April 2019

Study ideal gas law in chemistry and physics

A gas at equilibrium has a definite value of pressure, volume, temperature, and composition. These are called state variables and are determined experimentally.

The state of the gas can be defined by these variables. Boyle's in 1662, Charles's in 1787 and Avogadro laws give the birth of an equation of state for Ideal gas.

Derivation of Ideal gas law

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

Charles 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
Derive Ideal gas law and gas constant
Ideal gas law
This is called the ideal gas law of the state. This equation is found to hold most satisfactory when pressure tense to zero.
At ordinary temperature and pressure, the equation is found to deviated about 5%. Real gases attain ideal behavior only at low pressures and very high temperatures.

Value of universal gas constant

At NTP 1 mole gas at 1-atmosphere pressure occupied 22.4 lit of gas.

Ideal gas law
PV =RT
or, R = (PV)/(nT).

Putting the values of pressure, volume, and temperature in the above equation.

R = ( 1 atm × 22.4 lit)/(1 mole × 273 K)
= 0.082 lit atm mol⁻¹ K⁻¹
Problem
Derive the value of gas constant R when pressure is expressed in the atm, and volume in cm³ and pressure in dyne m⁻² and volume mm³.

Answer

  1. 82.05 atm cm³ mol⁻¹ K⁻¹
  2. 8.314 × 10¹⁴ dyne m⁻² mm³ mol⁻¹ K⁻¹

Universal gas constant in CGS and SI units

Pressure = 1 atm = 76 cm Hg
= 76 cm × 13.6 gm cm⁻² × 981 cm sec⁻²
= 76 × 13.6 × 981 dyne cm⁻²

Volume = 22.4 × 10³ cm³

Temperature = 273 K

∴ R = (76 × 13.6× 981 × 22.4 × 10³)/(1 × 273)
= 8.314 × 10⁷ dyne cm² mol⁻¹ K⁻¹

Work = force × displacement.
∴ erg = dyne cm².

R = 8.314 × 10⁷ erg mol⁻¹ K⁻¹

1 J = 10⁷ erg.
Thus universal constant of gas in SI units
= 8.314 J mol⁻¹ K⁻¹.

4.18 J = 1 calories,
∴ R = 8.314 / 4.18 calories mol⁻¹ K⁻¹
= 1.987 calories mol⁻¹ K⁻¹
≃ 2 calories mol⁻¹ K⁻¹
Problem
What is 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

What is the significance of gas constant?

For n mole ideal gas
PV = nRT
or, R = PV/nT.

∴ Unit of R = (unit of P × unit of V)/(unit of n × unit of T).

Pressure = (force/area)
= (force/ length²)
= force × length⁻².

Volume = length³.

∴ R = (force × length⁻² × length³)/(amount of gas × kelvin)
= (force × length)/(amount of gas × kelvin)

= (Work or Energy)/(amount of gas × kelvin).

R is energy per mole per kelvin or amount of work or energy that can be obtained from one mole of gas when its temperature is raised by one kelvin.

The molecular weight of ideal gas

n mole ideal gas
PV = nRT
or, PV= (g/M)RT

where g = weight of the gas in gram and M = molecular weight of the gas.

P = ( g/V) (RT/M)

Density = weight of the gas/volume
or, d = g/V

∴ P = dRT/M
From this equation, we can easily find out the molecular weight of a gas.

Problem
The volume of 12.8 grams of gas at 760 mm-Hg pressure and 27° C is 10 liter. Calculate the molecular weight of this gas.

Answer

Ideal gas law, PV = nRT.
or, PV = (g/M) × RT
∴ M = gRT/PV
= (12.8 × 0.082 × 300)/(1 × 10).
= 31.49 gm mol⁻¹
Problem
The density of ammonia at 5-atmosphere pressure and 30°C temperature 3.42 gm lit⁻¹. What is the molecular weight of ammonia?

Answer
The molecular weight of ideal gas,
M = dRT/P

∴ The molecular weight of ammonia
= (3.42 × 0.082 × 303)/5 gm mol⁻¹
= 16.99 gm mol⁻¹
≃ 17 gm mol⁻¹

Gas molecules present in an ideal gas

Ideal gas law for n mole gas,
PV = nRT.
or, PV= (N/N₀) RT
N = gas molecules present in an ideal gas,
N₀ = Avogadro number = 6.023 × 10²³.

∴ P = (N/V) × (R/N₀) × T


∴ P = N′ kT
N′ = gas molecules present per unit volume.
k = Boltzmann constant = R/N₀
= 1.38 × 10⁻¹⁶ erg molecules⁻¹ K⁻¹
Problem
Estimate the number of gaseous molecules left in a volume of 1 mi-liter if it pumped out to give a vacuum of 7.6 × 10⁻³ mm of Hg at 0°C.

Answer
Volume (V) = 1 ml = 10⁻⁶ dm³.

Pressure (P) = 7.6 × 10⁻³ mm Hg
= (7.6 × 10⁻³ mm Hg) (101.235 kPa/760 mm Hg)
= 1.01235 × 10⁻³ kPa.

Ideal gas law for n moles gas
PV = nRT
or, n = PV/RT
= (1.01235 × 10⁻³ × 10⁻⁶)/(8.314 × 273)
= 4.46 × 10⁻¹³ mole

Number of gaseous molecules
N = n N₀.
= (4.46 × 10⁻¹³ mole)(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 carbon dioxide contained in a vessel of 5-liter capacity at 50°C.

Answer
Moles of ethane
n₁ = 2/30 = 0.0667

Moles of carbon dioxide
n₂ = 3/44 = 0.0682

Total moles (n₁ + n₂) = (0.0667 + 0.0682)
= 0.1349

∴ Total pressure (P) = (n₁ + n₂)RT/V
or, P = (0.1349 × 0.082 × 323)/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 gas constant R by x.

Answer
Ideal gas law for 1-mole gas,
PV = nRT
or, V = (nR/P) × T.

The slope of the V vs T graph = nR/P
where P = 2 atm.

∴ (R/2 atm) = x L mol⁻¹ K⁻¹
Value of gas constant R = 2x L atm mol⁻¹ K⁻¹

Particles in a solid-liquid and gas

The common behavior of different gases suggested that the internal structure in all gases must be similar. Study in the seventeenth-century explained Gassendhi and Hooke explained the physical phenomena of gases on the assumption of the existence of rapidly moving independent minute particles. Bernoulli in 1738 the first time explained the properties of gases on a mechanical basis.

In solid the molecules or particles are held very closely together and are entirely devoid of any translatory motion. If heat is supplied to solid, it takes the form of vibrational motion with the rise of temperature. With the further increases the thermal energy, the vibrational motion rises to such extent the molecules break down to transform into a liquid.

Further absorption of heat causes the particles to brack away from the restraining forces holding together and the particles move away from the liquid to the gaseous state. The gases then are essentially composed freely moving particles.

These basic ideas were at the root of the theory to explain the behavior of gases called the kinetic theory of gases. In the nineteenth century, the effort of Joule, Kronig, Clausius, Boltzmann, and Maxwell, the theory succeeded in attaining a rigid mathematical form.

Fortunately, such theory has been developed for the formulation of the kinetic gas equation based upon certain postulates which are supposed to be applicable to an ideal gas.

Postulates of kinetic theory for gaseous molecules

  1. Gas molecules are composed of very small discrete particles. In any one gas, all the molecules are of the same size and mass, but these differ gas to gas.
  2. Gas molecules within the container are moving in all directions with a verity of speeds. Some are very fast while others are slow.
  3. Due to random motion, the gas molecules are executing collision with the walls of the container called wall collision and with themselves called 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. 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 at a constant temperature, wall collision becomes more frequent and pressure is increased.
  8. Through the molecular velocity constantly changing due to the intermolecular collision, the average kinetic energy of the gas molecules remains fixed at a given temperature.

Charles law from kinetic theory

When temperature raised, the molecules would move more vigorously resulting in a larger number of impacts on the wall of the container at constant volume. That is why we find an increase of pressure with a rise in temperature at constant volume. This is Charles's law.

    What is RMS for gas molecules?

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

    Formulate kinetic gas equation for the ideal gases

    Let us take a cubic gas container with edge length l containing N molecules of gas of mass m and RMS speed CRMS at temperature T and pressure P.

    The molecules are moving constantly with different velocities in different directions bombarding on the walls of the cube.

    Gas molecules, N₁ have velocity C₁, N₂ have velocity C₂, N₃ have velocity C₃, and so on.
    Concentrate our discussion on a single molecule among N₁ that has resultant velocity C₁ and the component velocities are Cx, Cy, Cz.
    Component velocity of gas molecules in cubic container
    Component velocity of gas molecules
    ∴ C₁² = Cx² + Cy² + Cz²

    The molecule will collide walls A and B of the container 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 for this collision,
    = 2 mCx × (Cx/l)
    = 2 mCx²/l.

    Similarly, along Y and Z directions, the rate of change of momentum for the molecule 2 mCy²/l and 2 mCz²/l respectively.

    Hence the net momentum imparted on all six walls of the container by the collision of the gas molecules,
    = 2 mCx²/l + 2 mCx²/l +2 mCz²/l
    = 2 (m/l) (Cx² + Cy² + Cz²)
    = 2 mC₁²/l

    For N₁ molecules, change of momentum
    = 2 mN₁C₁²/l.

    Consider all the molecules of the gas present in the cubic container, 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 CRMS = root means square velocity of the gases.

    According to Newton's second law of motion, the rate of change of momentum due to wall collision is equal to force developed within the gas molecules.

    P × 6l² = 2 mN CRMS²/l
    or, P × l³ = 1/3 m N CRMS²

    ∴ PV = 1/3 m N CRMS²

    where l³ = volume of the cubic container.
    Kinetic gas equation and kinetic energy of the gas molecules
    Kinetic gas equation

    Another form of the equation

    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 valid for any shape of the gas container.

    Problem
    Calculate the pressure exerted by 10²³ gas molecules each of the mass 10⁻²² gm in a container of volume 1 dm³. Given RMS speed of the gas molecule 10⁵ cm sec⁻¹.

    Solution

    Numer of gas molecules (N) = 10²³
    Mass of the gas molecule (m) = 10⁻²² gm
    = 10⁻²⁵ Kg
    Volume (V) = 1 dm³ = 10⁻³ m³
    CRMS = 10⁵ cm sec⁻¹ = 10³ m sec⁻¹.

    Kinetic gas equation,
    PV = 1/3 mN CRMS²
    or, P = (m × N × CRMS²)/(3 × V)

    ∴ P = (10⁻²⁵ × 10²³ × 10⁻⁶)/(3 × 10³)
    ∴ P = 0.333 × 10⁷ Pascal.

    Velocity formula for gas molecules

    The kinetic relations and ideal gas law may be used to formulate the velocity of the gas molecules
    Kinetic relation for a 1-mole ideal gas
    PV = 1/3 m N CRMS²
    where mN = mN₀ = M.

    Ideal gas law for one-mole gas
    PV = RT.

    ∴ PV = 1/3 m N CRMS²
    or, RT = 1/3 M CRMS²

    or, CRMS² = 3RT/M

    ∴ CRMS = √3RT/M
    Thus root means square velocity depends on the molecular weight and temperature of the gas molecules.

    Problem
    Calculate the root mean square speed of oxygen gas at 27⁰C.

    Solution

    RMS velocity of a gas molecule
    CRMS² = (3RT/M).

    M = 32 gm mol⁻¹.
    T = 27° C = (273+27)K = 300 K.

    CRMS² = (3 × 8.314 × 10⁷ × 300)/(32)

    ∴ CRMS = 48356 cm sec⁻¹

    Problem
    Calculate the RMS speed of ammonia at N.T.P.

    Solution
    At N.T.P,
    V = 22.4 dm³ mol⁻¹ = 22.4 × 10⁻³ m³ mol⁻¹
    P = 1 atm = 101325 Pa
    M = 17 × 10⁻³ Kg mol⁻¹.

    CRMS² = 3RT/M
    = (3 × 101325 × 22.4 × 10⁻³) /(17 × 10⁻³)
    ∴ CRMS = 632 m sec⁻¹

    Compare RMS velocity of hydrogen and oxygen

    CRMS² = 3RT/M
    Hence at a given temperature,
    (CRMS of hydrogen)²/(CRMS of oxygen)² = MO₂/MH₂
    = 32/2 = 16

    ∴ RMS velocity of oxygen = 4 × RMS velocity of hydrogen.

    The kinetic energy of the gas molecules

    The average kinetic energy defined as,
    Ē = 1/₂ m CRMS².

    Kinetic gas equation
    PV = 1/3 m N CRMS²
    or, PV = ²/3 N × 1/2 (m CRMS²)
    or, PV = ²/3 N Ē.

    Ideal gas law for 1-mole gas
    PV = RT and N = N₀.

    ∴ RT = ²/3 N₀ Ē

    or, Ē = ³/₂ (R/N₀) × T
    ∴ Ē = ³/₂ kT
    k = R/N0 = Boltzmann constant of gas
    = 1.38 × 10-23 J K⁻¹.

    Total kinetic energy for 1 mole of the gas molecules

    ETotal = N₀ × (Ē)
    = ³/₂ RT

    Average kinetic energy dependent on temperature only and average kinetic energy independent of the nature of the gas.

    Problem
    Calculate the kinetic energy of translation of 8.5 gm ammonia at 27⁰C.

    Solution
    Kinetic energy for 1 mole of the gas molecule,
    Etotal = (3/2)RT.

    = (3/2)(2 × 300 K)
    = 900 cal mol⁻¹

    8.5 gm ammonia = (8.5/17) mol
    = 0.5 mol

    The kinetic energy of 8.5 gm ammonia at 27⁰C
    = (0.5 × 900) calories
    = 450 calories

    Application of dipole moment in chemistry

    When a covalent bond is formed between two identical atoms, the two electrons forming the covalent bond may be regarded symmetrically disposed between the two atoms. The centers of gravity of the two electrons and nuclei therefore coincide.

    Two dissimilar atoms two electrons are not symmetrically disposed of because each atom has a different attraction for electrons.

    When chlorine and hydrogen combine o form covalent hydrogen chloride, the electrons forming the covalent bond displaced two-word the chlorine atom without any separation of the nucleus and the bond is a polar bond.

    Application of dipole moment for study polar bond is an important article for all school and college courses. Today we study the different applications of dipole moment for different polar molecules.

    How to find the percent ionic character of a bond?

    Let us consider compound hydrogen chloride having the dipole moment μobs and the bond length l cm. The ionic character can be calculated from the covalent bond formula.

    Chlorine is more electronegative than hydrogen. Chlorine and hydrogen will carry a unit negative charge and uni-positive charged respectively. Thus the polarity of bond arises in the molecule hydrogen chloride.

    ∴ μionic = e × ℓ
    = (4.8 × 10⁻¹⁰) ℓ esu cm.

    But the dipole moment of hydrogen chloride greater than zero. This data can use to calculate the percentage of ionic character of hydrogen chloride.
    How to find the percent ionic character of a bond?
    Ionic character of a bond
    Question
    The dipole moment of hydrogen chloride 1.03 Debye and bond length is 1.27 A.

    1. How to calculate the charge on the constituent atom?
    2. What is the percentage of the ionic character of a bond?
    Answer
    1. Charge on the constituent atom = 0.8 × 10⁻¹⁰ esu.
    2. Percentage of the ionic character of the bond = 16.89%.

    What is the polarization in chemical bonding?

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

    For gas molecules,
    Pi = {(D₀ - 1)/(D₀ + 2)} M/ρ
    D൦ close to unity under this condition.

    {(D0 - 1)/3} × 22400 = ( 4/3 ) π N₀ r³

    At NTP, M/ρ = molar volume = 22400cc/mole.
    For spherical molecule, αi= r³.

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

    Chemical properties of noble gases

    Mono-atomic noble gases are non-polar, and it indicates the symmetrical charge distribution in the molecule.

    Bonding in homonuclear diatomic molecules

    The homonuclear diatomic molecules are contained a covalent bond between the atom and largely non-polar.
    Nitrogen, oxygen, and chlorine are examples of such a molecule with symmetrical charge distributions. The bonding electron pair equally shared by the two bonding atoms.

    Heteronuclear diatomic molecules

    Hydrogen bromide and hydrogen iodide have non zero values of dipole moment. This indicates the unsymmetrical charge distribution between two bonding atoms.

    H⁺ ㄧ I⁻

    Due to the difference in electronegativity of the constituent atoms in heteronuclear diatomic molecules always polar. Electron pair is not equally shared and shifted to the more electronegative atom. Hydrogen chloride, hydrogen bromide, and hydrogen iodide are examples of such types of molecules.

    Hydrogen chloride1.03 Debye
    Hydrogen bromide0.79 Debye
    Hydrogen iodide0.38 Debye
    Hydrogen fluoride2.00 Debye


    Question
    The difference between the electronegativity of carbon and oxygen large but the dipole moments of carbon monoxide are very low - why?

    Answer
    The difference in electronegativity between carbon and oxygen large in carbon monoxide is very large but the dipole moment of carbon monoxide is very low. This suggested that the charge density in the oxygen atom is somehow back-donated to the carbon atom.

    Which explains by forming a coordinate covalent bond directing towards carbon atom.

    Examples of triatomic molecules

    Carbon dioxide, barium chloride, stannous chloride have zero dipole moment indicating that the molecules have a symmetrical charge distribution between the bond.

    Carbon dioxide molecule electric moment of one carbon-oxygen bond or bond polarity cancels the electric moment of the other carbon-oxygen bond.

    Online college chemistry courses
    The electric moment associated with the bond arising from the difference of electronegativity called the bond moment. In molecules, the vectorial addition of the bond moments gives the resultant dipole moment of the molecule.

    ∴ μ2 = m12 + m22 + 2m1m2Cosθ
    where m₁ and m₂ are the bond moments.

    Bond moments help to calculate the bond angle of the molecule. Carbon dioxide molecule has μ = 0 and m₁ = m₂.

    ∴ 0 = 2m²(1 + cosθ)
    or, θ = 180°

    The polarity of hydrogen sulfide, water, sulfur dioxide

    Water, hydrogen sulfide, and sulfur dioxide have μ ≠ 0 indicating that they have non-linear structures. The bond angle can be calculated from the polarity of the molecules.

    Why is the polarity of water important?

    Why is the polarity of water important?
    Polarity of water
    The dipole moment of water
    μ = 1.84 D and bond moment = 1.60 D.

    ∴ 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 included within the bond moment.

    Question
    The bond angle in hydrogen sulfide = 97° and dipole moment = 0.95 D. What is the polarity of bond in hydrogen sulfide? (Cos97° = - 0.122).

    Answer
    The dipole moment of hydrogen sulfide = 0.95 D
    and bond angle = 97°.
    m = mS-H

    μ² = 2 m² (1 + cosθ )
    ∴ (0.95)² = 2 m² (1 + cos97°)
    or, 0.9025 = 2 m² (1 - 0.122)
    or, m² = 0.9025/(2 × 0.878) = 0.5139
    ∴ m = 0.72D

    Bond moment of the S - H link = 0.72 D

    Boron trifluoride polarity

    Boron trichloride, boron trifluoride are the tetratomic molecules having 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 boron atom at the intersection of the molecules.
    Polarity of boron trichloride
    Polarity of boron trichloride
    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 hydrogen atoms are on the plane and nitrogen atom at the apex of the pyramid in ammonia.

    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.

    Examples of Penta atomic molecules

    Methane, carbon tetrachloride, platinum chloride are examples having zero dipole moment. This suggests that they are either regular tetrahedral or square planer. But polar molecules of this type have pyramidal structures.

    Methane chemical formula and polarity

    Let us discuss the structure of methane that has regular tetrahedral structure and the angle of each H-C-H = 109°28ˊ.

    The electric moment associated with a group 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 methyl group identical to the bond moment of a carbon-hydrogen bond. Two moments cancel each other resulting in zero dipole moment of the molecule in methane.
    ∴ mCH₃ = 3 mCH Cos(180° -109°28՛)
    = 3 mCH Cos 70°32՛
    = 3 mCH × (1/3)
    = mCH

    Dipole moment of methane
    = mCH (1 + 3 Cos 109°28՛)
    = mCH {1 - (3 ×(1/3)}
    = 0

    Compare dipole moment of CH₃Cl and CHCl₃


    Compare dipole moment of CH₃Cl and CHCl₃
    CCl₄, CHCl₃, and CH₃Cl molecules

    Polarity of methyl chloride

    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

    Polarity of chloroform

    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 carbon-hydrogen.

    The identical value of the dipole moment of the homologous alcohols and saturated hydrocarbons confirms the above statement.

    Radiation measurement by Curie and Becquerel

    The rate at which a radioactive sample disintegrates can be determined by counting the number of particles emitted in a given time. 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.

    3.7 × 10¹⁰ curie = 3.7 × 10⁷ millicurie
    = 3.7 × 10⁷ microcurie.

    Millicurie and microcurie respectively correspond to 3.7 × 10⁷ and 3.7 × 10⁴ disintegration per second.

    On this basis, the radioactivity of radium 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.

    How to convert becquerel to curie?

    SI unit of radioactivity is Becquerel or simply Bq. Becquerel has expressed one disintegration per second.
    3.7 × 10¹⁰ disintegrates per second = 1 curie.
    1 disintegration per second = 1 Bq.
    ∴ 3.7 × 10¹⁰ Bq = 1 curie.

    Rutherford or simply Rd is the practical unit of radioactivity.

    Problem
    What is the radioactivity of this element in curie if the radioactive isotopes have x number of disintegration per second?

    Answer
    3.7 × 10¹⁰ disintegrations per second = 1 curie.
    Thus x number of disintegration per second = x/(3.7 × 10¹⁰) curie.

    Cause of radioactive disintegration

    Radioactivity is a nuclear phenomenon it must be connected with the instability of the nucleus. The nucleus of an atom composed of two fundamental particles namely 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.

    Stability of the nucleus of radioactive isotopes

    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.

    1. Nuclei with an even number of protons and even number of neutrons are the most abundant and most stable isotopes of elements. 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. 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 neutrons ratio in stable isotopes

    The stability of the nucleus of an atom also influenced by the relative numbers of protons and neutrons along with the odd or even number of protons and neutrons. The neutron/proton or n/p ratio helps to predict the stability of the nucleus of radioactive elements.
    Graphical presentation of neutron proton ratio in stable isotopes
    Neutron proton ratio in stable isotopes
    The above graph is obtained by plotting the number of neutrons in the nucleus of the stable isotopes against the respective number of protons.
    1. 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.
    2. After atomic number 20, the line rises rather steeply. 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 serve as binding martial inside the nucleus and the way an unstable nucleus disintegrates will be resided by its position with respect to the actual n/p plot of the nucleus.
    3. The isotope located above this actual n/p plot with too high an n/p ratio, 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.
    Problem
    Why gold-197 is non-radioelement but radium-226 radioactive?

    Answer
    The number of neutron and protons 188 and 79 respectively in an isotope of gold-197.
    n/p ratio = 118/79 =1.49.
    n/p values for gold-197 less than 1.5. Thus isotope of a gold-197 stable.

    The number of neutron and protons 138 and 88 respectively in isotopes radium-226.
    n/p ratio = 226/88 = 1.57.
    n/p values for radium-226 less than 1.5. Thus isotope of a radium-226 stable.

    Isotopes with too many neutrons in the nucleus

    Isotopes with too many neutrons in the nucleus can attain greater nuclear stability if one of the decays of the neutron to the proton. Such disintegration leads to the emission of an electron from inside the nucleus.

    ₀n¹ → ₁H¹ + ₋₁e⁰ (electron)
    Thus n/p ratio higher than the expected n/p value for stability and beta ray emission will occur.
    n/p ratio is high when the mass number of radioactive isotopes greater than the average atomic weight of the element.

    n/p ratio for stable isotopes of carbon - 12 is 1.0 with six protons and six neutrons. But for carbon - 14 is 1.3 with eight neutrons and six protons. It can be predicted caron - 14 will be radioactive and will emit beta rays.

    ₆C¹⁴ → ₇N¹⁴ + ₋₁e⁰
    Similarly, the n/p ratio for the stable isotope of iodine-127 is 1.4 with seventy-four neutrons and fifty-three protons. Iodine-133 the n/p ratio equal to 1.5 with eighty neutrons and fifty-three protons and it is a beta emitter.

    ₅₃I¹³³ → ₅₄Xe¹³³ + ₋₁e⁰

    Isotopes with deficient in neurons in the nucleus

    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)

    Positron emission occurs with light radioactive isotopes of the elements of a low atomic number. Nitron-13 with n/p ratio of 0.86 disintegrate by positron emission.

    ₇N¹³ → ₆C¹³ + ₊₁e⁰

    Similarly, iodine-121 has n/p ratio 1.3 which is below the n/p ratio of 1.4, a stable isotope of iodine. Thus iodine expectedly decays by positron emission.

    ₅₃I¹²¹ → ₅₂Te¹²¹ + ₊₁e⁰

    Problem
    Which of the following elements are beta emitter and why?
    1. Isotopes of carbon-12 and carbon-14.
    2. Isotopes of iodine-127 and iodine-133.
    Answer
    The n/p ratio for stable carbon-12 is 1.0(6n + 6p) but that for carbon-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)

    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

    Electron capture reaction by light isotopes

    Orbital electron capture reaction occurs with too light isotopes of radioactive elements of relatively high atomic numbers. The nucleus captures an electron from the nearest orbital. Orbital electron capture changes one proton to a neutron.

    ₃₇Rb⁸² + ₋₁e⁰ → ₃₆Kr⁸²
    ₇₉Au₂¹⁹⁴ + ₋₁e⁰ → ₇₈Pt¹⁹⁴

    Heaviest nucleus, the total proton-proton repulsion large. The binding effect of the neutron is not enough to lead a stable nonradioactive isotope. For such a nucleus, alpha particle emission common mode of disintegration.

    The species thorium-232, thorium-234, uranium-235, and uranium-238 disintegrate to produce more stable isotopes of lead-82.
    ₉₀Th²³² → ₈₈Ra²²⁸ + ₂He⁴
    ₉₂U²³⁵ → ₉₀Th²³¹ + ₂He⁴
    ₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴

    Balancing oxidation and reduction equation

    Oxidation and reduction are always found to go hand in hand during a redox reaction. When an element or compound is oxidized, another element or another compound must be reduced.

    An oxidant is reduced and simultaneously the reductant is oxidized. Here we study balancing chemical equations on the basis of oxidation and reduction.
    1. Balancing chemical equations by electron
    2. Balancing chemical equations by oxidation number

    Balancing chemical equations by electron

    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 solution use a requisite number of hydrogen ion for balancing the number of atoms involved in the partial equation and for alkaline solution use hydroxyl ions.
    4. Balancing chemical partial equations by adding a suitable number of electrons. These electrons indicate the electrons involves in the oxidation and reduction half-reactions.
    5. Multiply each partial equation by a suitable number 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 chemical equations.

    Potassium permanganate in dilute sulfuric acid

    Potassium permanganate in dilute sulfuric acid oxidizes iron ferrous state to the ferric state. In this chemical equation permanganate ions are the oxidant and ferrous ion the reductant.

    The left-hand side of the ultimate equation will carry, permanganate ion (potassium permanganate), hydrogen ion (sulfuric acid) and ferrous ion (ferrous sulfate).

    The Right-hand side will have as products, manganese(II) ion (manganese sulfate), water and ferric ion (ferric sulfate). The partial equation representing the reduction of the oxidant

    MnO₄⁻ → Mn²⁺

    Since the reaction occurs in an acid solution we utilize eight hydrogens to balance the four oxygen atoms in permanganate ion.

    MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

    The above partial equation still unbalanced from the viewpoint of charge, the equation balanced by bringing in five electrons.

    MnO₄⁻ + 8H⁺ + 5e ⇆ Mn²⁺ + 4H₂O

    The partial equation representing the oxidation of the reductant.

    Fe⁺² ⇆ Fe⁺³ + e

    This equation multiplies by 5 and then adding to the partial equation representing the reduction of the oxidant.

    MnO₄⁻ + 8H⁺ + 5e ⇆ Mn²⁺ + 4H₂O
    5×( Fe⁺² ⇆ Fe⁺³ + e)

    MnO₄⁻ + 8H⁺ + 5Fe⁺² → Mn²⁺ + 5Fe⁺³ + 4H₂O

    Potassium iodide in dilute sulfuric acid

    Oxidation of potassium iodide by potassium dichromate in dilute sulfuric acid. In this reaction, chromium is reduced +6 state to +3 state and iodide ion oxidized to elementary iodine.

    Taking the dichromate side of the partial equation of reduction of oxidant and balancing the atoms and electrons.

    Cr₂O₇⁻² + 14H⁺ + 6e ⇆ 2Cr⁺³ + 7H₂O

    The partial equation representing the oxidation of the reductant.

    2I⁻ ⇆ I₂ + 2e

    This equation multiplies by 3 and adding with partial equations of the dichromate side.

    Cr₂O₇⁻² + 14H⁺ + 6e ⇆ 2Cr⁺³ + 7H₂O
    3 (2I⁻ ⇆ I₂ + 2e)

    Cr₂O₇⁻² + 14H⁺ + 6I⁻→ 2Cr⁺³ + 3I₂ + 7H₂O

    Permanganate ion in alkaline solution

    Permanganate ion in alkaline solution oxidizes sodium stannite (Na₂SnO₂) to sodium stannate (Na₂SnO₃) with the formation of manganese dioxide.

    The partial equation representing the reduction of oxidant in alkaline solution and balancing the chemical equations with a requisite number of hydroxyl ion.

    MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻

    The above partial equation still unbalanced from the viewpoint of charge, the equation balanced by bringing in three electrons.

    MnO₄⁻ + 2H₂O + 3e ⇆ MnO₂ + 4OH⁻

    The partial equation representing the oxidation of the reductant.

    SnO₂⁻² + OH⁻ ⇆ SnO₃⁻² + H₂O + 2e

    Balancing this chemical equation first partial equation multiplies 2 and the second partial equation multiply 3 and adding to given final balanced chemical equation.

    2MnO₄⁻ + 4H₂O + 6e ⇆ 2MnO₂ + 8OH⁻³
    SnO₂⁻² + 3OH⁻ ⇆ 3SnO₃⁻² + 3H₂O + 6e

    2 MnO₄⁻ + 3 SnO₂⁻² + H₂O → 2 MnO₂ + 3 SnO₃⁻² + 2 OH⁻

    Oxidation of manganese by lead dioxide in acid solution

    Lead dioxide oxidizes manganese (II) to permanganate ion in sulfuric acid solution with the formation of lead (II) ion. Balancing chemical equation for this reaction.
    Balancing chemical equations of oxidation reduction by electron
    Balancing chemical equation by electron
    Online college chemistry courses
    Problem
    Balancing chemical equation of permanganate to manganous by hydrogen peroxide in acid solution.

    Solution
    The partial equation for the reduction of oxidant

    MnO₄⁻ + 8H⁺ + 5e ⇆ Mn⁺² + 4H₂O

    Hydrogen peroxide will give oxygen to an acid solution.

    H₂O₂ ⇆ 2H⁺ + O₂ +2e

    The first equation is multiplying by 2 and the second is 5 for balancing the chemical equation.

    2MnO₄⁻ + 16H⁺ + 10e ⇆ 2Mn⁺² + 8H₂O
    5H₂O₂ ⇆ 10H⁺ + 5O₂ +10e

    2MnO₄⁻+5H₂O₂+6H⁺⇆2Mn⁺²+8H₂O+5O₂

    Oxidation of manganese by NaBiO₃ in acid solution

    NaBiO₃ oxidizes manganese (II) to permanganate ion in sulfuric acid solution with the formation of BiO⁺ ion. Balancing chemical equation for this reaction.
    Balancing chemical equation by electron
    Balancing chemical equation
    Problem
    Balancing chemical equation of reduction of nitrate ion to ammonia by aluminum in aqueous sodium hydroxide.

    Answer
    The partial equation for the reduction of oxidant in the sodium hydroxide solution.

    NO₃⁻ + 6H₂O + 8e ⇆ NH₃ + 9OH⁻

    Metallic aluminum will go over to aluminate ion in alkaline solution

    Al + 4OH⁻ ⇆ AlO₂⁻ + 2H₂O + 3e

    Multiplying by right factors for electron balancing the chemical equation.

    3NO₃⁻ + 18H₂O + 24e ⇆ 3NH₃ + 27OH⁻
    8Al + 32OH⁻ ⇆ 8AlO₂⁻ + 16H₂O + 24e

    3NO₃⁻ + 8Al + 2H₂O + 5OH⁻ → 3NH₃ + 8AlO₂⁻

    Balancing chemical equations by oxidation number

    The oxidation number of oxygen remains unchanged manganese in permanganate has an oxidation number +7. Decreases and increases in oxidation numbers provide an idea about oxidation-reduction reactions.
    Putting the right factors the decreases and increases in oxidation numbers and balancing chemical reactions.

    MnO₄⁻ + 5Fe⁺² → Mn⁺² + 5Fe⁺³ 

    Acidic solution the changes are not equal on the two sides of the above expression, hydrogen ions are added and the requisite number of water written.

    MnO₄⁻ + 8H⁺ + 5Fe⁺² → Mn⁺² + 5Fe⁺³ + 4H₂O

    Iodide ion in dilute sulfuric acid

    Cr₂O₇⁻² + I⁻ → 2Cr⁺³ + I₂

    The oxidation number of chromium decreases 2 and the oxidation number of Sn increases 1. Equalizing the increase and decrease in oxidation number.

    Cr₂O₇⁻² + 6I⁻ → 2Cr⁺³ + 3I₂
    Ionic charges are not equal on the two sides in an acid solution of the above expression. Fourteen hydrogen ions are added to the left-hand side and the requisite number of water added to the right-hand side.

    Cr₂O₇⁻² + 6I⁻ + 8H⁺ → 2Cr⁺³ + 3I₂ + 7H₂O

    Sodium stannite to stannate in alkaline solution

    MnO₄⁻ + SnO₂⁻² → MnO₂ + SnO₃⁻²

    The oxidation number of manganese decreases 3 and the oxidation number of Sn increases 2. Equalizing the increase and decrease in oxidation number.

    2MnO₄⁻ + 3SnO₂⁻² → 2MnO₂ + 3SnO₃⁻²

    Ionic charges are not equal on the two sides in the alkaline solution of the above expression. Two hydroxyl 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, water is added to balancing the chemical equations.

    2MnO₄⁻ + 3SnO₂⁻² + H₂O → 2MnO₂ + 3SnO₃⁻² + 2OH⁻

    Iodide ion and iodate ion in acid solution

    I⁻ + IO₃⁻ → I₂

    Change of oxidation number of iodine in the above chemical reaction.

    I⁻(-1) + IO₃⁻(+5) → I₂(0)

    Since I⁻ (-1) oxidation number of iodide increases 1 and IO₃⁻(+5) oxidation number decreases 5. Putting the right factors the decrease and increase in oxidation number for balancing equation.

    5I⁻ + IO₃⁻ → 3I₂

    Charges are unequal on two sides of the above expression in acid medium six hydrogen ion added on the left-hand side and write three water on the right-hand side.

    I⁻ + IO₃⁻ + 6H⁺ → I₂ + 3H₂O

    Sulfurous acid and dichromate in acid solution

    SO₃⁻² + Cr₂O₇⁻² → SO₄⁻² + 2Cr⁺³

    Oxidation of two chromium decreases 2 × (+3) = 6 and the oxidation number of sulfur increases 2. equalizing the above chemical equation.

    3SO₃⁻² + Cr₂O₇⁻² → 3SO₄⁻² + 2Cr⁺³

    Ionic charges are not equal on the two sides in acid medium, eight hydrogen 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 water added to balance the chemical equation.

    3SO₃⁻² + Cr₂O₇⁻² + 8H⁺ → 3SO₄⁻² + 2Cr⁺³ + 4H₂O
    Balancing chemical equation by oxidation number
    Balancing chemical equation by oxidation number

    Discovery of radioactivity by Becquerel

    The history of nucleus of atom induces 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 obvious 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 of radioactivity.
    1. Properties of these radiations were very similar to those of x -rays.
    2. These rays highly penetrating, they affected photographic plates, ionization of gases and would also induce the fluorescence in some substances.
    3. These rays are not influenced by heat, light or chemical composition.

    Radioactive Radium, Polonium, and actinium

    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. 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.

    Alpha-beta-gamma radiation

    The radiations emitted by naturally radioactive elements were shown to split by an electric or magnetic field into three distinct parts are known alpha (α), beta (β) and gamma (ɣ) rays.
    Radioactive artificial transmutations reaction
    Radioactive artificial transmutations

    Radiation of alpha particle or alpha rays

    Alpha particle consists of a stream of positively charged particle which carries +2 charge and has mass number 4.

    These particles are identical to the nuclei of the helium atom shown by Rutherford. Thus alpha particle is doubly charged helium ion (He⁺²) with atomic number 2 and mass number 4.

    When an Alpha particle ejected from within the nucleus of an atom, the mother element loss two units of atomic number and four units of mass number.

    ₉₂U²³⁸ → ₉₂U²³⁴ + ₂He⁴

    Radiation of beta particle or beta rays

    Beta particle made up of a stream of negatively charged particles. The beta particle is identical with electrons from a study of their behavior in electric and magnetic fields and from the study of their e/m values.

    e/m = 1.77 × 10⁸ coulombs/gm.

    The ejection of a beta particle with mass number 0 and charge 1, results in the transformation of a neutron into a proton.

    ₀n¹ → ₁H¹ + ˗₁e⁰

    When a beta particle 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 or gamma particle

    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 emission from the nucleus

    Works of the Curies and Rutherford yet another mode of nuclear transformation has been discovered. This involves the ejection of a positron 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 emission from the nucleus

    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 ± ½ (h/2π) each.
    ₀n¹ → ₁H¹ + ₋₁e⁰

    Angular momentum not balanced on the radioactive reaction. If the angular momentum of the proton and the electron are +½ (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 neutrino also ejected.

    Neutrino has a spin angular momentum ± ½ (h/2π). Sum of angular momentum of the particles ejected
    +½ (h/2π) + {- ½ (h/2π)} +1/2 (h/2π) = +1/2 (h/2π)
    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

    Neutron → proton + electron + neutrino.

    Charge and mass of alpha, beta, and gamma particles

    Charge and mass of alpha-beta-gamma radiation
    Charge and mass of alpha-beta-gamma

    Nuclear and chemical reaction


    Nuclear reactions are different from chemical reactions in many respects:

    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 the emission of alpha, 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.

    The nuclear reactions are accompanied by energy changes which far exceed the energy changes in chemical reactions.

    The energy evolved in the radioactive transformation of one gram of radium five hundred thousand times as large as the energy released when one gram of radium combine with chlorine to form RaCl₂.

    Bohr's model of the hydrogen atom

    Explanation of the hydrogen spectrum, Neils Bohr's in 1913 adopted the Rutherford model of the hydrogen atom in which an electron revolves around the single proton at the central nucleus.

    According to classical mechanics, when a charged particle is subjected to the acceleration it emits radiation and loses energy and the orbital radius must also be changed.
    An electron revolving around the nucleus would, therefore, be continually accelerated towards the center of the orbit and consequently emitting radiation.

    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 energy of the electron loses continuously, the observed atomic spectra should be continuous, consisting of broad bands merging one into the other. The observed emission spectrum consists of well-defined lines of definite frequencies. To resolve the anomalous position Niels Bohr's proposed a new atomic model of the hydrogen atom.

    Bohr's model of hydrogen atom postulates

    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 of an atom.

    The energy levels are numbered as 1, 2, 3, ........ starting from the nucleus of an atom 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.
    E₁, E₂, E₃ ...... denotes the number of energy levels of 1 (K - Shell), 2 (L - Shell), 3 (M - Shell) ......., these are in the order,
    E1ㄑE2ㄑE3ㄑ......

    An electron can jump from one energy level to another higher energy level on the absorption of energy and one energy level to another lower energy orbit with the emission of energy.
    Hydrogen energy levels in Bohr's model
    Hydrogen energy levels
    The angular momentum of an electron moving in an orbit or energy level is an integral multiple of h/2π. This integral multiple is known as the principal quantum number of an atom.

    ∴ mvr = n × (h/2π)

    where m = mass of an electron, v = tangential velocity of an electron in an energy level, r = distance between the electron and nucleus of an atom and n = whole number which has been given the principal quantum number of an atom.

    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 radiation emitted or absorbed by an electron and h is the Plank constant.
    Online college chemistry courses

    Ground state energy of hydrogen atom formula

    The nucleus has a mass m' and the electron has mass m. The radius of the circular orbit = r and the linear velocity of the electron = v. Evidently, on the revolving electron, two types of forces are acting, centrifugal force and electric force of attraction.
    The energy of an electron in Bohr's model
    The energy of an electron

    Centrifugal and electric force of an atom

    Energy levels stable when the centrifugal force exerted by the moving electron must equal to the attractive force between the electron and nucleus of the hydrogen atom.

    Centrifugal force = mv²/r.

    Two attractive forces are in the operation, one being the electric force of attraction between the nucleus and the electron, the other being the gravitational force. Gravitational force comparatively weak and can be neglected.
    The electric force of attraction between two opposite charges is given by Coulomb's law.

    Electric force = e × (e/r²) = e²/r².

    Centrifugal force and electric force has acted in the opposite direction. The electron may keep on revolving in its orbit, these two forces, which acts in opposite direction must balance each other.

    ∴ mv²/r = e²/r²
    or, mv² = e²/r.

    How do you calculate the atomic radius of hydrogen?

    Bohr's remarkable suggestion for the angular momentum of the electron is an integral multiple of h/2π. Angular momentum of an electron = mvr.

    ∴ mvr = nh/2π

    Where n = integer called quantum number indicating hydrogen energy levels having the values 1,2,3, ......∞.
    v = n × (h/2π) × (1/mr)

    e²/r = mv²
    Putting the value of v on this equation,
    e²/r = m × n² × (h/2π)² × (1/mr)².

    or, e²/r = n²h²/4π²mr²

    ∴ r = n²h²/4π²me²

    The radius of orbit of the hydrogen atom

    Solution for the radius of the permitted energy levels of the hydrogen atom in terms of the quantum number. When n = 1, the radius of the first stationary orbit of hydrogen.

    ∴ 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.

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

    ∴ rn = n² × r₁ = r₁ × a₀

    Question
    Calculate the radius of the second orbit of a hydrogen atom if the radius of the first orbit of hydrogen = 0.529 Å.

    Answer
    The radius of the second orbit of a hydrogen atom
    r₂ = n² × r₁ = 2.12 Å

    The velocity of an electron of the hydrogen atom

    mvr = nh/2π
    where v = velocity of an electron.
    or, v = (nh/2πm) × (1/r)

    Putting the values of r = n²h²/4π²me².

    v = (nh/2πm) × (4π²me²/n²h²)

    ∴ v = 2πe²/nh

    Thus the velocity of the second orbit will be one half of the first orbit and one-third of the first orbit and so on.
    v₂ = v₁/2
    v₃ = v₁/3.
    ∴ vn = v₁/n = velocity of first orbit/principal quantum shell.

    Question
    Calculate the velocity of the hydrogen electron in the first and third energy levels of an atom. How to calculate the number of rotation of an electron per second in the third energy level?

    Answer
    v = 2πe²/nh
    where n = 1, 2, 3, ........
    ∴ The velocity of an electron in the first energy level
    v₁ = 2πe²/1² × h
    = 2πe²/h
    = {2×(3.14)×(4.8×10⁻¹⁰)²}/(6.626×10⁻²⁷)
    = 2.188 × 10⁸ cm sec⁻¹.

    The velocity of the third energy level
    v₃ = v₁/3.
    ∴ v₃ = (2.188 × 10⁸ cm sec⁻¹)/3
    = 7.30 × 10⁷ cm sec⁻¹.

    ∴ The radius of the third energy level
    = 3² × 0.529 × 10⁻⁸ cm.

    Circumference of the third energy level
    = 2πr = 2 × 3.14 × 0.529 × 10⁻⁸ cm

    ∴ Rotation of an electron per second in the third energy level of the hydrogen atom
    = (7.30 × 10⁷)/(2 × 3.14 × 9 × 0.529 × 10⁻⁸ )
    = 2.44 × 10¹⁴ sec⁻¹.

    The kinetic and potential energy of an electron

    The energy of an electron moving in one particular energy level can be calculated by the total energy or the sum of the kinetic and the potential energy of an electron.

    ∴ The kinetic energy of a moving electron
    = ½ mv².

    Potential energy is the energy due to electric attraction and is given by:

    V = ∫(e²/r²)dr = - (e²/r)

    ∴ Total energy = E = ½ mv² - e²/r.

    Putting the value e²/r = mv².

    ∴ E = ½ mv² - mv²
    = - ½ mv²
    = - ½ e²/r

    Energy of an electron in hydrogen energy levels

    Energy of an electron, E = - ½ e²/r
    where r = n²h²/4π²me².

    ∴ E = - 2π² me⁴/n²h².
    When n = 1, that is the energy of the first orbit of the hydrogen atom

    E₁ = - 2π²me⁴/h²
    ∴ En = E₁/n².

    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.

    If energies associated with 1st, 2nd, 3rd,...., nth orbits are E1, E2, E3 ... En, these will be in the order,

    E₁ㄑE₂ㄑE₃ㄑ........ㄑEn.

    The energy of the moving electron in the first energy level obtained by putting n=1 in the energy expression of the hydrogen atom.

    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

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

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

    The kinetic energy of moving electron of an atom

    The kinetic energy of a moving electron
    T = ½ mv²
    = ½ m × (2πZe²/nh)²
    = (2π²mZ²e⁴)/n²h².

    where the charge of an electron (e) = 4.8 × 10⁻¹⁰ esu
    Plank's constant (h) = 6.626 × 10⁻²⁷ erg sec
    mass of an electron (m) = 9.1 × 10⁻²⁸ gm.

    ∴ The kinetic energy of hydrogen atom in first energy level
    = 13.6 eV
    Question
    The energy of an electron in the first energy level of the hydrogen atom = -13.6 eV. What is the energy value of the electron in the excited level of lithium-ion?

    Answer

    The energy of an electron in the excited level of lithium-ion
    = - 30.6 eV

    Chemistry 1

    Contact Form

    Name

    Email *

    Message *