De Broglie equation

source :http://switkes.chemistry.ucsc.edu

According to De Broglie, a moving small particle like electron, proton, neutron, dust particle, a small ball etc. has the properties of a wave (Dual nature of matter).

The wave length of a moving particle may be calculated as

λ=h/mv =h/p

De Broglie equation

λ =wave length

h= planck’s constt.

m=mass of particle

v=velocity of particle

p=momentum of the particle

Dual nature of electrons – Moving electrons in an orbit have the properties of both particle & wave. The following facts confirm the dual nature of electron.

1. Cathode rays (beam of electrons) rotate a pin whell placed in their path. This property shows the particle nature of electron.
2. A beam of electrons (Cathode rays) show diffraction & interference phenomenon (wave nature). This confirms the wave nature of moving electrons.

Radiations consist of photons (Packets of energy). These photons possess the property of the wave & particle energy of photon,

According to Planck,

E= hv               (1)

According to Einstein’s equation,

E=mc2                 (2)

m= mass of photon

From eq. (1) & (2)

hv    =mc2

v   =c/ λ

hc/ λ =mc2

λ =h/mc

Solved Numerical Problems

Question 1.     An electron of mass 9.1×10–28 gm. is moving with velocity of 3×1010 cm/sec. Find its wave length?

Ans.    m=9.1×10–28 gm.                     h= 6.6×10–27 erg-Sec

v= 3×1010 cm/Sec

λ =h/mc

=6.6×10–27/9.1×10–28 ×3×1010 = 0.2417 ×10–9

λ =2.42×10–10cm  Ans.

Question 2.   Mass of Electron is 9.1×10–31kg. If its kinetic energy is 3×10–25 Joule, then calculate its wave length?

Ans.    De Broglie Equation ,

λ =h/mc

m=9.1×10–31 kg

E= 3×10–25 Joule

K.E. or E =mv2/2

putting the values of m & E,

v = √ 3×10–25 x 2 / 9.1×10–31

v = 0.812 x 103

h= 6.6×10–34 Joule-Sec

λ =h/mc

λ =6.6×10–34/ (9.1×10–31 x 0.812 x 103 )

= 0.893×10–6

wave length = 8.93×10–7 m