According to this, matter like radiation posses dual behavior i.e. Wave nature as well as particle nature.
This means that photons, electrons possess both momentum as well as wavelength .
De Broglie, gave the following relation between wavelength ($\lambda$) and momentum (p) of a material particle.
$ \lambda =\frac {h}{mv}= \frac {h}{p}$
Here λ is wavelength , p is the momentum Important Note
de Broglie's prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction, a phenomenon characteristic of waves.
Every object in motion has a wave character. The wavelengths associated with ordinary objects are so short (because of their large masses) that their wave properties cannot be detected.
De-Broglie Equation Derivation
For a photon of light of frequency , We have from Einstein's Mass energy relation
$E=pc$
Where
E = energy
p = momentum
c = speed of light
Now from Planck's equation for wave nature of light
$E= h \nu = \frac {hc}{\lambda}$
Where
h= Planck constant
E=energy
c=speed of light
Now as per de-Broglie,these energy should be equal
$\frac {hc}{\lambda}= pc$
$\lambda = \frac {h}{p}$
De,Broglie suggested the above equation is a general one that applies to material particles as well as to photons.Now the momentum of the particle of mass m and velocity v is p=mv,so its de-broglie wavelength
$ \lambda =\frac {h}{mv}$
Solved Examples
Question 1
If velocity of electron in this microscope is 1.6 X 106 m/s. Calculate the de-Broglie wavelength. Solution
Given:- 1.6 X 106 m/s.
m = 9.1 X 10-31 kg
h = 6.6 X 10-34 J/s
λ=?
$ \lambda =\frac {h}{mv}= \frac {h}{p}$
=6.6 X 10-34 /( 9.1 X 10-31 X 1.6 X 106 )
= 0.4 X 10-9 m Question 2
The velocity associated with proton moving in a potential difference of 1000 V is 4.37 X 105 m/s. If the hockey ball of mass 0.1 kg is moving with this velocity; calculate λ associated with this velocity. Solution
m = 0.1kg = 100 gm
v = 4.37 X 105 m/s
h = 6.6 X 10-34 J/s
λ=?
$ \lambda =\frac {h}{mv}= \frac {h}{p}$
=6.6 X 10-34 /(.1 X 4.37 X 105 )
=15.10 X10-39 m
Question 3
What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m/s ? Solution
According to de Brogile equation
$ \lambda =\frac {h}{mv}= \frac {h}{p}$
$\lambda = \frac {6.626 \times 10^{-34}}{.1 \times 10}$
$= 6.626 \times 10^{-34} m ($J = kg m^2 s^{-2}$)
Question 4
Calculate the mass of a photon with wavelength 3.6 A0? Solution
$\lambda = 3.6 A^0 = 3.6 \times 10^{-10}$ m
Now According to de Brogile equation
$ \lambda =\frac {h}{mc}$
or
$m= \frac {h}{\lambda c}$
Substituting the values
$m=6.135 \times 10^{-29}$ kg
Relation between Circumference of Bohr orbit for Hydrogen atom and be de- Broglie Wavelength
According to Bohr theory
$(L) = n \frac {h}{2 \pi}$
where n = no. of shell
This can be written as
$mvr = n \frac {h}{2 \pi}$
$2 \pi r= n \frac {h}{mv}$
According to De- Broglie relation
$ \lambda =\frac {h}{mv}= \frac {h}{p}$
So, $2 \pi r = n \lambda$
Solved Examples
Question 1
If velocity of electron in this microscope is 2.19 X 106 m/s. Calculate the de-Broglie wavelength. Solution
Given:- v=2.19 X 106 m/s.
m = 9.1 X 10-31 kg
h = 6.6 X 10-34 J/s
λ=?
$ \lambda =\frac {h}{mv}= \frac {h}{p}$
=6.6 X 10-34 /( 9.1 X 10-31 X 2.19 X 106 )
= 0.3 X 10-9 m
