Question

Part B. Now we will repeat the calculation for states of energy E> 1. Write down solutions for the wave function in each of the regions. (use coefficients use coefficient and B) and D ) V should be written in terms of the parameter should be written in terms of the parameter (2 [ 2 E- 2. Apply the boundary condition vax t eliminate one of the coefficients (Which one will be eliminated depends on how you wote Ps) 3. Apply the boundary condition on y atx=2 to eliminate one of the coefficients Cor 4. Apply the boundary condition on y at- to express your remaining coefficient in Vieither Cor D) in terms of your remaining coefficient in either or 5). 5. Your entire wave function in the region should now depend on only one coefficient (either or B). Explain how the normalization condition allows you to determine this remaining coefficient. Don't carry out the calculation, just show how it can be done. 6. There is still one boundary condition you haven't yet applied - the condition on did atx Explain how applying this boundary condition allows you to determine the energy E. Don't carry out the calculation, just show how it can be done.

question 5, please show all the work/ steps so i can break it down for better understanding studying please, thank you

Answer 1

I Page 1 a Lo Hw or -3 chrödinger worksheet Region 1 : u(x) = 0 Region 2 : u(a)=0 ocast Region 3 : 012-4. Region 4 : U(a)=0 LE RE2L ny2L AU12) UFO uso 1 U-U. MI 72U 20 to nel M=2L baut :B Here energy of incident particle (E) > θα į Region to Schrödingas equation is given by: -ho they are trt, =ET, Immass of the - incident particle Now, V=U (%) =0, in region i. 80, [t, zo aco, Here v denotes potential. Region 2% schrödinges equation is : The one da U61942 = €72 =) - th hon den har to = Eta 22o em? 1. dne 27 20 Y d24 dn 2t + 2m G 72 52 d24 d242 + x²42=0 where ,k=/ 2me 1, 2 дек* +ве-"- от L 02x 4) Jo or, He = A sinkat B Coska

2 loge-2 / becour tag Im region ② , Schrödinger equation becomes- cha dhata me - tum V₂ = €13. Om 2 de t2 lesz + 2m (E-V.) Y3 -0 =) d2 4 + 41 242=0 where, u's | 2 m (EU) Eltz = cekla + Derina LE2%2L O or Y = C Coska+ D linkly i Im sesion Schrot und=0, so, neare function will be W(O) = a>,2L

In the expression of Yz or 4. I wrote two alternatives. Actually toro are the Rome thing only difference is in contant term. lo don't worry. However we will consider here, ( Y,=0, aco Y₂ = Alinkt B colka ,obhEL Uz - ecosk's to fink'se, L35 <2L 1 44 -0, >22 2 Boundary condition at a=o gives, - Y (220) = 4₂(x=0) [B=0 80,1%, - Alin ka 022EL 3 Boundary condition at x=2L gives, 44(X=21) = 43 (=21) »o=ecosk' (2L) & Drink (24) A » e cos (2W/L) + Dsin (26'L)=0 - D=-c.cot (26'L) so, 43=ccolt'n-c. cot(24'L) Linklal ② Boundary condition at n= L gives, 42 (22=L) = 43 (1=L) Arinul = c cosk'L-e cot(26/L) in K'La

foret Again, even the last . 1 des aut [in a kinite I since potential - at n=17 => Akces KL = cek' link'e - (KC0(26'L) COSM'L – simplifying equention ® we get, AlinkLze cosa'l - c.cat(rk'L) linkil =C cosk'L - c. cos (2K'L) einkl lin (2K'L) achin (2K'L) cos (K'L)- cos (24'L) sin() sin (2«'L) e lin(K'L) = Rin (26'L) 2) AsinkL = e. alintaL) cos(K'L) 2) Asinkh-e sinal) A linkt - e pockets 80, Y; becomes Yz = C[ cos (kn) - cot (2KL) sin kla] = & 298ink L.) COS (KL) [casckim) – - cot(2K/L) Linkix)] 14 = 2A sin (4L) cos (KL) [cos (K's) _eot (24'L) 8P1 sin(kx) ? روم) Co I now normalisation condition To w t4 ds = 1 [In general for too wty ase=1 any l I leave function 47. Here the above condition simplifies to

- Page 5 Sykla da + ( a to da = 1 can find the from this . equation we constant Sest te don = { Ar Rinkka ae =A2 szinten de = A ( - cas 2kx) dhe = FATE - met 1} RL And, is estes da af 4 sinh L) or (KL) - ł ( cos(xls) - eat (2412) Sinthiast Similarly by calculating this you can du find out Ń very easily. i e