Notes on Atomic Structure For BSc, GE and Entrance Examinations

Notes on Atomic Structure

Bohr's Atomic Model

    In 1913, Niels Bohr brought an atomic model that could manage an excellent balance between classical and quantum mechanical interpretation. A model that could compensate and explain the shortcomings and imperfections in the previous atomic models.

    The important postulates of the Bohr's atomic model are:

1. The electrons in an atom revolve round the nucleus only in certain permitted circular path called the shells or energy levels or stationary states.

2. Each energy level is  denoted with a principal quantum number 'n'. Where n = 1, 2, 3, .....

3. The shell or energy level closest to the nucleus is denoted with n = 1 or the K shell, the next one with n = 2 or L shell and so on .....

4. These energy levels are associated with definite amount of energies, given by the formula:

    E_n = -KZ² / n² where K = 13.6 ev, Z = atomic number and n = number of orbit from the nucleus.

5. The distance r of a shell from the nucleus is given by the formula,

r_n = a₀ n² / Z , where a₀ = Bohr’s first orbit radius = 0.529 A⁰

6. So long as the electron revolves in a particular orbit, it neither absorbs nor emit energy. This is the reason why it does not fall into the nucleus. In this way Bohr compensated the shortcoming of the Rutherford's Atomic Model

7. An electron gets excited to the higher energy levels when ever it absorbs energy. This energy is equal to the difference between the energies of two concerned energy levels.

ΔE = Ef - Ei , where Ef = energy of final orbit and Ei = energy of initial orbit

Also when an electron jumps from a higher orbit to the lower orbit, then difference in energy (Ef - Ei) is emitted.

This is the key explanation of the hydrogen spectrum.

8. Only those orbit are allowed for the electrons to revolve round the nucleus, for which the angular momentum (mvr) of the electron is a whole number multiple of h/2π. Mathematically, mvr = nh/2π. This principle is called quantisation of angular momentum.

9. The above postulates can be developed to reach at the conclusion that the circumference of an orbit (in which an electron revolves) is equal to the whole number multiple of the wavelength of the electron. mathematically, 2πr = nλ

10. The most important of all the postulates is the force balance equation. The centrifugal force of a revolving electron acting outward is counterbalanced by the electrostatic force (centripetal force) between electron and nucleus acting inward. i.e., mv² /r = Ze²/4πε₀ r²

This is the reason why electron does not fall into the nucleus and moves in a definite circular path.

Calculation of radius of nth orbit:

Applying force balance equation:

mv² /r = Ze²/4πε₀ r²

=> v² = Ze²/4πε₀ mr .......... eq. 1

since mvr = nh /2π

=> v = nh /2πmr

=> v² = n²h² /4π²m²r² …… eq.2

From eq 1 and 2,

n²h² /4π²m²r² = Ze²/4πε₀ mr

=> r = n²h² ε₀ / πme²z

=> r = (h² ε₀ / πme²) (n²/Z)

=> r = a₀ (n²/Z) where a₀ = Bohr’s first orbit radius for hydrogen atom = h² ε₀ / πme²

Putting the value of constants, we get a₀ = 0.529 A⁰ = 0.0529 nm

For hydrogen atom, Z = 1

If n =1, r₁ = a₀ = 0.529 A⁰

If n =2, r₂ = 4a₀ = 4 X 0.529 A⁰

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Problem 1: Determine the second Bohr's orbit of He+ ion.

Solution: The atomic number of He+ ion = z = 2

Second Bohr's orbit => n = 2
=>r₂ = a₀ n² / Z = 0.529 (2² / 4) = 2 X 0.529 A⁰

Problem 2: Establish a relationship between the nth Bohr's orbit of Hydrogen and He+ ion.
Solution: r = a₀ (n²/Z)
r(H) = a₀ (n²/Z) = a₀ n²
r (He+) = a₀ (n²/Z) = a₀ (n²/2)
r(H)/r(He+) = 2 
=> r(H) = 2 r(He+)

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Velocity of an electron in nth Bohr’s Orbit:
We know mvr = nh / 2π
=> v = nh/2πmr
Putting the value of  r = a₀ n² / Z
V = (h/2 πm a₀) (z/n)
Putting the values of the constants we have,
Vn = velocity of Bohr's nth orbit = 2.18X10⁶ (z/n)m s^-1

Problem: Determine the speed of an electron moving in the second orbit of unipositive helium ion.
Solution: We have Vn = 2.18X10⁶ (z/n)m s^-1
For He+ ion, z = 2, Second Bohr's orbit => n = 2
=> Vn = 2.18X10⁶ m s^-1

Note: The formula of velocity of electron in nth Bohr's orbit indicates that, all hydrogen like species having same z and n will have velocity equal to 2.18X10⁶ m s^-1

Assignment: Determine the Bohr's orbit of Li2+ ion in which the electron moves at a speed equal to the speed of electron in the first Bohr's orbit of  hydrogen.

Revolution per second (RPS) in nth Bohr's Orbit around nucleus:
The number of revolutions per second (RPS) ∝ Velocity of electron (V)
The number of revolutions per second (RPS) ∝ 1/circumference of orbit (2πr)
=> The number of revolutions per second (RPS) = V/2πr
Putting the value of V = nh/2πmr 
RPS = (nh / 4π²mr²)
and r = a₀ n² / Z
=> RPS = (nh/4 π²m) (z²/ a₀ n²)²
=> RPS = (h/4π²m a₀²) (z² /n³)
Putting the values of the constants,
=> RPS = 6.584X10¹⁵ (z² /n³) revolutions /s

Assignment: Calculate the number of revolutions per second made by an electron in the second Bohr's orbit of He+ ion.

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Energy of an electron in nth Bohr's Orbit: 

The sum of the kinetic and potential energy of the electron moving in an orbit gives its total energy. E(Total) = K.E + P.E

K.E  = ½ (mv²)

Since mv² /r = Ze²/4πε₀ r²

½ (mv²) = Ze²/8πε₀ r = K.E

    Potential energy of the electron is the energy possessed by it due to its position w.r.t. the nucleus. Consider an electron present at infinite distance apart from the nucleus, where its energy is considered to be zero. As it comes closer to the nucleus (into the influence of nucleus) it looses energy, hence its energy becomes negative when it is placed in any orbit.

    Its now clear from the above logic that the potential energy of an electron in a given Bohr's orbit is the negative of the work done required to move that electron from the said Bohr's orbit to the infinity Bohr's orbit.

Since work done = Force X Displacement
An infinitesimal work done by an infinitesimal displacement of the electron is given by, dW = F dr

Thus the total work done in displacing the electron to the infinity orbit should be,

Hence total energy (E) = K.E + P.E
=  (Ze²/8πε₀ r) + (- Ze²/4πε₀ r) = - Ze²/8πε₀ r
Putting the value of r = a₀ n² / Z
E (Total) = - Z²e²/8πε₀ n² = (- e²/8πε₀)( Z²/n²)
E (Total) = - Z²e²/8πε₀ n² = (- e²/8πε₀) (Z²/n²)
Putting the values of constants we get,
En =Energy of nth Bohr's orbit = - K (Z²/n²)
Where K = 13.6 ev = 2.18 X 10^-18 J

Problem: Calculate the energy of electron in the Bohr's first orbit of hydrogen atom in kilo Joule per mole.

Solution: For hydrogen atom, Z = 1 and n =1
E1 =  - K (Z²/n²) = - 2.18 X 10^-18 Joule / atom = - 2.18 X 10^-18 J X 6.02 X 10²³ Joule/ mole
= - 2.18 X 10^-18 J X 6.02 X 10²³ x 10^-3 K Joule / mole = - 1312 K Joule / mole

Transition Energy and Ionisation Energy: When ever an electron jumps from any initial orbit (ni) to the final one (nf), energy is absorbed or lost depending on whether it jumps up or down respectively.

Mathematically, the energy absorbed or lost = ΔE = Enf - Eni 
=> ΔE = - K (Z²/nf²) - {- K (Z²/ni²)} = KZ² {(1/ni²) - (1/nf ²)}

When an electron absorbs energy and jumps to such an orbit (or such a distance from the nucleus) so that the nucleus will have no more influence on the concerned electron, then the amount of energy absorbed by the electron is called the ionisation energy. The final orbit in this case, nf = ∞
    Thus ionisation energy, ΔE = KZ² {(1/ni²) - (1/nf²)} = KZ² {(1/ni²) - (1/∞²)} = KZ²/ni²

Problem: Calculate the energy required to excite an electron of hydrogen atom from first Bohr's orbit to the third orbit and also calculate the amount of energy emitted when the same electron jumps down from that 3rd Bohr's orbit to the first one.

Solution:  ΔE = KZ² {(1/ni²) - (1/nf²)}

    For hydrogen atom, Z = 1, when electron jumps up from first to the third orbit, ni = 1 and nf = 3. Thus ΔE = K {(1/1²) - (1/3²)} = 8/9 K = 8/9 X 13.6 eV amount of energy is gained.

    When the electron jumps down from third Bohr's orbit to the first one in hydrogen atom, ni = 3 and nf = 1. Thus ΔE = K {(1/3²) - (1/1²)} = - 8/9 K = -8/9 X 13.6 eV That means the same amount of energy is released.

Problem: Determine the third ionisation energy of lithium atom.

Solution:  After the loss of two electron lithium becomes a hydrogen like species (having only one electron). The amount of energy required to eject the third electron out of the influence of lithium nucleus is called its third ionisation 

energy.
Z = 3, ni = 1,    ΔE = KZ²/ni²
    Thus ionisation energy = 9K = 9 X 13.6 eV = 122.4 eV

Emission Spectrum of Hydrogen atom: Electrons of hydrogen atoms when absorb energy, get excited to the next higher shells. These excited electrons emit energy (in form of light), when they jump down to the lower energy levels. These energy (emitted light) are observed on a fluorescent screen which appear as group of lines. The number of lines observed on the screen indicates the number of possible transitions made by excited electrons.

    If the energy levels are shown as their increasing energy and quantum numbers, we get the following.

    The actual line spectra of hydrogen looks like the following:

    Hydrogen being the simplest atom (containing one electron and one proton) has the simplest emission spectrum. The energy of transition for any line is expressed in terms of wave number as follows:

ΔE = KZ² {(1/ni²) - (1/nf²)}

=> ΔE = hν = hc/λ = KZ² {(1/ni²) - (1/nf²)}

=> 1/λ = ṽ = wave number = (KZ²/hc) {(1/ni²) - (1/nf²)} 

= RH Z² {(1/ni²) - (1/nf²)} ,  

Where RH = K/hc = Rydberg's constant = 1.0973 X 10⁷ m^-1

    As is shown in the above figure,  the group of closely spaced lines (called a series) represent the transition of electrons from any higher energy levels to a fixed lower energy level.  The series of lines placed to the extreme left around 100 nm is called the Lyman Series. This series corresponds to the shortest wavelength which fall in the ultraviolet region. For this series, ni = 1 and nf = 2, 3, 4, .... The first line of the series is represented as Ly-α for which ni = 1 and nf = 2.   

    In Balmer series, ni = 2 and nf = 3, 4, 5, .... This series falls in the visible and near infrared region. 

    In Paschen series, ni = 3 and nf = 4, 5, 6, .... This series falls in the infrared region of the electromagnetic spectrum. 

    In Brackett series, ni = 4 and nf = 5, 6, 7, .... This series falls in the infrared region.

   In Pfund series, ni = 5 and nf = 6, 7, 8,  .... This series falls also in the infrared region.

    The maximum number of lines that can be obtained when an electron jumps down from nth orbit to the ground state = n (n-1)/2. For example if an electron jumps down from 6th orbit (n=6) the number of spectral lines = 6(6-1)/2 = 15

Problem: What will be the frequency and wave number of the second line of Balmer series.

Solution: For Balmer Series, ni = 2 and its second line has nf = 4

=> 1/λ = ṽ = wave number = (KZ²/hc) {(1/ni²) - (1/nf²)} = RH Z² {(1/ni²) - (1/nf²)} , 

Since C = νλ =>  ν = C/λ = C . ṽ = C . RH Z² {(1/ni²) - (1/nf²)} 

and put value for ni and nf

Now put value for C = 3 X 10⁸ m/sec

RH = 1.0973 X 10⁷ m^-1

and value for ni and nf and then get answer.

Correctness or Validity or Advantages of  Bohr's Atomic Model:

1. It explains the spectrum almost clearly including emission and absorption spectrum along with hydrogen spectrum. Electrons in atoms absorb energy and get excited to the higher energy levels and when they return back to the lower energy levels they emit energy.

2. As long as electrons move in a definite orbit they do not gain or loose electrons. This makes an atom stable. Thus Bohr explained the stability of atoms.

3. The theoretical values of the spectral lines as predicted by Bohr is almost equivalent to the experimental values obtained by eminent scientists Rydberg and Ritz.

4. Ionisation energy can be calculated with accuracy from Bohr's transition energy formula.

5. Since there is no force available to counter balance the force of attraction of the nucleus on the electron, a centrifugal force must be there due to the circular motion of the electron. This proves that the electrons are not stationary but they move.

Disadvantages or Drawbacks or Limitations of Bohr's Atomic model:

1. It fails to explain the spectra of multi electron species.

2. By Bohr's prediction the the path of the electrons are circular whereas modern study indicates 3 dimensional motion of the electrons. In addition to this the stationary state concept has been discarded by Heisenberg's uncertainty principle.

3. It is unable to explain the fine lines present inside a spectral lines when seen under powerful microscope and also fails to explain the splitting of spectral lines under external magnetic field (the finer spectral lines as shown by Zeeman).

    

Image Credit:

1. Emission of energy from electron 1:

By JabberWok, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=2639910

2. Emission of energy from electron 1:

By A_hidrogen_szinkepei.jpg: User:Szdoriderivative work: OrangeDog (talk • contribs) - A_hidrogen_szinkepei.jpg, CC BY 2.5, https://commons.wikimedia.org/w/index.php?curid=6273602

3. Energy level diagram as straight lines

By Rajettan - Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=41466201

4. Actual obsevation of line spectrum

By OrangeDog, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=6278485