Unit-I Quantum Mechanics and Atomic structure

Review of Bohr’s atomic model, Derivation of expressions for radius, energy and

ionization energies of hydrogen and hydrogen like species. Numerical Problems.

Limitations of classical mechanics. Wave particle duality, Uncertainty principle-

statement (both in words and mathematical form).New quantum mechanics: Sinusoidal

wave (explain sinusoidal wave) equation (classical wave mechanics); Schrodinger wave

equation- derivation. Postulates of quantum mechanics. Significance of terms: (i)

Hamiltonian operator (ii) Eigen function (significance of ψ and ψ2); (iii)Eigen values.

Application of Schrodinger equation to the (i) particle in one dimensional box

(derivation required) (ii) hydrogen atom (detailed solution not required). Expressing

the solution as a product of ψn, l. m(r, Ө, Φ) =n, l. (r) ψ l. m(Ө, Φ ), Explanation of

quantum numbers (only qualitative): definition and significance. Calculation of l, m and

s for a given values of n (1, 2 and 3). Radial probability distribution and angular

probability distribution. Orbitals-definition and difference between orbitals.

13 hours

Review of Bohr’s Atomic model:

Neil Bohr, a Danish physicist put forward his theory of structure of the atom

using the Planck’s concept and Rutherford’s model. The theory is based on the

following postulates (Bohr’s postulates of atomic model).

1. The electron moves around the nucleus in a circular orbit, the centripetal

force for this motion is provided by the electrostatic attraction.

2. The electron rotates in certain fixed orbits around the nucleus called

stationary state orbits or quantized orbits. Each such stationary states is

associated with specific value of energy. As long as the electron is in these

orbits, it cannot emit any radiation.

3. The angular momentum of the moving electron in the circular orbit is

quantized, it is the integral multiple of h/2π; that is mvr = nh/2π.

4. When an electron transits from higher to lower energy level, it emits

radiation, which is the difference between energies of two states following

Planck’s equation:

∆E = E2 – E1 = hv Where E2 and E1 are the energies of the final and initial

states respectively.

To Derive an expression for the radius of Bohr orbit of hydrogen atom

Bohr assumed hydrogen atom as a system consisting of a single electron with a charge

‘e’. The electron rotates in a circular orbit of radius ‘r’ around the nucleus of charge ’Ze’

where z is the effective charge of the nucleus or atomic number.

For the electron to remain its orbit, the centripetal force (attractive force between

nucleus and the electron) must be equal to the centrifugal force (repulsive force).

Centripetal force= columbic attractive force = Ze2/ 4πϵor2 -------- (1)

Where ϵo is a constant called permittivity of the medium being vacuum in this case.

Centrifugal force = mv2/r -----------(2)

Following the Bohr’s concept; Centripetal force = Centrifugal force;

Equating both the equations: Ze2/ 4πϵor2 = mv2/r;

Rearranging we get v2 = Ze2/ 4πϵo m r ------(3)

Using the concept of quantization of angular momentum; mvr = nh/2π

v= nh/2πmr; v2 = n2h2/4π2m2r2----(4)

Comparing the two expressions for v2; we get

Ze2/ 4πϵo m r = n2h2/4π2m2r2 ; Hence,

r = ϵon2h2/ Ze2πm This is the expression for radius of the orbit in which the electron moves in terms of an

integer ’n’ which defines energy level of the electron and the quantum number.

Calculation of the radius of 1st Bohr orbit for Hydrogen

For hydrogen atom Z=1, and for the first orbit nearest to the nucleus n=1;

Substituting the constant values in the above equation we get

r = ϵon2h2/ Ze2πm ; = 8.85 x10-12C2J-1m-1 x (6.63x10-34J.S.)2 = 5.292 x10-11m

3.14 x 9.109x10-31kg x (1.602 x 10-19)2x1 0.05292nm.

Radius of the first orbit of hydrogen atom = 0.05292nm, this value is in agreement with the

experimental results. Radius of any other orbit is in good agreement with the values obtained

by the experimental of methods. The radius of any orbit of quantum number n for H atom is:

r=n2r1, If n=2, r=22 r1=4r1, If n=3, r=32r1 = 9r1, If n=4, r=42r1 = 16r1, where r1 is the radius

of the first orbit.

To derive an expression for Energy of an electron in Bohr orbit

The total energy E of the electron in a Bohr orbit of a hydrogen atom is the sum of its

potential and kinetic energies. Potential energy of the electron is the columbic forces of

attraction between the

electron and the nucleus multiplied by the distance r from the nucleus.

Potential energy = -Ze2/4πϵor --------(1)

Kinetic energy of the electron = ½ mv2 ---------------(2)

Hence Total energy of the electron E = K.E + P.E

E = ½ mv2 -Ze2/4πϵor -------------(3)

We know from the previous steps that v2 = Ze2/ 4π2ϵom r

Substituting the value of v2 to equation (3) we get

E = ½ m(Ze2/ 4πϵom r) - Ze2/4πϵo r

E = Ze2/4πϵor( ½ - 1) = -Ze2/8πϵor

Substituting the value of r from the previous step equation that r = ϵon2h2/ Ze2πm

E = -Ze2/8πϵo(ϵon2h2/ Ze2πm)

Hence E = -Z2e4πm/8πϵo2n2h2

Or E = - Z2e4m/8ϵo2n2h2, it is the expression of the electron in a hydrogen atom in its

nth orbit. The significance of negative symbol in this equation is that there is a decrease

in energy of the electron when it moves closer to the nucleus.

Transition between Energy Levels

According to Bohr’s concept, when an electron in a higher energy level E2 transits to a lower

energy level E1, the electron loses energy. The loss in energy is emitted as a photon. The

energy of the photon is given by the difference in the energies of the two energy levels.

Energy of the emitted photon = E2-E1. Frequency of the emitted photon is given by

∆E = E2- E1 = hν.

By substituting the values of energy levels in the hydrogen atom, Bohr calculated the

frequency of the spectral line emitted.

∆E = E2-E1 = hν = -Z2e4m/8ϵo2n2h2 =- Z2e4m/8ϵo

2h2 [1/n22 - 1/n1

2] or

hν = Z2e4m/8ϵo2h2 [ 1/n1

2 - 1/n22] Hence ν = Z2e4m/8ϵo

2h3[ 1/n12 - 1/n2

2]

Since Ῡ = ν/c = Z2e4m/8ϵo2c h3 [1/n1

2 - 1/n22] = Ry [1/n1

2 - 1/n22] ------- (1)

Where Ry is called Rydberg constant = Z2e4m/8ϵo2c h3, Since for Hydrogen Z=1;

Ry has the value =1.096x107m-1, the equation (1) is called Rydberg equation.

Hydrogen spectrum and Energy level diagram

Hydrogen atom consists of only one electron in the ground state. On absorbing energy this

electron is raised to higher energy level which is less stable. When this electron transits to lower

energy level energy is being released in the form of photons or different frequencies.

E2-E1 = hν = hc/λ , It is found that greater the energy emitted, then shorter will be the wavelength

of the spectral line observed. Hydrogen spectrum is the simplest to analyze. It consists of discrete

lines in the UV-Visible, and Infrared regions of the electromagnetic spectrum. The lines in the

atomic spectra are grouped into several series known as spectral series. They are the lines

obtained due to the transitions between two energy levels applied to

Rydberg equation: Ῡ = Ry [ 1/n12 - 1/n2

2], the lines are

1. Lyman series obtained when the electron jumps from n2= 2, 3, 4, 5….. To n1= 1

found in the UV region.

2. Balmer series obtained when the electron jumps from n2= 3, 4, 5, 6….. To n1= 2

found in the Visible region.

3. Paschen series obtained when the electron jumps from n2= 4, 5, 6, 7….. To n1=

3 found in the Infra-red region.

4. Brackett series obtained when the electron jumps from n2= 5,6,7,8….. to n1= 4

found in the Infra-red region.

5. Pfund series obtained when the electron jumps from n2= 6,7,8,9….. to n1= 5

found in the Infra-red region.

Balmer series of spectral lines were the first to be discovered. Balmer showed that the

wave number Ῡ of any spectral line in the visible region of hydrogen spectrum is

governed by the empirical formula Ῡ = Ry [ 1/22 - 1/n22], where n2 = 3,4,5……Further he

also predicted that all the lines in the spectrum of atomic hydrogen including those

which appear in the infrared and uv regions.

Ritz combination principle:

According to Ritz, the wave number of any spectral line may be represented as a

combination of two terms, one of which is a constant and the other a variable

throughout each spectral series. The general formula of Ritz combination principle is:

Ῡ = R [ 1/x2 - 1/y2] ; where x is a constant term. The values of x and y are integers for

hydrogen but may not be for other atoms. Ritz combination principle is applicable for

the entire range of spectroscopic investigations. This is very much similar to Rydberg

equation.

Calculation of Ionization energies of hydrogen like atoms

In the case of hydrogen like atoms such as He+, Li2+ in which the number of electrons

is one but the number of protons is more than one, the charge on the nucleus will be +2e

and +3e respectively or in general it is Z is the atomic number of the element concerned.

The energy of the electron in such cases can be calculated using the equation:

E = - Z2e4m/8ϵo2n2h2, by assuming the nucleus to be stationary. If nuclear spin is taken

into consideration, the mass is replaced by the reduced mass µ = m1 m2 /m1 + m2 where

m1 and m2 are the masses of the nucleus and electron.

Limitations of Bohr’s theory: Bohr’s theory fails to explain the following

1. The spectra and energy of atoms containing more than one electron.

2. Fine structure and occurrence of spectral lines that is characteristics of atomic spectra.

3. Effect of magnetic (Zeeman Effect) and electric fields (Stark effect) on spectral lines

4. Formation of molecules.

Limitations of Classical mechanics:

Classical mechanics fails to explain the following 1. Black body radiation 2. Photo electric effect 3. Atomic and molecular spectra 4. Heat capacities of solids.

Wave mechanical concept of atom

In order to provide satisfactory explanation for the failures of classical mechanics, In 1923

an important concept that electrons like light has dual particle-wave nature was introduced by

Louis de Broglie, later in 1926 a new approach called wave mechanics or quantum

mechanics was developed independently by Heisenberg and Schrodinger to explain the

structure of atoms.

Dual nature of matter- de-Broglie equation

Louis de-Broglie suggested that electron may be associated with wave-particle dual character. Accordingly, if light waves can have particle like behavior then particles of matter can exhibit wave characteristics under certain conditions. Particles of extremely small mass like electron show wave characteristics.

The photons obey the equation: E = hν ------(1);

Einstein’s equation is E = mc2---(2)

Equating both the equations; we get hν = mc2 ; Or ν = mc2/h

Since ν = c/λ; Substituting in the above equation c/λ = mc2/h

Rearranging we get λ = h/mc, where λ is the wavelength, h is the Planck’s constant, m is

mass of the photon, and c is the velocity of light.

Substituting the velocity of light ‘c’ by the velocity of electron ‘v’ the above equation

modifies to λ = h/mv; or λ = h/p, where p is the momentum of the electron. This

equation is called de-Broglie equation.

de-Broglie concept reveals that the wave character is an inherent property of all matter

and it is not that matter behaves like particles at some time and waves at other times. The

wave nature is predominant in micro particles where mass is less and wavelength is more

as evident from the de-Broglie equation. The particle nature is more predominant. de-

Broglie suggested that for an electron moving in a circular orbit around a nucleus in a

wave like propagation and to remain stationary and continuously in phase, the

circumference of the circle must be an integral multiple of wavelength.

Circumference of the circle = 2πr = nλ; substituting λ from de-Broglie equation we get

2πr = nh/mv; rearranging mvr = nh/2π;

This equation is same as Bohr’s postulate where the electrons move only in those orbits

whose angular momentum is an integral multiple of h/2π. Hence the quantization of

angular momentum assumed by Bohr is evident later from the wave nature of the

electron.

Heisenberg’s Uncertainty principle

The wave particle duality of an electron has shown that an electron can no longer be

considered to occupy a specific orbit. If momentum is specified, its position cannot be

accurately specified. This concept has been stated by Heisenberg’s Uncertainty

principle.

Statement:

‘It is not possible to determine precisely both the position and the momentum (or

velocity) of a small moving particle (such as electron) simultaneously’.

If ∆x is the uncertainty in position measurement and ∆p is the uncertainty in the

momentum measurement, then ∆x . ∆p ≈ h/4π -----Mathematical statement of the Un-

certainty principle. This means that if ∆x is very small, then ∆p would be large and vice-

versa. This relationship is an inherent relationship in nature.

It is thus not possible to identify an electron in an atom in quantum mechanics. An

electron may be associated with a definite amount of energy that is said to belong to a

definite energy level and not a particular orbit. Hence in quantum mechanics, the term

orbital is introduced in place of orbit.

An Orbital is defined as the three dimensional region in space around the nucleus where

the probability of finding an electron having a certain amount of energy is maximum.

Schrodinger’s wave equation

Erwin Schrodinger described the behavior of microscopic particles such as subatomic

particles in the form of an equation. This equation includes both the particle behavior

in terms of mass ‘m’ and wave behavior in terms of a wave function ’Ψ’. This equation is

useful in solving all types of problems involving atoms, electrons and molecules. The

electron is associated with a dual nature of wave-particle duality, its behavior is best

described by considering standing waves (it is the resultant of two simple harmonic

waves travelling with equal speed but in opposite directions, which is stationary).

Derivation of Schrodinger’s wave equation

Consider the simple type of wave motion like the vibration of a stretched string

travelling along the x-axis with a velocity ‘u’. ω is the amplitude of the wave at any

point whose co-ordinate is ‘x’ at any time, then the equation for such a wave motion is

given by

∂2ω/∂x2 = 1/u2 ∂2ω/∂t2 …….(i)

This differential equation indicates that the amplitude ω

of the wave at any time travelling with a particular

velocity depends upto n the displacement x and the time

t. Hence it implies that ω is a function of x and time,

then we can write:

ω = f(x) f’(t) …….(ii)

Where f(x) is a function of the co-ordinate x only and f’(t) is a function of time t only.

But for the stationary waves, it follows the equation: f’(t) = A sin 2πνt …..(iii)

Where ν is the frequency of vibration and A is a constant, equal to the maximum

amplitude of the wave.

Substituting the value f’(t) from the above equation to equation (ii) we get

ω= f(x) A sin 2πνt;-------(iv). Differentiating this equation twice w.r.t to ‘t’ we get

∂2ω/∂t2 = -f(x) 4π2ν2 (A sin 2πνt) = -4π2ν2 f(x) f’(t)………(v)

Differentiating the equation (ii) twice w.r.t ‘x’ we get

∂2ω/∂x2 = ∂2f(x)/∂x2 f’(t) ---------(vi)

Substituting the values of ∂2ω/∂t2 and ∂2ω/∂x2 into the equation (i) we get

∂2f(x)/∂x2 f’(t) = -1/u24π2ν2 f(x) f’(t) Or ∂2f(x)/∂x2 = -4π2ν2 f(x)/ u2 ……..(vii)

This equation is time-independent and hence it represents the variation of amplitude

function of f(x) and x only.

We know that the velocity (u), frequency (ν) and wavelength λ is related by the

equation u=νλ. Substituting this relationship in equation (vii), we get

∂2f(x)/∂x2 = -4π2f(x)/λ2…….. (viii)

this equation represents the wave motion for one direction only. Hence, applying the

same to three dimensions, the amplitude function f(x) is replaced by Ψ(x, y, z) for three

co-ordinate systems, the equation becomes

∂2Ψ/∂x2 + ∂2Ψ/∂y2 + ∂2Ψ/∂z2 = - 4π2Ψ/λ2---------(ix)

This equation is applicable to all sub-atomic particles like electrons, nuclei etc..

Applying the de-Broglie equation λ = h/mv; the above equation modifies to:

∂2Ψ/∂x2 + ∂2Ψ/∂y2 + ∂2Ψ/∂z2 = - 4π2 m2 ν2 Ψ/h2-------(x)

Further we know that: Total energy (E) = Kinetic energy + Potential energy

E = ½ mv2 + V; or E-V = ½ mv2; Where V is the potential energy of the particle.

Substituting the value of mv2 from the above equation to equation(x); we get

∂2Ψ/∂x2 + ∂2Ψ/∂y2 + ∂2Ψ/∂z2 + 8π2m (E-V) Ψ/h2 = 0

This equation is called Schrodinger wave equation which represents the wave motion

of the particle in three dimensions.

Postulates of Quantum Mechanics:

1. The physical state of a system at time t is fully described by the wave function Ψ

(x.y.z.t), it is single valued, continuous and finite throughout the configuration of

space.

2. The possible wave function Ψ (x.y.z.t) are obtained by solving appropriate

Schrodinger wave equation.

3. Every dynamical variable corresponding to a physically observable quantity can

be represented by a linear operator.

4. Quantum mechanical operators corresponding to physical properties are

obtained from the classical expression in terms of the variable and converting

them to operators using certain set of procedures.

Operators in Quantum Mechanics

An operator is a mathematical instruction or a procedure to be carried out on a function so as to get another function. That is: (operator). Function = another function. There are two types of operators used in Quantum Mechanics, they are 1. Laplacian Operator: It is represented as 𝛁2= ∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2

Schrodinger wave equation is…

∂2Ψ/∂x2 + ∂2Ψ/∂y2 + ∂2Ψ/∂z2 + 8π2 m(E-V) Ψ/h2 = 0

May be written in terms of Laplacian operator as: 𝛁2Ψ + 8π2 m(E-V) Ψ/h2 = 0

2) Hamiltonian Operator: The total energy of a system is equal to the sum of Kinetic

and potential energies in classical mechanics. This is generally written as H

H = Ek + EP,

The corresponding operator is H’ called Hamiltonian operator.

H’Ψ = EΨ; Where the operator H = -h2∇2/8π2 m + Ep

{-h2𝛁2/8π2 m + Ep}Ψ = EΨ;

H’ is a way of representing the total energy of the system and E is the numerical value of

that energy. In the above equation ∇2 is a Laplacian operator and is equal to

∂2/∂x2 + ∂2/∂y2+∂2/∂z2 hence

H’Ψ = -h2/8π2 m(∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2)Ψ + VΨ = EΨ

Where VΨ is the potential energy term. When H operates on the wave function Ψ, a set

of allowed energy values are obtained.

Eigen Values and Eigen functions

The Schrodinger equation can have several solutions. Out of these only some are

acceptable. The solutions that give acceptable values of the total energy are called Eigen

values. The corresponding wave functions for which these solutions are applicable are

called Eigen functions or wave functions. The wave function (Ψ) must satisfy the

following conditions:

1. It must have definite values.

2. It must be single valued.

3. Its first derivative must be continuous with respect to its variables. When these

conditions are satisfied, it is a well behaved function.

Significance of Ψ and Ψ2

The wave function Ψ is an amplitude function and has no physical meaning. Ψ2 gives

the probability of finding the particle at a point at any given moment. Since the

probability of finding the particle at a point at any given point in space must be real and

not imaginary, in such cases ΨΨ* is generally taken. Ψ* is the complex derivative of Ψ.

The product of ΨΨ* will always be a real and not a negative quantity. Thus ΨΨ* gives

the probability of locating an electron within a certain region of space (It is length for

one dimension, area in two dimensions and volume in three dimensions).

Application of Schrodinger equation to Particle in One dimensional Box Consider a single particle say an electron of mass ‘m’,

confined to one dimensional box of length ‘a’. The walls of

this box are formed by potential energy barriers, infinitely

high that the particle cannot escape. The potential energy

out-side is infinite but inside the box, it is zero: i.e., V(x)=0

Schrodinger equation is:

d2Ψ/dx2 + 8π2 m (E-V) Ψ/h2 = 0;

since V(x)=0; d2Ψ/dx2 + 8π2 mEΨ/h2 = 0

or d2Ψ/dx2 + k2 Ψ = 0; where k2 = 8π2 mE/h2 is a constant independent of x.

The solutions of the second order differential equation of the above type is

Ψ = A Cos kx + B Sin kx; where A and B are the two arbitrary constants.

The value of these constants can be determined by considering the fact outside the box;

V=∞; the Schrodinger equation then becomes:

d2Ψ/dx2 + 8π2 m(E-∞) Ψ/h2 = 0 This is possible if the particle is outside the box, but

it cannot be, since it is present or confined inside the box.

By one of the postulates of quantum mechanics, Ψ must be continuous function of x.

Hence Ψ must be zero at the walls of the box, that is x=0 and x=a, as there can be no

variation in the values of Ψ at the walls of the box.

.. . Ψ=0 at x=a , At x=0, the equation Ψ = A Cos kx + B Sin kx becomes

0 = A Cos k.0+ B Sin k.0; 0=A, Since Cos 0=1.

Hence, Schrodinger equation is : Ψ = B Sin kx

Again, at the other wall, that is x=a, Ψ=0, therefore 0=B Sin ka; or Sin ka =0

B cannot be zero as it leads to Ψ being equal to zero everywhere which is not acceptable

Hence, Sin Ka =0 = Sin nπ; that is ka = nπ; Or k = nπ/a where n=1,2,3…..

The solution of Schrodinger equation is:

Ψ = Ψn = B Sin[nπx/a], where n is the quantum number.

Hence k = nπ/a ……(1) and k2 = 8π2 mE/h2 -------(2) from earlier equation equating

both; 8π2 mE/h2 = n2 π2 /a2; rearranging we get

E = n2 h2/8ma2

n=1, 2, 3…, E corresponds to the energy of the particle in one dimensional box.

Application of Schrodinger equation to the Hydrogen atom

The hydrogen atom is a two particle system consisting of an electron and atomic

nucleus. The nucleus is stationary as compared to the fast moving electron. The

Schrodinger equation corresponds to that of an electron of mass m, charge–e moving

around the nucleus of charge Ze, Z is the atomic number. Hence reduced mass µ id used

instead of m, where µ = memn/ me + mn where me and mn are the masses of electron and

nucleus respectively.

The Potential energy of the electron from a distance r from the nucleus is:

V = -Ze2/4πϵ0r;

Schrodinger equation for hydrogen atom is then written as

𝛁2Ψ + 8π2 µ/h2 [E + Ze2/ 4πϵo r]Ψ=0

Here the Cartesian co-ordinates(x,y and z) are converted into polar co-ordinates(r,θ,Φ).

The two are related as; x= r sinθ cos Φ, y= x= r sinθ sin Φ, z= r cosθ. The ranges of the

Co-ordinates are x from 0 to ∞, θ from 0 to π and Φ from 0 to 2π.

The Schrodinger equation changes to

1/r2∂/∂r( r2∂Ψ/∂r) + 1/r2 sinθ ∂/∂θ( sinθ∂Ψ/∂θ) + 1/r2 sin2 θ ∂2Ψ/∂Φ2 8π2 µ/h2 [E +

Ze2/ 4πϵo r]Ψ=0

Potential energy term is dependent only on r and therefore the wave function Ψ may be

written as a product of three functions; that is Ψ = R(r) θ(θ) Φ(Φ), R is a function r. θ is

a function of θ, and Φ is a function of Φ only. This equation is separated into three

differential equations, also partial derivatives are replaced by ordinary derivatives as

each function is mono variable.

R equation: 1/r2∂/∂r( r2dR/dr) -β/r2 R + 8π2 µ/h2 [E + Ze2/ 4πϵo r]R=0

where m and β are constants

θ equation: 1/ sinθ d/dθ( sinθ dθ/dθ) – m2 θ /sin2 θ + βθ =0

Φ equation: d2Φ/dφ2 + m2Φ = 0

The above equations are solved properly to obtain the allowed values of R, θ, Φ and

Energy E. On the basis of the solutions it is found that the acceptable solutions are n =

1,2,3………..; l = 0,1,2,………..(n-1); m= -l ,–l+1, ……0,1,2,…………l for each value of l.

The value of energy E is given by: E = -Ze4µ/8ϵo2n2h2. Where n can have integral

values starting from 1 and is called the Principal quantum number. This value is

identical with Bohr’s expression for energy. The values of n, l and m are the same as

given earlier for the principal, azimuthal and magnetic quantum numbers of electrons.

These values now has theoretical basis as requirements for the solution of hydrogen

and hydrogen like atoms. The solutions of Schrodinger equation do not have spin

quantum number of an electron. Later this was introduced to account for the spin and

values are + ½ and -½for the clockwise and anticlockwise spin about its own axis.

Quantum numbers

Quantum mechanics signifies quantum numbers are used to describe the distribution of

electrons in an atom. These are useful to describe atomic orbitals and the

distribution of electrons in an atom. From the solution of Schrodinger equation three

quantum numbers are indicated, the fourth quantum number spin quantum number,

refers to spin and magnetic property of the electron, arises from spectral evidence.

1. Principal quantum number (n): This quantum number represents the size of the electron shell. It also indicates the distance between the nucleus and the electron. It can have integral values such as 1, 2, 3………. n=1 is the nearest energy level to the nucleus and having the least energy. These energy levels are named as K, L, M, and N...shells. As n increases, the energy associated with it also increases.

2. Azimuthal quantum number (l): It denotes the angular momentum of the electron in its motion around the nucleus. It determines the shape of the orbital. It is also called subsidiary or orbital or angular momentum quantum number. A collection of orbitals having the same value of n is frequently called a shell. Orbitals with the same value of n and l are referred to as subshells. The number of values of l depends on the value of n. l can have integral values from 0 to (n-1). Orbital with ‘l=0’ is designated as ‘s’ orbital; Orbital with ‘l=1’ is designated as ‘p’ orbital with ‘l=2’ is designated as ‘d’ orbital; Orbital with ‘l=3’ is designated as ‘f’ orbital ( the letters designates sharp, principal, diffuse and fundamental).

3. Magnetic quantum number (m): This quantum number describes the orientation of the orbital in space in a strong magnetic field. Within a subshell, the value of m depends on the value of l. For an orbital there are 2l+1 types of orientation in space. In the absence of an electric or magnetic field, these orientations are degenerate (having equal energy). It can have values –l to +l. For example for n=1, l=0, m=2l+1, 2x0+1=1 value that is it has one value that is 0. The number of m values indicate the number of orbitals in a subshell of value l.

4. Spin quantum number(s): This quantum number arises from spectral evidence that the rotating electron spins about its own axis. It is important to describe the spin of the electrons. It has two values +1/2 and -1/2 for its clockwise and anticlock wise spinning motions for each orientations.

n l m s Designation of orbitals No.of orbitals

1 0 0 +1/2 & -1/2 1s 1 2

0 1

0 -1,0,1

+1/2 & -1/2 for each

2s 2px,2py,2pz

1 2

3 0 1 2

0 -1,0,1 -2,-1,0,1,2

+1/2 & -1/2 for each

3s 3px,3py,3pz 3dxy,3dyz,3dxz,3dx2y2,3dz2,

1 3 5

The wave function for an electron in an atom is called an atomic orbital. The orbital for any electron is determined by the values of the quantum numbers n, l and m. The Schrodinger equation for hydrogen and hydrogen like atoms reveals the resolution of the wave function into two factors:

Ψ(r,θ,Φ) = R(r), Y(θ,Φ). R(r) is the radial wave function. Y(θ,Φ) is the angular wave function.

Radial wave function: It is the R function in Schrodinger equation for Hydrogen

and hydrogen like atoms. The allowed energies depends only on the solutions of R

equation that is only on the principal quantum number. The radial wave functions gives

a distribution of electrons as a function of distance r from the nucleus.

Radial Distribution function:

The probability of distribution of electrons in an orbital is represented by plotting a

graph of Ψ2 Vs distance from the nucleus ’r’. This way of representation is prominent

and informative by radial distribution functions. It gives the total probability of finding

electrons in a spherical shell between r and r+dr instead of a point location. The

spherical shell is the space between two concentric spheres one with a radius ’r’ and the

other with a radius of r+dr. The volume of this shell is 4πr2dr. Hence the probability of

finding the electron in this small volume is = Ψ2 x Volume = Ψ2 x 4πr2dr. The function

4πr2Ψ2 is called the radial distribution function. A plot of radial distribution function

(Probability) against r is given below for electrons in1s, 2s orbitals.

From this plot, it is seen that for 1s orbital there is a maximum of 0.0529 nm which is exactly

same as Bohr radius. The s orbital is spherically symmetrical. For 2s orbital(n=2,l=0) there

are two maxima in the plot, which implies that there is probability of finding the electron at a

distance corresponding to the highest peak, and there is also a lesser probability of finding it

at the distance corresponding to smaller peak. Such a region of zero electron density is called

node or nodal region.

Angular wave function

This wave function corresponds to Y ((θ, Φ) depends on the angles θ and Φ. It

is a function of the quantum numbers l and m. The wave functions for p atomic

orbitals depends on the angular co-ordinates θ and Φ in addition to r. From

the probability plots, it is observed that the probability of finding a Pz electron

is maximum along z axis, it is zero in the X-Y plane which is called the nodal

plane. A nodal plane separates the two lobes. The atomic orbital shape as

dumb-bell. Similarly Px orbital has two lobes along x axis with y-z as the nodal

plane. Py orbital has two lobes along y axis with x-z as the nodal plane. The

three p orbitals are in similar in shape, size and are mutually perpendicular.

Nodal plane

Numerical Problems

1. Calculate the wavelength of the wave associated with mass of 1 kg moving with a

speed of 10m/s. h=6.63x10-34J.S.

Solution: Given data: m=1kg; v= 10m/s. h=6.63x10-34J.S. λ=?

From de Broglie equation we have λ = h/mv = 6.63x10-34J.S./1kgx 10m/s

= 0.663 x 10-34 m = 6.63 x 10-35m.

2. Calculate the velocity of electron in Bohr’s first orbit of hydrogen atom. Given

h=6.63x10-34J.S., m= 9.11x10-31kg. r = 0.053nm or 0.053x10-9m.

Solution: According to the principle of quantization of angular momentum

mvr = nh/2π; or v= nh/2πmr = 6.63x10-34J.S _________________ = 2.188x108 m/s

2x 3.14 x 9.11x10-31kg. x 0.053x10-9m

3. Calculate the energy of an electron in the second Bohr orbit in hydrogen atom.

given h=6.63x10-34J.S, electronic charge= 1.602x10-19C. m= 9.11x10-31kg

ϵo= 8.85 x10-12C/J/m.

+

-

Solution: We Know that E = -Z2e4m/8ϵo2n2h2 ; Since Z=1 for Hydrogen atom and n=2,

hence, E = -12e4m/8ϵo222h2

E = -e4m/32ϵo2h2 = -(1.602x10-19C)4x 9.11x10-31kg______ = -5.528x10-19J

32 x(8.85 x10-12C/J/m)2x(6.63x10-34J.S)2

3. Calculate the wave number of the spectral line when an electron jumps from n=3

to n=2, R= 1.097x107/m. or Calculate the wave number of the first line in the

Balmer series.

Solution: By Rydberg equation we have: Ῡ = Ry [ 1/n12 - 1/n22],

Substituting the values of R, n1=2, n2=3.

Ῡ = 1.097x107/m[1/22 - 1/32], = 1.524 x 106 /m.

4. Calculate the frequency and wavelength of the second line in the Lyman series,

given R=1.097x107/m.

Solution: Second line in Lyman series means n1=1, n2=3, Substituting in the Rydberg

equation we get: Ῡ = Ry [ 1/n12 - 1/n22]

Ῡ = 1.097x107/m[ 1/12 - 1/32], = 9.751 x 106 /m.

By definition: Ῡ=ν/c ; or ν = Ῡxc = 9.751 x 106 x 3x108 = 2.9253 x 1015 Hz. and

λ = 1/ Ῡ = 1/9.751 x 106 = 1.025 x 10-7m or 102.5nm.

5. Calculate the ionization energy of the hydrogen atom using Bohr’s theory. Energy

of the electron in the first Bohr orbit = -2.17 x 10-18J.

Solution: The ionization energy is the energy required to remove an electron from a

neutral gaseous atom to an orbit n=∞. In hydrogen atom the electron is in the orbit n=1.

Hence the IE corresponds to energy difference for the transition from n=1 to n=∞.

At n= ∞, the energy is taken as zero. At n=1, energy is -2.17 x 10-18J.

Ionization energy ∆E = (E∞ - E1)Z2 = 0-(-2.17 x 10-18) 12 = 2.17 x 10-18J

Ionization energy of hydrogen atom = 2.17 x 10-18J.

Ionization energy for one mole = 2.17 x 10-18 x 6.02 x 1023J = 13.06 x 105J.

6. Calculate the ionization energy per mole of Li2+ ions.

Solution: Energy of the first Bohr orbit is -2.17 x 10-18J. Li2+ ion has only one electron

(Similar to H atom) and this electron is in the first orbit has n=1.

Ionization energy ∆E = (E∞ - E1)Z2 = 0-(-2.17 x 10-18) 32 = 19.53 x 10-18J

Ionization energy for per mole = 19.53 x 10-18 x 6.02 x 1023

= 1.176 x 107J = 1.176 x 104kJ.

Questions 2 marks questions: 1. State Heisenberg’s uncertainty principle.

2. Derive de-Broglie’s equation.

3. How ionization energy of hydrogen atom in the first Bohr orbit is calculated.

4. Mention any limitations of classical mechanics.

5. Give the significance of Ψ and Ψ2.

6. What are quantum numbers?

7. Give all the values of l and m when n=2.

8. Write the shapes of the orbital when l=0 and l=1

9. Differentiate between orbit and orbital.

10. What is meant by wave particle duality of a particle?

11. Give any two limitations of Bohr’s theory.

12. Write the mathematical form of uncertainty principle.

13. Define Eigen values and Eigen functions.

14. State Ritz combination principle.

15. What is spin quantum number? Give its values.

16. Write radial probability distribution curve for 1s electron.

17. Explain wave particle duality

18. Differentiate between orbit and orbital

Questions more than 2 marks:

1. Derive an expression for the radius of nth orbit of hydrogen atom.

2. Derive an expression for the energy of the first Bohr in hydrogen atom.

3. Write the Rydberg equation and indicate the terms.

4. Give the postulates of quantum mechanics.

5. Write Schrodinger equation and indicate the terms.

6. Derive Schrodinger equation.

7. What is meant by Hamiltonian operator and Laplacian operator?

8. Derive Schrodinger equation for particle in one dimensional box and solve the equation.

9. Give the significance of i) Radial probability distribution and ii) Angular probability

distribution curves.

10. Write Schrodinger equation for hydrogen atom in three dimensions and indicate the

terms.

11. Give the relative equations for the conversion of Cartesian co-ordinates to polar co-

ordinates.

12. Mention the significance of the solutions of r, θ, Φ in Schrodinger equation for hydrogen

atom.

13. What are quantum numbers? Give the significance of each quantum number.

14. Explain hydrogen spectrum based on Rydberg equation.