Give the idea of De-Broglie matter waves
Answers
Albert Einstein’s theory of special relativity was first published in 1905[1] (discussed in Book I). This suggested that light has a momentum - which is classically equal to an object’s mass multiplied by its velocity - even if photons have no mass.
Special relativity shows that energy is related to mass via E2=p2c2+m2c4, where E refers to energy, p to momentum, c to the speed of light, and m to an object’s mass. An object with no mass, like a photon, will have an energy of E2=p2c2, or E=pc.
Arthur Compton proved that photons do have the momentum Einstein predicted in 1922.[2] Compton did this by firing X-rays at aluminium foil. When the X-rays hit the electrons in the outermost shell of the aluminium atoms, they transferred some of their angular momentum. The electrons gained enough energy to leave the atom, and the X-ray photon lost the same amount of energy. This process is now known as Compton scattering.
When a photon collides with an electron and gains energy the process is known as inverse Compton scattering. Compton scattering is now utilised in radiobiology,[3] and both Compton scattering and inverse Compton scattering are important in X-ray astronomy[4].
15.2 Electron waves
In 1924, Louis de Broglie used Einstein’s equations to show that electrons can act like waves, just as photons can act like particles.[5]
15.2.1 The wavelength of photons
The wavelength of a photon is calculated by combining Einstein’s equation for determining a photon’s energy with the Planck relation.[6] Using E = pc (which is the energy of photons according to special relativity) and E = hν (which is the energy of photons according to the Planck relation),
pc=hν (15.1)
Using c = λν,
λ = h /p
The wavelength of particles
De Broglie proposed that particles also have a wavelength and that this can be calculated using the same equation, except here the particle’s momentum is equal to mv, where m is the particles’ mass, and v is its velocity.
λ = h /mv
De Broglie realised that electrons must orbit the nucleus like a standing wave - a wave that is constrained at each end. This means that only a whole number of wavelengths fit exactly around each orbit,
λ = 2πr /n
Here, the orbit is assumed to be circular. 2πr is the circumference of a circle of radius r, and n is the shell number. This means that the angular momentum of each electron is quantised, in agreement with Niels Bohr’s theory of the atom[7] (discussed in Chapter 10).
Bohr had previously shown that the angular momentum of each electron (L) is equal to the shell number multiplied by a constant.
h /mv = 2πr /n
nh /2π = rmv
nh /2π = rp = L