de Broglie wavelength is an important concept while studying quantum mechanics. The wavelength (λ) that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength. A particle’s de Broglie wavelength is usually inversely proportional to its force.

## de Broglie Waves

It is said that matter has a dual nature of wave-particles. de Broglie waves, named after the discoverer Louis de Broglie, is the property of a material object that varies in time or space while behaving similar to waves. It is also called matter-waves. It holds great similarity to the dual nature of light which behaves as particle and wave, which has been proven experimentally.

**Also Read:** Photoelectric Effect

The physicist Louis de Broglie suggested that particles might have both wave properties and particle properties. The wave nature of electrons was also detected experimentally to substantiate the suggestion of Louis de Broglie.

The objects which we see in day-to-day life have wavelengths which are very small and invisible, hence, we do not experience them as waves. However, de Broglie wavelengths are quite visible in the case of subatomic particles.

## de Broglie Wavelength for Electrons

In the case of electrons going in circles around the nuclei in atoms, the de Broglie waves exist as a closed-loop, such that they can exist only as standing waves, and fit evenly around the loop. Because of this requirement, the electrons in atoms circle the nucleus in particular configurations, or states, which are called stationary orbits.

## de Broglie Wavelength Formula and Derivation

de Broglie reasoned that matter also can show wave-particle duality, just like light, since light can behave both as a wave (it can be diffracted and it has a wavelength) and as a particle (it contains packets of energy hν). And also reasoned that matter would follow the same equation for wavelength as light namely,

**λ = h / p**

Where p is the linear momentum, as shown by Einstein.

### Derivation

de Broglie derived the above relationship as follows:

1) E = hν for a photon and λν = c for an electromagnetic wave.

2) E = mc^{2}, means λ = h/mc, which is equivalent to λ = h/p.

Note: m is the relativistic mass, and not the rest mass; since the rest mass of a photon is zero.

Now, if a particle is moving with a velocity v, the momentum p = mv and hence λ = h / mv

Therefore, the de Broglie wavelength formula is expressed as;

** λ = h / mv**

## Applications of de Broglie Waves

1. The wave properties of matter are only observable for very small objects, de Broglie wavelength of a double-slit interference pattern is produced by using electrons as the source. 10 eV electrons (which is the typical energy of an electron in an electron microscope): de Broglie wavelength = 3.9 x 10^{-10} m.

This is comparable to the spacing between atoms. Therefore, a crystal acts as a diffraction grating for electrons. The diffraction pattern allows the crystal structure to be determined.

2. In a microscope, the size of the smallest features we can see is limited by the wavelength used. With visible light, the smallest wavelength is 400 nm = 4 x 10^{-7} m. Typical electron microscopes use wavelengths 1000 times smaller and can be used to study very fine details.

## Thermal de Broglie Wavelength

The thermal de Broglie wavelength (λ_{th}) is approximately the average de Broglie wavelength of the gas particles in an ideal gas at the specified temperature.

The thermal de Broglie wavelength is given by the expression:

λ_{D} = h / √ 2 π m k_{B}T

where,

h = Planck constant,

m = mass of a gas particle,

k_{B }= Boltzmann constant,

T = temperature of the gas,

λ_{D }= λ_{th} = thermal de Broglie wavelength of the gas particles.

**Also Read:**

## Bohr’s model for Hydrogen

The electrons move in circular orbits around the nucleus in atoms. The electrons have the form of disk-shaped clouds. In the hydrogen atom, the electron in the ground state with the minimum energy can be modelled by a rotating disk, the inner edge of which has the radius ½ rB(1) and the outer edge has the radius 3/2 rB (2) where rB is the Bohr radius.

If we assume that the electron’s orbit in the atom includes ‘n’ of de Broglie wavelengths, then in case of a circular orbit with the radius, for the circle perimeter and the angular momentum L of the electron we will obtain the following:

2 πr = n λ_{B}, L = rp = nh / 2π, λ_{B} = h / p (3)

This is exactly the postulate of the Bohr’s model for the Hydrogen atom. According to postulate, the angular momentum of the hydrogen atom is quantized and proportional to the number of the orbit ‘n’ and the Planck constant.

## Solved Problems

**Question 1: An electron and a photon have the same wavelength. If p is the momentum of the electron and E is the energy of the photon. The magnitude of p/E in S.I unit is**

**(a) 3.010 ^{8} (b) 3.3310^{-9}**

**(c) 9.110 ^{-31} (d) 6.6410^{-34}**

Answer: b

As we know that, for electron, λ = h/p

Or

p = h/λ

And for photon E = hc / λ

Thus, p / E = 1 / c = 1 / (310^{8} m/s) = 3.33 10^{-9} s/m

**Question 2: What is the energy and wavelength of a thermal neutron?**

Answer:

KE = 3/2 kT = 3/2 (1.38 × 10^{–23}) (293) = 6.07 × 10^{–21} J

λ = h / p

λ = h / √2m0 (KE)

λ = 6 / 63 × 10^{–34} / √2 (1.67 × 10^{–27}) (6.07 × 10^{–21})

λ = 0.147 nm

**Question 3: A particle of mass m is confined to a narrow tube of length L. Find**

**(a) The wavelengths of the de-Broglie wave which will resonate in the tube,**

**(b) The corresponding particle momenta, and**

**(c) The corresponding energies.**

Answer:

(a) The de Broglie waves will resonate with a node at each end of the tube.

A few of the possible resonance forms are listed below:

λn = 2L / n; = 1, 2, 3, …

(b) Since de-Broglie wavelengths are

λn = h / pn

∴ pn = h / λn = nh / 2L n = 1, 2, 3, …

(c) (KE) n = Pn^{2} / 2m = n^{2}h^{2}/8L^{2}m, n = 1, 2, 3, …

**Question 4: What is the wavelength of an electron moving at 5.31 x 106 m/sec?**

Given: mass of electron = 9.11 x 10-31 kg h = 6.626 x 10-34 J·s

Answer:

de Broglie’s equation is

λ = h/mv

λ = 6.626 x 10-34 J·s/ 9.11 x 10-31 kg x 5.31 x 106 m/sec

λ = 6.626 x 10-34 J·s/4.84 x 10-24 kg·m/sec

λ = 1.37 x 10-10 m

λ = 1.37 Å

The wavelength of an electron moving 5.31 x 106 m/sec is 1.37 x 10-10 m or 1.37 Å.

**Question 5: Which of the following is called as non-mechanical waves?**

**a) Magnetic waves**

**b) Electromagnetic waves**

**c) Electrical waves**

**d) Matter waves**

Answer: b

The waves which travel in the form of oscillating electric and magnetic waves are called electromagnetic waves. Such waves do not require any material for their propagation and are called non-mechanical waves.

**Question 6: Which of the following is associated with an electron microscope?**

**a) Matter waves**

**b) Electrical waves**

**c) Magnetic waves**

**d) Electromagnetic waves**

Answer: a

The waves associated with microscopic particles when they are in motion are called matter waves. Electron microscope makes use of the matter waves associated with fast-moving electrons.

**Question 7: Calculate the de-Broglie wavelength of an electron which has been accelerated from rest on application of a potential of 400volts.**

**a) 0.1653 Å**

**b) 0.5125 Å**

**c) 0.6135 Å**

**d) 0.2514 Å**

Answer: c

de-Broglie wavelength = h/√ (2×m×e×V)

de-Broglie wavelength = (6.625×10-14)/√(2×9.11×10-31×1.6×10-17×400)

Wavelength = 0.6135 Å.

**Question 8: The de Broglie wavelength of a particle is the same as the wavelength of a photon. Then, the photon’s energy is:**

**(a) Equal to the kinetic energy of the particle.**

**(b) less than the kinetic energy of the particle.**

**(c) greater than the kinetic energy of the particle.**

**(d) Nothing can be specified.**

Answer: c

The photon’s energy depends only on the frequency of the photon. Hence, the photon’s energy is greater than the kinetic energy of the electron.

**Question. 9: An electron and a proton have the same de Broglie wavelength. Then the kinetic energy of the electron is?**

**(a) Zero **

**(b) Infinity**

**(c) Equal to the kinetic energy of the proton**

**(d) Greater than the kinetic energy of the proton**

Answer: d

The electron and the proton have the same de Broglie wavelength, which means their momenta are the same. Since the mass of the proton is greater than the mass of the electron, its speed is less than the speed of the electron. Hence, KE = (½) pv, the kinetic energy of the electron is greater than the kinetic energy of the proton.

**Question. 10: A nucleus of mass M at rest emits an α-particle of mass m. The de Broglie wavelengths of the α-particle and residual nucleus will be in the ratio**

**(a) m : M (b) (M+m) : m**

**(c) M : m (d) 1 : 1**

Answer: d

The nucleus is initially at rest. By the conservation of momentum principle, we can assume that the momenta of the α-particle of mass m and the residual nucleus will be equal and opposite. The de Broglie wavelength is inversely proportional to momentum. Hence, the de Broglie wavelengths of the α-particle and residual nucleus will be in the ratio = 1:1.

**Question 11: The operation of the electron microscope depends on the:**

**(a) The very short wavelength of X-rays**

**(b) Wave nature of electrons**

**(c) Electromagnetic theory of waves**

**(d) Photoelectric effect**

Answer: b

The wave nature of the electron particles is utilized in the electron microscope.