Various methods for solving the resulting wave equation are developed and the solutions found are then used to illustrate a number of important properties of waves. A wave has different properties like frequency, wavelength, period, speed etc. This section gives an idea about wavelength and the equation to find out the wavelength and problems related to that.

De Broglie wavelength concept arises from the dual nature of matter. De Broglie wave is also known as matter wave. The dual nature of matter indicates that, it shows both particle as well as wave nature.
The equation for De Broglie wavelength is given by,
where,
$\lambda$ is the de Broglie wavelength,
h is the Planck's constant,
p is the momentum which is given by, p = mc.
The equation for De Broglie wavelength is given by,
$\lambda$ = $\frac{h}{p}$
where,
$\lambda$ is the de Broglie wavelength,
h is the Planck's constant,
p is the momentum which is given by, p = mc.
Frequency and wavelength are two important parameters in the case of a wave. These two parameters are connected to each other in terms of light velocity. The expression is given as,
$\lambda$ is the wavelength,
c is the velocity of light,
f is the frequency of wave.
$\lambda$ = $\frac{c}{f}$
or
f = $\frac{c}{\lambda}$
where, or
f = $\frac{c}{\lambda}$
$\lambda$ is the wavelength,
c is the velocity of light,
f is the frequency of wave.
Energy and wavelength relation between waves are mentioned below:
E is the energy of a photon,
h is the Planck's constant,
c is the velocity of light,
$\lambda$ is the wavelength.
This equation is derived from Einstein's mass energy relation.
Problems related to wavelength of a wave are described here.E = $\frac{hc}{\lambda}$
where, E is the energy of a photon,
h is the Planck's constant,
c is the velocity of light,
$\lambda$ is the wavelength.
This equation is derived from Einstein's mass energy relation.
Solved Examples
Question 1: Calculate the wavelength of a wave if the frequency is 2500Hz?
Solution:
Solution:
Given that,
Frequency, f = 2500Hz = 2500 s^{1}
We know that, light velocity, c = 3 $\times $ $10^8$m/s
Wavelength can be calculated using the below equation,
$\lambda$ = $\frac{c}{f}$
$\lambda$ = $\frac{3 \times 10^8}{2500}$ = 12 $\times $ $10^4$m
Question 2: Find out the wavelength of a photon whose energy is 15J?
Solution:
Given that,
Energy E = 15J
We know that,
c = 3 $\times$ $10^8$ m/s, h = 6.626 $\times$ $10^{34}$Js
The energywavelength relation is,
E = $\frac{hc}{\lambda}$
So, $\lambda$ = $\frac{hc}{E}$
$\lambda$ = $\frac{6.626 \times 10^{34} \times 3 \times 10^8}{15}$ = 1.325 $\times$ $10^{26}$m
Solution:
Given that,
Energy E = 15J
We know that,
c = 3 $\times$ $10^8$ m/s, h = 6.626 $\times$ $10^{34}$Js
The energywavelength relation is,
E = $\frac{hc}{\lambda}$
So, $\lambda$ = $\frac{hc}{E}$
$\lambda$ = $\frac{6.626 \times 10^{34} \times 3 \times 10^8}{15}$ = 1.325 $\times$ $10^{26}$m