Question

Derive an expression for de Broglie wavelength of matter waves.

Answer 1

Hint: Here, we will proceed by equating Einstein's equation relating matter and energy with the planck's equation with the hypothesis that the energies in both the cases are equal.
Formulas Used: ${\text{E}} = {\text{m}}{{\text{c}}^2}$, \[{\text{E}} = {\text{h}}\nu \] and $\nu = \dfrac{{\text{v}}}{\lambda }$.

Complete Step-by-Step solution:
De Broglie derived his equation using well established theories through the following series of substitutions:
According to Einstein's famous equation relating matter and energy,
${\text{E}} = {\text{m}}{{\text{c}}^2}{\text{ }} \to {\text{(1)}}$ where E denotes energy, m denotes mass and c denotes the speed of light
Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation
\[{\text{E}} = {\text{h}}\nu {\text{ }} \to {\text{(2)}}\] where E denotes the energy, h denotes the planck's constant and \[\nu \] denotes the frequency
The value of the planck's constant is given by ${\text{h}} = 6.62607 \times {10^{ - 34}}$ Joule-sec
Because real particles do not travel at the speed of light, de Broglie used v for the velocity of the particle instead of the speed of light (c)
By replacing c with v in the equation (1), we have
${\text{E}} = {\text{m}}{{\text{v}}^2}{\text{ }} \to {\text{(3)}}$
Since de Broglie believed particles and waves have the same traits, he hypothesized that the two energies would be equal.
Clearly, the LHS of both equations (2) and (3) are the same (i.e., energy E). So, the RHS of both these equations can be equated and will be equal.
$
   \Rightarrow {\text{m}}{{\text{v}}^2} = {\text{h}}\nu \\
   \Rightarrow {\text{m}}{{\text{v}}^2} = {\text{h}}\nu \\
   \Rightarrow \nu = \dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{h}}}{\text{ }} \to {\text{(4)}} \\
 $
According to the relation between the wavelength, velocity and the frequency of any wave, we can write
$\nu = \dfrac{{\text{v}}}{\lambda }$
By substituting $\nu = \dfrac{{\text{v}}}{\lambda }$ in equation (4), we get
$
   \Rightarrow \dfrac{{\text{v}}}{\lambda } = \dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{h}}} \\
   \Rightarrow \dfrac{{\text{1}}}{\lambda } = \dfrac{{{\text{mv}}}}{{\text{h}}} \\
   \Rightarrow \lambda = \dfrac{{\text{h}}}{{{\text{mv}}}}{\text{ }} \to {\text{(5)}} \\
 $
As we know that the momentum p is the product of mass m and the velocity v
i.e., p = mv
By substituting p = mv in equation (5), we get
$ \Rightarrow \lambda = \dfrac{{\text{h}}}{{\text{p}}}$ where $\lambda $ denotes de Broglie wavelength of matter waves, h is planck's constant (${\text{h}} = 6.62607 \times {10^{ - 34}}$ Joule-sec) and p denotes the momentum of the matter wave.
The above equation represents the required expression for de Broglie wavelength of matter waves.

Note- De Broglie's equation states that a matter acts as waves such as light and radiation and can also act as waves and particles. The equation further describes the probability of diffracting a beam of electrons much like a beam of light. Very small particles of mass traveling at speeds less than that of light operate like a particle and a wave.