<< Chapter < Page | Chapter >> Page > |
This requires {(√(2mE))/ ћ}W = 0 radians or
{(√(2mE))/ ћ}W = π radians or
{(√(2mE))/ ћ}W = n π radians where n= 0,1,2,3,4…….
Squaring both sides we get:
{(2mE)/ ћ ^{2} }W ^{2} = (n π) ^{2}
or E = n 2 h 2 /(8mW 2 ) 1.49
Eq.(1.49) implies that the permissible energy states of an electron in an infinite potential well are quantized.
Why electrons cannot occupy a continuum energy states as they do in free space? The answer is the following:
We had assumed at the beginning of the analysis that V(x) = 0. This implies that potential energy is zero and electron possesses only Kinetic Energy.
Therefore total energy E = p ^{2} /(2m) = n ^{2} h ^{2} /(8mW ^{2} )
Therefore p = (nh)/(2W) 1.50
From de Broglie postulate: λ = h/p = h/[(nh)/(2W)]
To satisfy the standing wave condition in bounded space which an infinite 1-D potential well is, following boundary condition must be satisfied
W = n λ/2 1.51
Eq.(1.51) is the necessary condition for Standing Wave pattern. This standing wave pattern requirement causes the quantization of energy states.
Here we digress briefly to the chapter of light to fully understand the behavior of matter wave.
Fig.(1.25) A plane wavefront light ray perpendicularly incident on an interface of two optical mediums.
Whenever light travels from one optical medium of refractive index n _{1} to the other optical medium of refractive index n _{2} , the incident wave Transverse Electromagnetic Wave (TEM) experiences partial reflection at the interface of the two media and partial transmission into the second medium.
Let us assume that medium 1 is absolute vacuum hence its refractive index = n _{1} = 1 and medium 2 is a solid medium of refractive index n _{2} = n. The mathematical form of the incident wave, reflected wave and the transmitted wave is given in Fig(1.25).
Wave vector in medium 1 is k _{1} = 2π/λ _{1} and wave vector in medium 2 is k _{2} = 2π/λ _{2} ;
And ν λ 1 = c, ν λ 2 = v; 1.52
Therefore c/v = λ _{1} / λ _{2} = n/1;
Therefore λ 1 = n. λ 2 1.53
We know that if the second medium is metal, the incident light is totally reflected and the reflected light experiences a phase change of 180°.
Also at the interface E _{incident} + E _{reflected} = 0;
But if the second medium is dielectric then we have partial reflection and partial transmission and at the interface we have: E _{incident} + E _{reflected} = E _{transmitted} ;
The incident wave is the forward wave:
E(z,t)= E xoincident Exp[j(k z1 z – ω.t)];
The reflected wave is the backward wave:
E(z,t)= E xoreflected Exp[j(k z1 z + ω.t)];
The transmitted wave is also moving in forward direction therefore it is:
E(z,t)= E xotransmitted Exp[j(k z2 z – ω.t)]; 1.54
Wave vectors have been defined in Eq.(1.52).and Eq.(1.53).
The incident forward and reflected backward wave interfere to form Standing Wave as they do on a mismatched transmission line. If a transmission line is not terminated in Characteristic Impedance then partial reflection takes place at the load and a partial standing wave pattern is formed on the transmission line. Standing Wave implies there is no transmission of energy. In case of metal there is total reflection. Hence we have 100% standing wave in medium 1 and there is no penetration of light into the metallic medium 2. Hence no transmission of light energy. For dielectric medium 2 , we have partial reflection hence only partial standing wave.
Notification Switch
Would you like to follow the 'Solid state physics and devices-the harbinger of third wave of civilization' conversation and receive update notifications?