Chemistry by Inam Jazbi: Derivation of Radius, energy, energy difference, frequency, wave no, and wavelength of the nth Bohr’s Orbit for Hydrogen-Like Atoms

Derivation of Radius of the n^th Bohr’s Orbit for Hydrogen-Like Atoms

Assumptions for simple atom

To derive an expression for radius, consider a hydrogen atom (or hydrogen-like atoms such as He⁺, Li²⁺, Be³⁺, B⁴⁺, C⁵⁺) with atomic number equal to z consisting of a single electron with charge –e and mass m revolving around the nucleus of charge +Ze (+e is charge of proton) with a tangential velocity v in the orbit whose radius is r.

Now revolving electron is being acted upon simultaneously by the following two types of forces;

(i) Electrostatic force of Attraction / Centripetal Force

According to Coulomb’s law, the electrostatic force of attraction (F_e) between the nucleus of charge ‘+Ze’ and electron of charge ‘–e’ separated by a distance ‘r’ is given by:

Where ‘K’ is proportionality constant. It is equal to 1/4πε_or²

Hence attractive force between nucleus and electron can be written as

(ii) Centrifugal Force

This Coulombic force of (F_c) supplies the centrifugal force to keep the electron in an orbit and is given by:

Equating F_e and F_c

To keep the electron in the same orbit, these two opposite forces must be equal to each other i.e.

Determination of v² of electron by using Bohr’s Postulate

According to Bohr’s postulate, angular momentum of electron revolving around the nucleus is an integral multiple of h/2p.

Calculation of Radius (r)

Substituting the value v² from equation (ii) in equation (i)

Where

h = Planck’s constant = 6.625 x 10⁻³⁴ J.s

m_e = Mass of electron = 9.11 x 10⁻³¹ kg

e = Charge of electron = 1.602 x 10⁻¹⁹ C

𝛆_o= Vacuum permittivity constant = 8.84 x 10⁻¹² C²/J.m

Calculation of n^th Bohr’s orbit (r_n)

Assembling all constants in equation (iii), we get

Where a is known as Bohr’s constant or Bohr radius and its value is 0.529 x 10⁻⁸ cm or 0.529 Å or 0.0529nm or 52.9 pm. This is the radius of the first orbit of H. This equation is used for the determination of n^th orbit of hydrogen atom and hydrogen like ions like He⁺, Li²⁺ etc.

The above equation shows that radius of orbit is directly proportional to the square of the principal quantum numbers (r α n²i.e. 1, 2, 3, ……..) and inversely proportional to atomic number. As the value of n increases, the radius of the orbit will increase.

Derivation of Energy of the n^th Bohr’s Orbit

Basic of Derivation

The total energy of an electron revolving in any orbit around the nucleus is the sum of kinetic energy and potential energy given by,

E_total = K.E + PE ……………….(i)

Calculation of K.E

The K.E. of electron with mass m revolving around the nucleus with velocity v is given by the following expression;

Now the centrifugal and centripetal forces upon the revolving electron are given as:

At uniform circular equilibrium motion, these two opposite forces must be equal to each other i.e.

Calculation of P.E.

P.E is the work done in bringing the electron from infinity to a point at a distance r from nucleus and can be calculate as

P.E =work done = −force x displacement = −Fe x r

Here negative sign indicates that P.E decreases when electron is brought form infinity to a point at a distance r. Here negative sign indicates a net attractive interaction, giving algebraically lower energy at shorter distance.

Calculation of Total Energy

E_total = K.E + PE

Here K is a factor assembled by various constant present in energy equation. Its value is 2.18 x 10⁻¹⁸ J/atom or 1312.8 kJ/mol

E is always negative. Negative sign shows that the electron is bound to the atom and energy must be spent in order to remove it from the orbit.

All energy states are bound states as the negative sign indicates. When n = 1; this corresponds to electron at the closest possible distance from the nucleus and at its lowest energy and is called ground state energy. All energy states with value of n higher than 1 are termed as excited states. When n = α then E = 0; which means that the system is unbound and the electron is free.

It should be noted that the energy is increasing as the n (orbits) increasing; however the difference of energy between two orbits is decreasing.

Conclusion

If total energy = − x

Then

KE = + x

PE = − 2x

2.12 Expression for ∆E of electronic transition between orbits

Calculation of ∆E

Let E₁ be the energy of n₁ orbit and E₂ is for n₂ orbit. To calculate the energy emitted by atom in the form of radiation when an electron jumps from a higher energy state n₂ to lower energy orbit n₁; let us make use of the postulate of Bohr’s model; according to which, the emitted energy is written as

Emitted energy = ∆E = E₂ – E₁