Expression for de-Broglie wavelength :
According to quantum theory, the energy of a photon is given by
$E = h\nu \qquad \dots (i)$
According to Einstein's mass-energy equivalence, the energy of the photon is
$E = mc^2 \qquad \dots (ii)$
From equations (i) and (ii), we have,
$mc^2 = h\nu$
$mc^2 = h\frac{c}{\lambda}( \because c = \nu \lambda)$
where, $\lambda$ is de-Broglie wavelength.
$mc = \frac{h}{\lambda}$
$ \lambda = \frac{h}{mc} \qquad \dots (iii)$
But $mc = p$, the momentum of photon, therefore,
$\lambda = \frac{h}{p} \qquad \dots (iv)$
If instead of a photon, we have a material particle of mass $m$ moving with velocity $v$, then eqn. (iii) becomes,
$\lambda = \frac{h}{mv} \qquad \dots (v)$
which is the expression for de-Broglie wavelength.
Conclusions. From eqn. (v), it is concluded that,
(i) de-Broglie wavelength is inversely proportional to the velocity of the particle, i.e. $\lambda \propto \frac{1}{v}$. If particle moves faster, the wavelength of the wave associated with it is smaller and vice-versa.
(ii) The particle at rest ($v=0$), has uniformed de-Broglie wavelength. Such a wave cannot be observed.
(iii) de-Broglie wavelength is inversely proportional to the mass of the particle, i.e. $\lambda \propto \frac{1}{m}$. Thus, the wavelength of a matter wave due to a heavier particle is smaller than that due to a lighter particle.
(iv) de-Broglie or matter wave is independent of the charge on the material particle. It means, matter wave or de-Broglie wave will result due to every moving particle (whether charged or uncharged).
(v) de-Broglie waves are not electromagnetic in nature because electromagnetic waves are produced by accelerated charged particles, whereas de-Broglie waves are produced by charged as well as uncharged particles.
(vi) Macroscopic objects like ball, stone, cars etc. do not show wave-like properties. This is because their mass is large and speed is small and the value of '$\lambda$' is extremely small. Hence, wavelength associated with macroscopic objects is extremely small and insignificant.
Note : If the material particle is moving with a velocity comparable to the velocity of light, then the mass of the particle changes according to the relation.
$m = \frac{m_0}{\sqrt{1 - v^2/c^2}}$
So de-Broglie wavelength is given by,
$\lambda= \frac{h}{mv} = \frac{h}{\frac{m_0}{\sqrt{1 - v^2/c^2}} v} = \frac{h\sqrt{1 - v^2/c^2}}{m_0 v}$
Derive an expression for the wavelength of a wave associated with an electron accelerated through a potential of 'V' volt.
Let us consider a beam of electrons travelling through a potential difference of $V$ volt. The beam of electrons acquires kinetic energy given by
$$ \frac{1}{2} mv^2 = eV $$
$$m^2 v^2 = 2meV $$
$$(mv)^2 = 2meV $$
$$ mv = \sqrt{2meV} \quad \cdots (i) $$
Using relation for de-Broglie wavelength, $\lambda = \frac{h}{mv}$ and substituting the value of eqn. $(i)$, in it we get,
$$ \lambda = \frac{h}{\sqrt{2meV}} $$
But $eV = E$
$$\therefore \lambda = \frac{h}{\sqrt{2mE}} \quad \cdots (ii) $$
Now, $h = 6.625 \times 10^{-34}J-s$
$m = 9.1 \times 10^{-31} \text{ kg}$ and $e = 1.6 \times 10^{-19} \text{ C}$
$$ \lambda = \frac{6.625 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times V}} \text{ m} $$
$$ \lambda = \frac{6.625 \times 10^{-34}}{\sqrt{29.12 \times 10^{-50} \times V}} \text{ m} $$
$$ \lambda = \frac{6.625 \times 10^{-34}}{5.396 \times 10^{-25} \sqrt{V}} \text{ m} $$
$$ \lambda = \frac{1.227 \times 10^{-9}}{\sqrt{V}} \text{ m} $$
$$ \lambda = \frac{12.27 \times 10^{-10}}{\sqrt{V}} \text{ m} $$
But, $10^{-10} \text{ m} = 1 \text{ Å}$
$$ \lambda = \frac{12.27 \times 10^{-10}}{\sqrt{V}} \times 10^{10} \text{ Å} $$
$$ \lambda = \frac{12.27}{\sqrt{V}} \text{ Å} = \frac{1.227}{\sqrt{V}} \text{ nm} \quad \cdots (iii) $$
The variation of de-Broglie wavelength ($\lambda$) associated with an electron with the variation of accelerating potential $V$ is shown in figure :
De-Broglie wavelength in terms of temperature :
According to kinetic theory, the average kinetic energy of a particle is given by,
$$ \frac{1}{2} mv^2 = \frac{3}{2} kT \quad \cdots (i) $$
where, $k = 1.38 \times 10^{-23} \text{ JK}^{-1}$ is the Boltzmann constant and $T$ is the absolute temperature of the particle.
From eqn. $(i)$ we get,
$mv^2 = 3kT$
On multiple both side with m
$m^2 v^2 = 3kT \times m$
$(mv)^2 = 3kT \times m$
$mv = \sqrt{3mkT}$
Since, de-Broglie wavelength is given by
$\lambda = \frac{h}{mv}$
$\therefore \lambda = \frac{h}{\sqrt{3mkT}}$.
Thus, $\lambda \propto \frac{1}{\sqrt{T}}$.