Quantum Physics Past Paper PDF

Part 2 (B) 1 Outline 2.5 Schrodinger Infinite Potential Energy Well (1D) 2.6 Schrodinger Equation: Potential energy step (1D) 2.7 Schrodinger Equation: Finite Potential Energy Well (1D) 2 ทบทวน Confining a particle: :...

Part 2 (B) 1 Outline 2.5 Schrodinger Infinite Potential Energy Well (1D) 2.6 Schrodinger Equation: Potential energy step (1D) 2.7 Schrodinger Equation: Finite Potential Energy Well (1D) 2 ทบทวน Confining a particle: : #! 𝐸! = 𝑛" $%&! Infinite Potential Energy Well 1) Quantization of energy is corrected for 1D direction. 2) The wave function in each is not clear. These are the allowed or quantized values of the energy of the3 electron at L=100 pm. 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) Assume: one particle is confined in the region between x = 0 and x = L with infinite walls and moves freely in this region and makes elastic collisions the walls. This phenomenon is sometimes called “a particle in a box.” 1D infinite potential well 4 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) The probability of detecting the particle (i) outside the well and (ii) at the boundaries, is zero (𝜓(𝑥) = 0) The Schrodinger Eq. inside the well, where U x = 0 is 𝑑!𝜓 𝑥 !𝜓 𝑥 , 2𝑚(𝐸 − 𝑈) = −𝑘 𝑘= ,U = 0 𝑑𝑥 ! ℏ ! (time independent 1D) The general solution is A and B are determined by boundary conditions. 5 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) Wave function 1. At x = 0 à 𝜓 𝑥 = 0 (𝒃) S𝒊𝒏𝒄𝒆 𝒔𝒊𝒏 𝟎 = 𝟎 𝒂𝒏𝒅 𝐜𝐨𝐬 𝟎 = 𝟏 (𝒂) Thus, to make Eq. (b) = 0, the value that is ‘0’ is 𝑩. 2. At x = L à 𝜓 𝐿 = 𝐴𝑠𝑖𝑛𝑘𝐿 + 𝐵𝑐𝑜𝑠𝑘𝐿 (𝒄) From above, 𝐵 = 0 𝝍 𝑳 = 𝑨𝒔𝒊𝒏𝒌𝑳 (𝑑) 𝐴𝑡 𝐿, 𝜓 𝐿 =0 (𝑒) To satify Eq. (e), if A = 0, applitude =0 è no particle anywhere, thus. 𝐬𝐢𝐧 𝒌𝑳 = 𝟎 (𝑓) Particle cannot exist at the boundaries. So, To satify Eq. (f), 𝒌𝑳 = 𝒏𝝅, 𝒏 = 𝟏, 𝟐, 𝟑, … the wave function is ”zero" at the boundaries. 𝑛𝜋 𝑘= (𝑔) !" 𝐿 B = 0 and 𝑘 = in Eq.(a) # 𝑛𝜋 𝜓 𝑥 = 𝐴 sin 𝑥 (ℎ) 𝐿 6 (Wave function of particle in the infinite well. (not complete)) 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) Energy If 𝑼 = 𝟎, 𝒂𝒏𝒅 p = ℏ𝑘, 𝑬= 𝑲 (2) From, 𝒌𝑳 = 𝒏𝝅, 𝒏 = 𝟏, 𝟐, 𝟑, … 𝑝! (ℏ𝑘)! ℏ! 𝒏𝝅 ! (See Part 2(A), slide 16) 𝒏𝝅 = = = ( ) 𝒌= (𝟏) 2𝑚 2𝑚 2𝑚 𝑳 𝑳 Y! X! h! 𝐸X = Zh! [ \! ! Y! 𝐸X = 𝑛 Z[\! excited states 𝑬𝟏 = 𝒆𝒏𝒆𝒓𝒈𝒚 𝒂𝒕 𝒈𝒓𝒐𝒖𝒏𝒅 𝒔𝒕𝒂𝒕𝒆 1st excited states ℎ& ground state 𝐸% = 8𝑚𝐿& 𝐸5 = 𝑛6 𝐸7 7 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) Calculate the energy difference ΔE between n=1 level and n=2 level of electron for a 1D infinite potential well of length L=1 nm. 43.90 𝑥10−68 ! Y! ∆𝐸 = 3 𝐸X = 𝑛 Z[\! 8(9.11×10−31)(1×10−18) ℎ& 43.90 𝑥10−19 𝐸% = =3 8𝑚𝐿& 72.88 𝐸5 = 𝑛6 𝐸7 = 3 n 0.602 𝑥10−19 ℎ ! = 1.807 𝑥 10"#$ J ∆𝐸 = (2! − 1! )( ! ) 8𝑚𝐿 (6.626×10−34)! = (3)( ! ) 8(9.11×10 )(1×10 ) −31 −9 8 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) For infinite potential well, A ? Home work From, Hint 1: 𝑛𝜋 𝜓 𝑥 = 𝐴 sin 𝑥 Determine ‘A’ using the normalization. 𝐿 𝟐 Prove that 𝑨 = 𝑳 𝜓(𝑥) & = 𝜓∗𝜓 Hint 2 (if need): 1 𝑠𝑖𝑛!𝑥 = (1 − cos 2𝑥) 2 1 $ 𝑐𝑜𝑥 2𝑥 𝑑𝑥 = sin 2𝑥 + 𝑐 2 9  (3) 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) energy & ,! 𝐸! = 𝑛 From, -.#! 𝑛𝜋 𝜓 𝑥 = 𝐴 sin 𝑥 𝐿 𝟐 𝑨= 𝑳 (2) & probability density 𝜓(𝑥, 𝑡) The complete wave function is 2 𝑛𝜋 𝜓 𝑥 = sin 𝑥, 𝐿 𝐿 𝑛 = 1,2,3, … (1) The wave functions (solid lines) and probability Several of the allowed energies Wave form for an electron confined to the densities (shaded regions) of the first four states in the one-dimensional infinite potential energy well. infinite well with width L =100 pm. 10 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) Example of finding probability density 2 𝑛𝜋 𝜓 𝑥 = sin 𝑥, 𝑛 = 1,2,3, … 𝐿 𝐿 At n =2, where the probability density is maximum ? & 𝜓(𝑥) & = 𝜓∗𝜓 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝜓(𝑥) & 2 𝑛𝜋 2 𝑛𝜋 𝜓(𝑥) =( sin 𝑥)( sin 𝑥) 𝐿 𝐿 𝐿 𝐿 11 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) At n =2, where the probability density is maximum ? From, 𝑠𝑖𝑛 𝜃 = 1 & 2 & 2𝜋𝑥 𝜓(𝑥) = 𝑠𝑖𝑛 ( ) 𝜋 𝐿 𝐿 𝜃=𝑚 , 𝑚 = 1, 3, 5, … 2 𝜓(𝑥) & Thus, 2𝜋𝑥 𝑠𝑖𝑛 =1 𝐿 𝜓(𝑥) 𝟐𝝅𝒙 𝝅 = 𝒎 , 𝒎 = 𝟏, 𝟑, 𝟓, … 𝑳 𝟐 At n =2, the maximum probability density is at (Firstly, think about how many loops should be for n =2) 𝐿 𝐿 3𝐿 2𝜋𝑥 𝜋 2𝜋𝑥 𝜋 0 𝐿 4 2 2 = (1) = (3) 𝐿 2 𝐿 2 Maximum means probability density is the positions where the sine function reaches its 𝐿 3𝐿 𝑥= Ans. 𝑥= Ans. maximum value = 1. 4 4 2𝜋𝑥 𝑠𝑖𝑛 =1 12 𝐿 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) For 1D infinite potential well of L=1 nm, at the energy From, 𝑠𝑖𝑛 𝜃 = 1 level n = 3. Find the position where the probability 𝜋 density is maximum. 𝜃=𝑚 , 𝑚 = 1, 3, 5, … 2 Thus, 2 𝑛𝜋 𝜓 𝑥 = sin 𝑥, 𝑛 = 1, 2, 3, … 3𝜋𝑥 𝐿 𝐿 𝑠𝑖𝑛 =1 𝐿 Sol. 𝟑𝝅𝒙 𝝅 = 𝒎 , 𝒎 = 𝟏, 𝟑, 𝟓, … 𝑳 𝟐 2 3𝜋𝑥 𝜓(𝑥) & & = 𝑠𝑖𝑛 ( ) (Firstly, think about how many loops should be for n =3) 𝐿 𝐿 3𝜋𝑥 𝜋 3𝜋𝑥 𝜋 3𝜋𝑥 𝜋 = (1) = (3) = (5) Maximum means probability density is the 𝐿 2 𝐿 2 𝐿 2 positions where the sine function reaches its maximum value, e.g. 1. 𝐿 𝐿 5𝐿 𝑥= 𝑥= 𝑥= 3𝜋𝑥 6 2 6 𝑠𝑖𝑛 =1 𝐿 1 1 5 L=1 nm, then 𝑥 = 𝑛𝑚 𝑥 = 𝑛𝑚 𝑥= 𝑛𝑚 6 2 6 13 Ans. Ans. Ans. 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) 1 1 5 L=1 nm, then 𝑥 = 𝑛𝑚 𝑥= 𝑛𝑚 𝑥= 𝑛𝑚 6 2 6 Ans. Ans. Ans. 𝐿 = 1 𝑛𝑚 14 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) 0& An electron trapped in an infinite potential well (1D) 2 & (2)𝜋𝑥 𝑃(𝑥% , 𝑥& ) = 𝑠𝑖𝑛 n dx with the length of 0.1 nm. 0% 𝐿 𝐿 Q: In the first excited state, what is the probability 2𝜋 𝑖𝑓 𝑎 = , 𝑡ℎ𝑒𝑛 𝐿 of finding the electron (P(x1, x2)) in the range between 𝑥 = 0 𝑛𝑚 𝑡𝑜 0.025 𝑛𝑚? 2 %! ! 𝑃(𝑥$, 𝑥!) = $ 𝑠𝑖𝑛 𝑎𝑥 A dx 𝐿 %$ 𝒙𝟐 1) Probability of detection 𝑷 𝒙𝟏 , 𝒙𝟐 = 𝝍(𝒙) 𝟐 𝒅𝒙 2 %! 1 − cos(2𝑎𝑥) = $ ( )dx 𝒙𝟏 𝐿 %$ 2 2) In the first excited state (n=2), wave function is 2 %! 1 %! cos 2𝑎𝑥 2 1 1 %! = [$ 𝑑𝑥 − $ dx] = ( 𝑥 − sin(2𝑎𝑥))F 𝐿 %$ 2 %$ 2 𝐿 2 2 A 2𝑎 %$ 2 𝑛𝜋 2 (2)𝜋 𝜓 𝑥 = sin 𝑥 𝜓 𝑥 = sin 𝑥 %! 𝐿 𝐿 𝐿 𝐿 2 1 1 2𝜋 2 1 𝐿 4𝜋 %! = ( 𝑥− sin(2( )𝑥))H = ( 𝑥− sin( 𝑥))F 𝐿 2 2𝜋 𝐿 𝐿 2 8𝜋 𝐿 2A2 𝐿 %$ 2 2𝜋𝑥 %$ 𝜓 𝑥 = sin 𝐿 𝐿 1 𝐿 4𝜋 &.&!( =( 𝑥− sin( 𝑥))F 𝐿 4𝜋 𝐿 & 15 2.5 Schrodinger Equation: Infinite Potential Energy Well (1D) 0 12.2&4 𝑥 1 4𝜋𝑥 ! 𝑃(𝑥% , 𝑥& ) = [ − n sin ]‚ 𝐿 4𝜋 𝐿 0 12 $ sin 0 = 0 %.%!' # ()(%.%!' ) % # ()(%) = [( − sin )−( − sin )] %.# () , , () , %.%!' # ()(%.%!' ) = [( − sin )] (b) In first excited state (n=2), what is the probability of %.# () %.# finding the electron in the region x = 0 nm to 0.025 nm. sin 𝜋 = 0 %.%!' # = − sin 𝜋 %.# () = 0.25 = 25% 𝐿 = 0.1 𝑛𝑚 𝑃(𝑥% , 𝑥& ) = 25% 𝑥=0 𝑥 = 0.025 𝑛𝑚 16 17 2.6 Schrodinger Equation for Potential energy step !! 𝐸 > 𝑈[ 𝐸 < 𝑈[ (A) (B) 18 2.6.1 Schrodinger Equation for Potential energy step, 𝐸 > 𝑈‡ Energy of electron (E) is greater than the height of the potential energy barrier (𝑼𝟎 ), the particle will experience both reflection and transmission. 𝐸 > 𝑈[ In this part, we use !! 1. Time-independent Schrodinger Eq. 2. 1D motion. 𝑑!𝜓 𝑥 !𝜓 𝑥 , 2𝑚(𝐸 − 𝑈) = −𝑘 𝑘= 𝑑𝑥 ! ℏ! Region I : the particle's energy is higher than the potential energy (𝑬 > 𝑼),where 𝑼 = 𝟎. Region II: the particle's energy is higher than the potential energy (𝑬 > 𝑼), 𝒘𝒉𝒆𝒓𝒆 𝑼 = 𝑼𝟎. 19 2.6.1 Schrodinger Equation for Potential energy step, 𝐸 > 𝑈‡ For Region 1: For Region 1: 𝑵𝒐𝒕𝒆 ∶ 𝑑!𝜓 𝑥 !𝜓 𝑥 , 2𝑚(𝐸 − 𝑈) = −𝑘 𝑘= 2𝑚(𝐸 − 0) 𝑑𝑥 ! ℏ! 𝑘# = ℏ! 2𝑚𝐸 𝑑 ! 𝜓# 𝑥 𝑘#! = ! + 𝑘#! 𝜓# 𝑥 = 0, 𝑘 = 𝑘# ℏ 𝑑𝑥 ! 2𝜋 𝜆# = 𝑘# 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑔𝑒𝑛𝑒𝑟𝑎𝑙): Region 1 Region 2 𝜓ˆ 𝑥 = A𝑒 ‰Š" ‹ + B𝑒 Œ‰Š" ‹ 20 2.6.1 Schrodinger Equation for Potential energy step, 𝐸 > 𝑈‡ For Region 2: For Region 2: 𝑵𝒐𝒕𝒆: 𝑑!𝜓 𝑥 !𝜓 𝑥 , 2𝑚(𝐸 − 𝑈) = −𝑘 𝑘= 2𝑚(𝐸 − 𝑈% ) 𝑑𝑥 ! ℏ! 𝑘! = ℏ! 𝑑 ! 𝜓! 𝑥 ! (𝑈 = 𝑈2 ) + 𝑘! 𝜓! 𝑥 = 0, 𝑘 = 𝑘! 𝑑𝑥 ! 2𝑚(𝐸 − 𝑈% ) 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑔𝑒𝑛𝑒𝑟𝑎𝑙): 0, 𝑁𝑜 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 𝑤𝑎𝑣𝑒. 𝑘!!= ℏ! Region 1 Region 2 𝜓! 𝑥 = 𝐶𝑒 ‰Š! ‹ + D𝑒 Œ‰Š! ‹ 2𝜋 𝜆! = 𝑘! 𝜓! 𝑥 = 𝐶𝑒 ‰Š! ‹ 21 2.6.2 Schrodinger Equation for Potential energy step, 𝐸 < 𝑈‡ Energy (E) of electron is less than the height of the potential energy barrier(U0). !! 𝐸 < 𝑈% In this part, we will use !! 1. Time-independent Schrodinger Eq. 2. One-dimensional motion. Region 1: The Schrödinger Eq. and the wave function behave the same as in the ‘region 1 of 𝐸 > 𝑈% ’. Note: In classical physics, If 𝑒 " arrives at the Region 2: The Schrödinger Eq. is different. step, it will stop at x = 0 , and then all will bounce back. Here, it is total reflection of the particles. 22 2.6.2 Schrodinger Equation for Potential energy step, 𝐸 < 𝑈‡ Region 1: The Schrödinger Eq. and the wave function behave the same as in the ‘region 1 of 𝐸 > 𝑈% ’. For Region 1: 𝑵𝒐𝒕𝒆 ∶ 𝑑!𝜓 𝑥 2𝑚(𝐸 − 𝑈) ! = −𝑘 ! 𝜓 𝑥 , 𝑘= 2𝑚(𝐸 − 0) 𝑑𝑥 ℏ! 𝑘# = ℏ! 2𝑚𝐸 𝑑 ! 𝜓# 𝑥 𝑘#! = ! + 𝑘 ! 𝜓# 𝑥 = 0, 𝑘 = 𝑘# ℏ # 𝑑𝑥 ! 2𝜋 Region 1 Region 2 𝜆# = 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑔𝑒𝑛𝑒𝑟𝑎𝑙): 𝑘# 𝜓ˆ 𝑥 = A𝑒 ‰Š" ‹ + B𝑒 Œ‰Š" ‹ 23 2.6.2 Schrodinger Equation for Potential energy step, 𝐸 < 𝑈‡ Region 2: The Schrödinger Eq. are different. For Region 2: (x ≥ 0) 𝑵𝒐𝒕𝒆 𝟏: 𝑑!𝜓 𝑥 2𝑚(𝐸 − 𝑈) ! = −𝑘 ! 𝜓 𝑥 , 𝑘= 2𝑚(𝐸 − 𝑈% ) 𝑑𝑥 ℏ! 𝑘! = ℏ! 𝑑 ! 𝜓! 𝑥 ! − 𝑘! 𝜓! 𝑥 = 0, 𝑘 ! = −𝑘!! 𝐸 < 𝑈2 𝑑𝑥 ! 𝐸 − 𝑈2 = − (𝑈2 − 𝐸) Region 1 Region 2 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑔𝑒𝑛𝑒𝑟𝑎𝑙): D = 0, as x increases the value is increased (it is impossible) 𝜓! 𝑥 = 𝐶𝑒 ŒŠ! ‹ + D𝑒 Š! ‹ 𝜓! 𝑥 = 𝐶𝑒 ŒŠ! ‹ 24 2.7 Schrodinger Equation: Finite Potential Energy Well (1D) (หน้ านีไ* ม่ ออกสอบ) Infinite potential well is an idealization of a technique for confining a particle, We should examine the solution when the barriers at the sides of the well are finite. well depth (𝑈5 ) The solution in the center region (between x = 0 and x = L) is exactly the same as it was for the infinite well. The region x < 0 and x>L shows that the energy E of the particle is less than the potential energy U0, 25 2.7 Schrodinger Equation: Finite Potential Energy Well (1D). (หน้ านีไ* ม่ ออกสอบ) !!" # 2)(+# − -) !!" # 2)(+# − -) = % "! " # , %′ = !!" # = −& ! " # , &= 2)(+ − ," ) = % "! " # , %′ = !# ! ℏ! !# ! ℏ! !# ! ℏ! !" !" 𝑦 𝑥 = 𝐶𝑒 ! !" + 𝐷𝑒 #! !" 𝑦 𝑥 = 𝐹𝑒 ! + 𝐺𝑒 #! )/ )/ 𝜓 𝑥 = 𝐶𝑒. 𝜓 𝑥 = 𝐺𝑒 ". There are now 4 coefficients to be determined (A, B, C, G) along with the energy E. 26 2.7 Schrodinger Equation: Finite Potential Energy Well (1D) Rather than attempting to solve this equation for the finite well, we simply state the results for particular numerical values of U0 and L. U0 =450 eV and width L = 100 pm. Only states n = 1, 2, 3, and 4 are allowed. If electron has an energy of 450 eV or greater, it is not trapped and its energy is not quantized. 27 Finite Potential Energy Well (1D) 2.7 Schrodinger Equation: Finite Potential Energy Well (1D) 28