The Quantum Theory

WAVE PARTICLE DUALITY NATURE OF MATTER
States that: "Matter has particle as well as wave nature". Meaning that matter has dual properties or two properties i.e. particle nature and wave nature.
The wave particle duality nature of matter was put forward by De Broglie scientist.
De Broglie derived an expression which was applied to find the De Broglie’s wave length.
De Broglie’s wave length is expressed in terms of mass and momentum.

See below expression;
From Einstein equation;
E = mc² …………..……….(i)

From Planck’s equation
E = hc  ………….......…… (ii)
        λ
Comparing equation (i) and (ii) gives
mc²= hc

            λ

     λ=   h     .................*

           mc 

     λ=  h

            p

De Broglie’s wave length in terms of energy

From:

λ =  h   

       mc

multiply by 1/c both sides.

λ =   h 

c     mc²

Since; E = mc²

λ =   h 

c      E

Problem # 07

a) Alpha particles emitted from radium have energy of 4.4MeV. What is the de-Broglie’s wave 

length?

b) The mass of moving particles is 9.01 x 10-19g. What is the de-Broglie’s wave length?

c) The momentum of particles is 2.0 x 10-¹ºgm/s. What is the de-Broglie’s wave length?

Hint:

h = 6.63 x 10^-34 Js

c = 3.0 x 10^8m/s

Solution

a) Given.

E = 4.4MeV ≈ 7.04×10^-¹³

Recall; 

λ =  hc

       E 

   = 6.63×10-³⁴Js × 3.0×10^8ms-¹

                7.04×10-¹³J

 .: De Broglie's wavelength, λ= 2.83×10-¹³ m

b)

  Given

Mass, m = 9.01×10^-19 g.

              λ=?

Recall;

 λ  = hc

       mc²

 λ  = h

       mc 

 =            6.63×10-³⁴Js          

   9.01×10-19g × 3×10^8ms-¹

 .:De broglie's wavelength, λ=2.45×10-²⁴ m.

c)

Given

Momentum, p=2.0 × 10-¹ºgms-¹

                   λ= ?

Recall; λ=h

                 p

               = 6.63×10-³⁴Js

                  2.0×10-¹ºgms-¹

 .: De Broglie's wavelength, λ = 3.32×10-²⁴Jg-¹m-¹s-²

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