = 0.693/5760 years
    = 1.20 ×10⁻⁴ yr⁻¹
    N₀/N = 1.00/0.70
    ∴ 2.303 log(N₀/N) = kt
    or, t = (2.303 log(N₀/N)/k
    Putting the value, above equation, we have,
    t = (2.303 × 0.155)/(1.20 × 10⁻⁴)
    = 2970 years
  • Problem 2:
    A sample of uranium (t₁/₂ = 4.5 × 10⁹) ore is found to contain 11.9 gm of uranium-238 and 10.3 gm of lead-206. Calculate the age of the ore.
  • Answer:
    11.9 gm of uranium-238 = 11.9/238
    = 0.05 mole of uranium
    10.3 gm of lead-206 = 10.3/206
    = 0.05 mole of lead -206
    Thus mole of uranium -238 present in the ore at zero time = (0.05+0.05) = 0.010 mole.
    And radioactive decay constant
    = 0.693/(4.5 × 10⁹)
    = 0.154 × 10⁻⁹ yr⁻¹.
    Then 2.303 log(0.10/0.05) = kt
    ∴ t = (2.303 log2)/(0.154 × 10⁻⁹ yr⁻¹)
    = 4.5 × 10⁹ year

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 neutron to proton of the unstable, radioactive nucleus is the factor responsible for radioactivity.
    Nuclear scientist studied this problems and concluded that the 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 even number of protons and even number of neutrons are most aboundent and most stable. It is observed that even number leads to spin pairing, and odd number leads to unpaired spins.
  2. Nuclei with even number of neutrons and odd number protons or odd number of neutrons and even number of protons slightly lees stable then even number of neutrons and protons.
  3. By far the least stable isotopes are those which have odd numbers of protons and 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, the 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 is cause of radioactivity
    The above graph is obtained by plotting the number of neutrons in the nuclei of a 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 of 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 disintegrate 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:

    An isotopes with too many neutrons in the nucleus ( that is, with more neutrons than it need for stability) can attain greater nuclear stability if one of the neutrons decay to proton. Such a disintegration leads to 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 then the n/p value expected for stability.
    This is almost always the case when the mass number of the radioactive isotopes is greater then 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 proton to a neutron and this will be achieved either by the emission of a positron 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 number. 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 neutron is not enough to lead to a stable non radioactive 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.
    Milicurie and microcurie respectively corresponds to 3.7×10⁷ and 3.7×10⁴ disintegration per second.
  • Explanation:
    On this basis, radioactivity of radium is 1 curie per gram. Phosphorus-32, a beta - emitter, has an activity of 50 milicuries 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 a alpha particle the new born 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 new born element has mass number 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 becomes 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 Radioactive Disintegration Series (4n + 2) Series:

    Uranium - 238 decays ultimately to an isotopes 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 (4n + 3 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 cause of radioactivity

Thorium- 232 Disintegration Series (4n 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 cause of radioactivity

Group displacement Low: 

    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 1:
    79Au197 is non radioactive but 88Ra226 is Radioactive-Why?
  • Answer:
    The number of neutron in gold(Au) is 118 and proton is 79.
    So, n/p ratio = 118/79
    Which is lees then 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 2:
    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 3:
    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. 

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