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  Chemistry 101Particle-in-a-Box (PIB)...The 2nd and 3rd Dimensions... Introduction - Results from 1-D In lecture, we solved the Schr¨odinger equation for the wave function ψ n ( x ) =   2 L  sin  nπxL   and for the energy  E  n  =  n 2 h 2 8 mL 2  for a ”bead-on-a-string”, i.e.,for a particle in a 1-D box. In the expressions for  ψ n ( x ) and for  E  n , n is thequantization parameter, i.e., the quantum number n = 1, 2, 3, ... We will see thatthe 3-D ”box” wave function is simply a multiplication of three 1-D wave functionssuch that each wave function ”piece” has a unique quantum number - denoted n x ,n y , or  n z  accordingly - and a unique length - similarly denoted  L x ,L y , or  L z accordingly. The energy also will have three contributions - each contribution of the same mathematical form as the 1-D energy formula - except that eachcontribution will have its unique n ( n x ,n y ,n z ) and L ( L x ,L y ,L z ). First we will setup the results for the 3-D case and then merely list the results for the 2-D case,hoping that you will  agree with the results  . Here goes. Extending the 1-D Results to a 3-D ”box” and to a 2-D ”corral” If one were to work out the details for the PIB in 3-D, it could be shown that eachdimension, x, y, z, could be solved independently and that the resultant quantizedenergy would simply be the sum of the separately-obtained quantized energies foreach dimension. In addition, the total wave function would simply be the productof the separately-obtained wave function for each dimension. [ For those interested,this is a property of the form of the differential equation, i.e., the sum of the derivatives - relating to the product rule from calculus, etc. ] Anyhow, here are theresults - which hopefully make sense to you. Putting it all together for a particle ina 3-D rectangular box, we have the following. First, the 3-D wave function: ψ n x ,n y ,n z ( x,y,z  ) =    2 L x    2 L y    2 L z sin  n x πxL x  sin  n y πyL y  sin  n z πz L z  .  (1) This can be simplified as follows: ψ n x ,n y ,n z ( x,y,z  ) =    8 L x L y L z sin  n x πxL x  sin  n y πyL y  sin  n z πz L z  .  (2)Since the volume of the rectangular solid box is merely the product of the threelengths, we have:1  V   =  volume   =  L x L y L z ,  (3)giving: ψ n x ,n y ,n z ( x,y,z  ) =   8 V  sin  n x πxL x  sin  n y πyL y  sin  n z πz L z  .  (4)Next, the energies. Analogously, the total energy for a 3-D box is merely the sumof the separate results for each dimension (x, y, and z), i.e., the sum of three 1-Denergies. Similar to the situation with the 3-D wave function, we now have threequantum numbers,  n x ,n y , and  n z , the quantum numbers for the x, y, and zdimensions, respectively. In addition, as with the wave function, the lengthconstraints in the x, y, and z directions are: 0  ≤  x  ≤  L x , 0  ≤  y  ≤  L y , 0  ≤  z   ≤  L z .Otherwise, each term in the three-term sum has the same form as the 1-D energyexpression derived in lecture. The result is shown below. E  n x ,n y ,n z =  E  n x +  E  n y +  E  n z (5) E  n x ,n y ,n z =  n 2 x h 2 8 mL 2 x + n 2 y h 2 8 mL 2 y +  n 2 z h 2 8 mL 2 z (6) E  n x ,n y ,n z =  n 2 x L 2 x + n 2 y L 2 y +  n 2 z L 2 z   h 2 8 m  (7)In the case of a cubical box, we have  L x  =  L y  =  L z  ≡  L . The result is: E  n x ,n y ,n z =  n 2 x  +  n 2 y  +  n 2 z   h 2 8 mL 2  ...Cubical Box   (8)Remember what was said in lecture about energy degeneracy, especially in the caseof the cubical box. Distinct combinations of ( n x ,n y ,n z ) give the same quantizedenergy. Each distinct ( n x ,n y ,n z ) has its own slot on the energy-level diagram.Thus, if more than one ( n x ,n y ,n z ) has the same energy, the slots will all be at thesame level on the energy-level diagram. For example, the following quantummechanical states will all have the same energy in a cubical box:( n x ,n y ,n z )  ⇒  (2 , 1 , 1);(1 , 2 , 1);(1 , 1 , 2) will all have the same energy, namely: E  2 , 1 , 1  =  { 2 2 + 1 2 + 1 2 }  h 2 8 mL 2 =  E  1 , 2 , 1  =  { 1 2 + 2 2 + 1 2 }  h 2 8 mL 2 =  E  1 , 1 , 2  =  { 1 2 + 1 2 + 2 2 }  h 2 8 mL 2 ⇒  E   =  6 h 2 8 mL 2 .2  For all three (3) states the energy is the  same  , i.e., the energy level is said to be three-fold degenerate. Some other examples for a  cubical   box  are shown below. Consider thepossible quantum mechanical states that can be constructed from one of the nvalues being assigned ”1”, another ”2”, and the third ”3”. The possibilities are:( n x ,n y ,n z )  ⇒  (1 , 2 , 3);(1 , 3 , 2);(2 , 1 , 3);(2 , 3 , 1);(3 , 1 , 2) : (3 , 2 , 1) . They will all have the same energy, namely: E  1 , 2 , 3  =  { 1 2 + 2 2 + 3 2 }  h 2 8 mL 2 E  1 , 3 , 2  =  { 1 2 + 3 2 + 2 2 }  h 2 8 mL 2 E  2 , 1 , 3  =  { 2 2 + 1 2 + 3 2 }  h 2 8 mL 2 E  2 , 3 , 1  =  { 2 2 + 3 2 + 1 2 }  h 2 8 mL 2 E  3 , 1 , 2  =  { 3 2 + 1 2 + 2 2 }  h 2 8 mL 2 E  3 , 2 , 1  =  { 3 2 + 2 2 + 1 2 }  h 2 8 mL 2 ⇒  E   =  14 h 2 8 mL 2 .For all six (6) states the energy is the  same  , i.e., the energy level is said to be six-fold degenerate. Similarly: if: ( n x ,n y ,n z )  ⇒  (1 , 1 , 1) or (2 , 2 , 2), each of these are  non-degenerate  since thereis only one state possible for each energy. The results are: E  1 , 1 , 1  =  { 1 2 + 1 2 + 1 2 }  h 2 8 mL 2 ⇒  E   =  3 h 2 8 mL 2 .Thus, there is only  one  state with this energy  ⇒  the (1 , 1 , 1) state is non-degenerate. E  2 , 2 , 2  =  { 2 2 + 2 2 + 2 2 }  h 2 8 mL 2 ⇒  E   =  12 h 2 8 mL 2 .Thus, there is only  one  state with this energy  ⇒  the (2 , 2 , 2) state is also non-degenerate. 2-D Case - Particle in a  ”Corral” If one were to work out the details for the PIB in 2-D, it could be shown that eachof the two dimensions, x and y, say, could be solved independently and the resultantquantized energy is simply the sum of the separately-obtained quantized energiesfor each dimension. Once again, the total wave function is simply the product of the separately-obtained wave function for each dimension. Here are the results -which now hopefully make sense to you. Putting it all together for a particle in a2-D rectangular corral, we have the following. First, the 2-D wave function:3  ψ n x ,n y ( x,y ) =    2 L x    2 L y sin  n x πxL x  sin  n y πyL y  .  (9) This can be simplified as follows: ψ n x ,n y ( x,y ) =    4 L x L y sin  n x πxL x  sin  n y πyL y  .  (10)Since the area of the rectangular corral is merely the product of the two lengths, wehave: A  =  area   =  L x L y ,  (11)giving: ψ n x ,n y ( x,y ) =   4 A sin  n x πxL x  sin  n y πyL y  .  (12)Next, the energies. Analogously, the total energy for a 2-D corral is merely the sumof the results for each dimension (x and y), i.e., the sum of two 1-D energies.Similar to the situation with the 2-D wave function, we now have two quantumnumbers,  n x  and  n y , the quantum numbers for the x and y dimensions,respectively. In addition, as with the wave function, the length constraints in the xand y directions are: 0  ≤  x  ≤  L x , 0  ≤  y  ≤  L y . As expected, each term in thetwo-term sum has the same form as the 1-D energy expression derived in lecture.The result is shown below. E  n x ,n y =  E  n x +  E  n y ,  (13) E  n x ,n y =  n 2 x h 2 8 mL 2 x + n 2 y h 2 8 mL 2 y ,  (14) E  n x ,n y =  n 2 x L 2 x + n 2 y L 2 y   h 2 8 m .  (15)In the case of a square corral, we have  L x  =  L y  ≡  L . The result is: E  n x ,n y =  n 2 x  +  n 2 y   h 2 8 mL 2  .  (16)4

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