In case of light some phenomenon like diffraction and interference can be explained on the basis of its wave character. However, the certain other phenomenon such as black body radiation and photoelectric effect can be explained only on the basis of its particle nature. Thus, light is said to have a dual character. Such studies on light were made by Einstein in 1905.

Louis de Broglie, in 1924 extended the idea of photons to material particles such as electron and he proposed that matter also has a dual character-as wave and as particle.

Derivation of de-Broglie Equation:

The wavelength of the wave associated with any material particle was calculated by analogy with photon.

In case of photon, if it is assumed to have wave character, its energy is given by

E = h$\nu$ …(i) (according to the Planck's quantum theory)

where $\nu$ is the frequency of the wave and 'h' is Planck's constant

If the photon is supposed to have particle character, its energy is given by

E = mc² … (ii) (according to Einstein's equation)

where 'm' is the mass of photon, 'c' is the velocity of light.

By equating (i) and (ii)

h$\nu$ = mc²

But $v=c/\lambda$

h $\frac{c}{\lambda }$ = mc²

(or) $\lambda =h$ /mc

The above equation is applicable to material particle if the mass and velocity of photon is replaced by the mass and velocity of material particle. Thus for any material particle like electron.

$\lambda$ = h/mv or $\lambda$ = $\frac{h}{p}$ where mv = p is the momentum of the particle.

Relation Between Kinetic Energy and Wavelength:

K.E (E)=1/2 mv²

v = $\sqrt{\frac{2E}{m}}$ Þ $\lambda =\frac{h}{mv}$

$\lambda =\frac{h}{\sqrt{2E.m}}$

Derivation of Angular Momentum from de Broglie Equation:

According to Bohr's model, the electron revolves around the nucleus in circular orbits. According to de Broglie concept, the electron is not only a particle but has a wave character also.

If the wave is completely in phase, the circumference of the orbit must be equal to an integral multiple of wave length ($\lambda$)

Therefore $2\pi r=n$$\lambda$

where 'n' is an integer and 'r' is the radius of the orbit

But $\lambda$ = h/mv

$\therefore$ $2\pi r$ = nh /mv or mvr = n $h/2\pi$

which is Bohr's postulate of angular momentum, where 'n' is the principal quantum number.

"Thus, the number of waves an electron makes in a particular Bohr orbit in one complete revolution is equal to the principal quantum number of the orbit".

Alternatively:

Number of waves 'n' = $\frac{2\pi r}{\lambda }$ = $\frac{2\pi r}{h/mv}$ = $\frac{2\pi mvr}{h}$

Where v and r are the velocity of electron and radius of that particular Bohr orbit in which number of waves are to be calculated, respectively.

The electron is revolving around the nucleus in a circular orbit. How many revolutions it can make in one second?

Let the velocity of electron be v m/sec. The distance it has to travel for one revolution $2\pi r$, (i.e., the circumference of the circle).

Thus, the number of revolutions per second is = $\frac{v}{2\pi r}$

Common unit of energy is electron volt which is amount of energy given when an electron is accelerated by a potential of exactly 1 volt. This energy equals the product of voltage and charge. Since in SI units coulombs x volts = joules, 1 eV numerically equals the electronic charge except that joule replaces coulombs.

Illustration 1. An e^-, a proton and an alpha particle have K.E of 16 E, 4 E and E respectively. What's the qualitative order of their Broglie wavelengths?

(A) ${\lambda }_e>{\lambda }_p>{\lambda }_a$ (B) ${\lambda }_e>{\lambda }_p={\lambda }_a$

(C) ${\lambda }_P<{\lambda }_e<{\lambda }_a$ (D) ${\lambda }_a<{\lambda }_e={\lambda }_P$

Solution: $\lambda$=$\frac{h}{\sqrt{2mK.E}}$. Hence (B) is correct.

Distinction between the wave- particle nature of a photon and the particle- wave nature of a sub atomic particle:

Illustration 2. Calculate the number of waves made by electron in 4^th Bohr's orbit.

Solution: Number of waves made by electron in n^th orbit = n

Hence the number of waves in 4^rd orbit = 4

Illustration 3. What is the de-Broglie wavelength of electron having K.E. of 5 eV?

Solution: K.E. = $\frac{1}{2}m{v}^2$

$v=\sqrt{\frac{2K.E.}{m}}$

$Now,\lambda =\frac{h}{mv}$

= $\frac{h}{\sqrt{2K.E.m}}$

= $\frac{6.62\times {10}^{-34}}{\sqrt{2\times 5\times 1.6\times {10}^{-19}\times 9.1\times {10}^{-31}}}$

= 5.486 x 10^-10 m

HEISENBERG'S UNCERTAINTY PRINCIPLE

All moving objects that we see around us e.g., a car, a ball thrown in the air etc., move along definite paths. Hence their position and velocity can be measured accurately at any instant of time. Is it possible for subatomic particle also?

As a consequence of dual nature of matter, Heisenberg, in 1927 gave a principle about the uncertainties in simultaneous measurement of position and momentum (mass x velocity) of small particles.

This Principle States:

"It is impossible to measure simultaneously the position and momentum of a small microscopic moving particle with absolute accuracy or certainty" i.e., if an attempt is made to measure any one of these two quantities with higher accuracy, the other becomes less accurate.

The product of the uncertainty in position $\left(\Delta x\right)$and the uncertainty in the momentum ($\Delta$p = $\left(m.\Delta x\right)$where m is the mass of the particle and $\Delta v$ is the uncertainty in velocity) is equal to or greater than h/4p where h is the Planck's constant.

Thus, the mathematical expression for the Heisenberg's uncertainty principle is simply written as

$\Delta x.\Delta p\ge h/4\pi$

Explanation of Heisenberg's uncertainty principle

Suppose we attempt to measure both the position and momentum of an electron, to pinpoint the position of the electron we have to use light so that the photon of light strikes the electron and the reflected photon is seen in the microscope. As a result of the hitting, the position as well as the velocity of the electron are disturbed. The accuracy with which the position of the particle can be measured depends upon the wavelength of the light used. The uncertainty in position is $\pm \lambda$. The shorter the wavelength, the greater is the accuracy. But shorter wavelength means higher frequency and hence higher energy. This high energy photon on striking the electron changes its speed as well as direction. But this is not true for macroscopic moving particle. Hence Heisenberg's uncertainty principle is not applicable to macroscopic particles.

Illustration 4. Why electron cannot exist inside the nucleus according to Heisenberg's uncertainty principle?

Solution: Diameter of the atomic nucleus is of the order of 10^–15m

The maximum uncertainty in the position of electron is 10^–15 m.

Mass of electron = 9.1 x10^–31 kg.

$\Delta x.\Delta p$= $\frac{h}{4\pi }$

$\Delta x$ x (m.Dv) = h/4p

$\Delta v$ = $\frac{h}{4\pi }\times \frac{1}{\Delta x.m}$ = $\frac{6.63\times {10}^{-34}}{4\times \frac{22}{7}}$ x$\frac{1}{{10}^{-15}\times 9.1\times {10}^{-31}}$

$\Delta v$ = 5.80 x 10¹⁰ ms^–1

This value is much higher than the velocity of light and hence not possible.

QUANTUM MECHANICAL MODEL OF ATOM

The atomic model which is based on the particle and wave nature of the electron is known as wave or quantum mechanical model of the atom. This was developed by Erwin Schrodinger in 1926. This model describes the electron as a three dimensional wave in the electronic field of positively charged nucleus. Schrodinger derived an equation which describes wave motion of an electron. The differential equation is

$\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}+\frac{{\partial }^2\psi }{\partial z^2}+\frac{8{\pi }^{2}m}{h^2}(E-V)\psi =0$

where x, y, z are certain coordinates of the electron, m = mass of the electron E = total energy of the electron. V = potential energy of the electron; h = Planck's constant and $\psi$ (psi) = wave function of the electron.

Significance of $\psi$: The wave function may be regarded as the amplitude function expressed in terms of coordinates x, y and z. The wave function may have positive or negative values depending upon the value of coordinates. The main aim of Schrodinger equation is to give solution for probability approach. When the equation is solved, it is observed that for some regions of space the value of $\psi$ is negative. But the probability must be always positive and cannot be negative, it is thus, proper to use ${\psi }^2$ in favour of $\psi$.

Significance of ${\psi }^2$: ${\psi }^2$ is a probability factor. It describes the probability of finding an electron within a small space. The space in which there is maximum probability of finding an electron is termed as orbital. The important point of the solution of the wave equation is that it provides a set of numbers called quantum numbers which describe energies of the electron in atoms, information about the shapes and orientations of the most probable distribution of electrons around nucleus.

Nodal Points and Planes:

The point where there is zero probability of finding the electron is called nodal point. There are two types of nodes: Radial nodes and angular nodes. The former is concerned with distance from the nucleus while latter is concerned with direction.

No. of radial nodes = n – $\ell$– 1

No. of angular nodes = $\ell$

Total number of nodes = n – 1

Nodal planes are the planes of zero probability of finding the electron. The number of such planes is also equal to $\ell$.

Illustration 5. Calculate radial nodes and angular nodes for the following type of orbitals.

(a) 1s, (b) 2p, (c) 3p, (d) 3d, (e) 4s and (f) 4d

Solution: (a) 0, 0 (b) 0, 1

(c) 1, 1 (d) 0, 2

(e) 3, 0 (f) 1, 2

PHOTOELECTRIC EFFECT

Sir J.J. Thomson, observed that when a light of certain frequency strikes the surface of a metal, electrons are ejected from the metal. This phenomenon is known as photoelectric effect and the ejected electrons are called photoelectrons.

A few metals, which are having low ionization energy like Cesium, show this effect under the action of visible light but many more show it under the action of more energetic ultraviolet light.

An evacuated tube contains two electrodes connected to a source of variable voltage, with the metal plate whose surface is irradiated as the anode. Some of the photoelectrons that emerge from this surface have enough energy to reach the cathode despite its negative polarity, and they constitute the measured current. The slower photoelectrons are repelled before they get to the cathode. When the voltage is increased to a certain value V₀, of the order of several volts, no more photoelectrons arrive, as indicated by the current dropping to zero. This extinction voltage (or also referred as stopping potential) corresponds to the maximum photoelectron kinetic energy i.e., eV_o = ½ mv²

The experimental findings are summarised as below:

$\bullet$Electrons come out as soon as the light (of sufficient energy) strikes

the metal surface.

$\bullet$The light of any frequency will not be able to cause ejection of electrons from a metal surface. There is a minimum frequency, called the threshold (or critical) frequency, which can just cause the ejection. This frequency varies with the nature of the metal. The higher the frequency of the light, the more energy the photoelectrons have. Blue light results in faster electrons than red light.

$\bullet$Photoelectric current is increased with increase in intensity of light of same frequency, if emission is permitted i.e., a bright light yields more photoelectrons than a dim one of the same frequency, but the electron energies remain the same.

Light must have stream of energy particles or quanta of energy (h$\nu$). Suppose, the threshold frequency of light required to eject electrons from a metal is $\nu$₀, when a photon of light of this frequency strikes a metal it imparts its entire energy (h$\nu$₀) to the electron.

"This energy enables the electron to break away from the atom by overcoming the attractive influence of the nucleus". Thus each photon can eject one electron. If the frequency of light is less than $\nu$₀ there is no ejection of electron. If the frequency of light is higher than $\nu$₀ (let it be $\nu$), the photon of this light having higher energy (h$\nu$), will impart some energy to the electron that is needed to remove it from the atom. The excess energy would give a certain velocity (i.e, kinetic energy) to the electron.

h$\nu$ = h$\nu$₀ + K.E

h$\nu$ = h$\nu$₀ + ½ mv²

½ mv² = h$\nu$–h$\nu$_o

where, $\nu$ = frequency of the incident light

$\nu$₀ = threshold frequency

h$\nu$₀ is the threshold energy (or) the work function denoted by $\varphi$ = h$\nu$₀ (minimum energy of the photon to liberate electron). It is constant for particular metal and is also equal to the ionization potential of gaseous atoms.

The kinetic energy of the photoelectrons increases linearly with the frequency of incident light. Thus, if the energy of the ejected electrons is plotted as a function of frequency, it result in a straight line whose slope is equal to Planck's constant 'h' and whose intercept is h$\nu$₀­.

Illustration 6. Work function of sodium is 2.5 eV. Predict whether the wavelength 6500$\stackrel{o}{A}$

is suitable for a photoelectron ejection or not.

Solution: Energy of incident light $=\frac{hc}{\lambda }=\frac{6.62\times {10}^{-34}\times 5\times {10}^8}{6500\times {10}^{-10}}$

= 3.055 x 10^-19 J

= 1.9 eV

Which is lower than work function. Hence no ejection will take place.