Dual Behaviour Of Matter (De-Broglie Equation)
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 10^{6} m/s. Calculate the de-Broglie wavelength.
Solution
Given:- 1.6 X 10^{6} 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 10^{6} )
= 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 10^{5} 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 10^{5} 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 10^{5} )
=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 A^{0}?
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 10^{6} m/s. Calculate the de-Broglie wavelength.
Solution
Given:- v=2.19 X 10^{6} 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 10^{6} )
= 0.3 X 10^{-9} m
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