Contents
From the study of subatomic particles to the laws of motion, Physics Topics offer insights into the workings of the world around us.
What are the Properties of Matter Waves? What is De Brogue Hypothesis?
Nature of Radiation: Wave Particle Duality
Electromagnetic radiations, if assumed to be streams of photon particles, can explain phenomena like photoelectric emission, blackbody radiation, atomic spectra, etc. But the theory fails to explain other optical phenomena like interference, diffraction, polarisation, etc. On the other hand, wave theory of radiation can interpret these phenomena successfully. Hence, depending on the type of experiment, radiation sometimes behaves like waves and sometimes like a stream of particles. Thus, wave theory and particle theory are not contradictory but complementary to each other. This is called wave particle duality.
Matter Wave
It has already been stated that radiation shows both wave nature and particle nature. But the fact that matter can show wave nature was unimaginable till 1924 when French physicist Louis de Broglie put forward the theory that a stream of material particles may behave as a wave. Most probably the following reasons led him to such a conclusion:
i) Nature prefers symmetry. Hence, two physical entities, matter and energy must co-exist in symmetry.
ii) If radiation can have both particle and wave nature, matter would also possess particle and wave nature.
iii) We know that a beam of light, which is a wave, can transfer energy and momentum at different points of a substance. Similarly, a stream of particles can also transfer energy and momentum at different points of a substance. Therefore, this stream of particles, may be a matter wave.
de Broglie’s hypothesis: Matter also consists of waves. For a radiation of frequency f, the energy of a photon (see section 1.3),
E = hf or, E = h\(\frac{c}{\lambda}\) [∵ c = fλ]
or, λ = \(\frac{h}{E / c}\) = \(\frac{h}{p}\) …… (1)
where p is the momentum of the photon.
As per de Broglie’s hypothesis, equation (1) is also applicable to an electron or any other particle. In this case, λ gives the wave length of the electron (or particle) of momentum p and is known as the de Broglie wavelength.
Thus, substituting the values of Planck’s constant h and momentum of the particle p in equation (1), we get de Broglie wavelength of the wave associated with the moving particle.
Hence a stream of any particle behaves like a beam of light, i.e., like a wave. The wave is known as matter wave. Wavelength of this matter wave,
λ = \(\frac{h}{p}\) = \(\frac{h}{m v}\) ….. (2)
where, m = mass of the particle, u = velocity of the particle, p = mu = momentum of the particle.
We can make the following inferences from the above relation connecting wavelength (a characteristic of wave) and momentum (a characteristic of particle):
- If v = 0, then λ = ∞, it means the waves are associated with moving material particles only.
- The de Broglìe wavelength does not depend on whether the moving particle is charged or uncharged. It means that matter waves are not electromagnetic waves because the electromagnetic waves are produced from accelerated charge particles.
- If the mass m and the velocity u of the particle are large, the associated de Broglie wavelength becomes very small.
- If the momentum of the particle increases, wavelength decreases.
- The wave nature and particle nature of any physical entity (matter or radiation) are mutually exclusive, i.e., if we consider the particle nature of radiation at any instant, the wave nature of radiation is to be excluded at that instant.
In 1927 C J Davisson and L H Germer of Bell Telephone Laboratories and George P Thomson of University of Aberdeen, Scotland, were able to show diffraction of electron streams and hence established experimentally the existence of matter waves.
Hence, any moving stream of particle or matter exhibits interference, diffraction, polarisation phenomena which can only be explained with wave theory.
Interference pattern, obtained by using a double slit type of experiment, using about 70000 moving electrons is shown in Fig.
de Broglie wavelength of moving electron:
Let an electron of mass m move with velocity v. Its de Broglie wavelength is given by
λ = \(\frac{h}{m v}\)
If the kinetic energy of the electron is K, then
K = \(\frac{1}{2}\)mv2 or, v = \(\sqrt{\frac{2 K}{m}}\)
So, de Broglie wavelength is
λ = \(\frac{h}{m \sqrt{\frac{2 K}{m}}}\) = \(\frac{h}{\sqrt{2 m K}}\) ……. (1)
The equation (1) is the expression for de Broglie wavelength of a moving particle in terms of its kinetic energy.
Now suppose an electron is at rest. It is accelerated through a potential difference V. The kinetic energy acquired by the electron is K = eV; e = charge of the electron.
On substituting the value of K ¡n equation (1), the de Broglie wavelength associated with the electron is given by
λ = \(\frac{h}{\sqrt{2 m e V}}\) …. (2)
Putting the values of h, e, m in equation (2) we have,
λ = \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times V}}\)
= \(\frac{12.27 \times 10^{-10}}{\sqrt{V}}\)m = \(\frac{12.27}{\sqrt{V}}\)Å
Wavelength of matter wave:
i) Let velocity of an electron (mass = 9.1 × 10-31 kg), v = 107 m ᐧ s-1. de Broglie wavelength of the electron,
λ = \(\frac{h}{m v}\) = \(\frac{6.63 \times 10^{-34}}{\left(9.1 \times 10^{-31}\right) \times 10^7}\)
= 7.3 × 10-11m = 0.73 Å
This wavelength is equivalent to the wavelength of X-rays.
ii) Let mass of a moving marble, m = 10 g = 0.01 kg and its velocity, v = 10 m ᐧ s-1. Then de Broglie wavelength of the marble,
λ = \(\frac{h}{m \nu}\) = \(\frac{6.63 \times 10^{-34}}{0.01 \times 10}\)
= 6.63 × 10-33 m
Value of this wavelength is too small to be measured or to be observed by any known experiment. Existence of such small wavelengths in electromagnetic radiation or any other real waves is still unknown to us.
From the above discussions we can infer that, de Broglie’s hypothesis is of no use in case of the macroscopic objects that we encounter in our daily life.
The concept of matter waves is only important in case of particles of atomic dimensions.
Numerical Examples
Example 1.
What is the de Broglie wavelength related to an electron of energy 100 eV? (Given, mass of electron, m = 9.1 × 10-31 kg; e = 1.6 × 10-19 C and h = 6.63 × 10-34 J ᐧ s)
Solution:
Let the velocity of electron be u. Its kinetic energy,
\(\frac{1}{2}\)mv2 = E or, m2v2 = 2mE
or, mv = \(\sqrt{2 m E}\)
∴ de Broglie wavelength related with the electron,
λ = \(\frac{h}{m v}\) = \(\frac{h}{\sqrt{2 m E}}\)
= \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 100 \times 1.6 \times 10^{-19}}}\)
= 1.23 × 10-10m = 1.23Å
Example 2.
Calculate the momentum of a photon of frequency 5 × 1013Hz. Given, h = 6.6 × 10-34J ᐧ s and c = 3 × 108m ᐧ s-1. [HPSEB 2000]
Solution:
de Broglie wavelength, λ = \(\frac{h}{m v}\) = \(\frac{h}{p}\)
∴ Momentum, p = \(\frac{h}{\lambda}\)
Therefore, momentum of the photon,
p = \(\frac{h}{\lambda}\) = \(\frac{h f}{c}\) = \(\frac{\left(6.6 \times 10^{-34}\right) \times\left(5 \times 10^{13}\right)}{3 \times 10^8}\)
= 1.1 × 10-28 kg ᐧ m ᐧ s-1
Example 3.
The wavelength λ of a photon and the de Broglie wavelength of an electron have the same value. Show that the energy of the photon is \(\frac{2 \lambda m c}{h}\) times of the kinetic energy of the electron. Here m, c and h have their usual meaning. [CBSE’ 11, ‘03]
Solution:
The energy of the photon, E = hf = \(\frac{h c}{\lambda}\)
The kinetic energy of the electron,
E’ = \(\frac{1}{2}\)mv2 = \(\frac{p^2}{2 m}\) [where p = mv = momentum]
As p = \(\frac{h}{\lambda}\) So, E’ = \(\frac{h^2}{2 m \lambda^2}\)
∴ \(\frac{E}{E^{\prime}}\) = \(\frac{h c}{\lambda} \cdot \frac{2 m \lambda^2}{h^2}\) = \(\frac{2 \lambda m c}{h}\) or, E = \(\frac{2 \lambda m c}{h}\) ᐧ E’
Example 4.
An electron and a photon have same de Broglie wave length λ = 10-10m. Compare the kinetic energy of the electron with the total energy of the photon.
Solution:
The kinetic energy of an electron having mass m and velocity v, K = \(\frac{1}{2}\)mv2
Wavelength, λ = \(\frac{h}{m v}\) or, v = \(\frac{h}{m \lambda}\)
∴ K = \(\frac{1}{2}\)m ᐧ \(\frac{h^2}{m^2 \lambda^2}\) = \(\frac{h^2}{2 m \lambda^2}\)
The energy of the photon of wavelength λ, E = \(\frac{h c}{\lambda}\)
∴ \(\frac{K}{E}\) = \(\frac{h^2}{2 m \lambda^2} \cdot \frac{\lambda}{h c}\) = \(\frac{h}{2 m \lambda c}\)
= \(\frac{6.6 \times 10^{-34}}{2 \times 9.1 \times 10^{-31} \times 10^{-10} \times 3 \times 10^8}\) = 0.012 < 1
Hence, the kinetic energy of the electron is lower than the total energy of the photon.
Example 5.
Calculate the de Broglie wavelength of an electron of kinetic energy 500 eV.
Solution:
The kinetic energy of an electron 500 eV means the electron is accelerated by the potential 500 V.
So, the de Broglie wavelength associated with the electron,
λ = \(\frac{12.27}{\sqrt{500}}\) = 0.55 Å