# Dual Character , Particle and Wave Character of Matter and Radiation

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 ν      ….  (i)   (according to the Planck’s quantum theory)

where n 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)

hν  = mc2

But ν= c/λ

$\large \frac{h c}{\lambda} = m c^2$

$\large \lambda = \frac{h}{m c}$

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.

$\large \lambda = \frac{h}{m v} = \frac{h}{p}$

where  m v = p is the momentum of the particle.

### Derivation of Angular Momentum:

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 (l)

Therefore 2πr = nλ

where ‘ n ’ is an integer and  ‘ r ’ is the radius of the orbit

But l = h/mv

∴ 2πr = nh /mv

(or) mvr = n h/2π

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 ’   = 2πr /λ

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 2πr , (i.e., the circumference of the circle).

Thus, the number of revolutions per second is = v/2πr

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 × volts = Joules, 1 eV numerically equals the electronic charge except that joule replaces coulombs.

Question: Two particles A and B are in motion. If the wavelength associated with particle A is 5 × 10–8 m, calculate the wavelength associated with particle B if its momentum is half of A.

Solution: According to de-Broglie equation

$\lambda_A = \frac{h}{p_A}$ and $\lambda_B = \frac{h}{p_B}$

$\frac{\lambda_A}{\lambda_B} = \frac{p_B}{p_A}$

But pB = (1/2) pA (given)

$\frac{\lambda_A}{\lambda_B} = \frac{1}{2}$

λB = 2 λA

λB = 2 × 5 × 10–8 m

= 10–7 m

Question: Calculate the de Broglie wavelength of a ball of mass 0.1 kg moving with a speed of 60 ms–1.

Solution: $\lambda = \frac{h}{m v} = \frac{6.6 \times 10^{-34}}{0.1 \times 60}$

λ = 1.1 × 10–34 m

This is apparent that this wavelength is too small for ordinary observation.
Although the de Broglie equation is applicable to all material objects but it has significance only in case of microscopic particles.
Since, we come across macroscopic objects in our everyday life, de Broglie relationship has no significance in everyday life.

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

 Photon Sub Atomic Particle 1. Energy  = h ν Energy  = (1/2) mv2 2. Wavelength = c/ν Wavelength = h/mv