Physics 220. Exam #2. May 20, 2016

Transcription

1 Physics 22 Exam #2 May 2, 26 Name Please rea an follow these instructions carefully: Rea all prolems carefully efore attempting to solve them. Your work must e legile, an the organization clear. You must show all work, incluing correct vector notation. You will not receive full creit for correct answers without aequate explanations. You will not receive full creit if incorrect work or explanations are mixe in with correct work. So erase or cross out anything you on t want grae. Make explanations complete ut rief. Do not write a lot of prose. Inclue iagrams. Show what goes into a calculation, not just the final numer. For example! p m v " = 5kg ( 2 m s ) = kg m s Give stanar SI units with your results unless specifically aske for a certain unit. Unless specifically aske to erive a result, you may start with the formulas given on the formula sheet incluing equations corresponing to the funamental concepts. Go for partial creit. If you cannot o some portion of a prolem, invent a symol an/or value for the quantity you can t calculate (explain that you are oing this), an use it to o the rest of the prolem. Each free-response part is worth 2 points Prolem # /6 Prolem #2 /6 Total /2 I affirm that I have carrie out my acaemic eneavors with full acaemic honesty.

2 . Consier the potential V(x) =!2 a 2 m sech2 (ax), where a is a positive constant an sech(θ) is the hyperolic secant function. The potential is graphe on the right. -Sech( ) a. Show that the groun state wave function can e written as ψ = Asech(ax). Determine the energy of the groun state E an the normalization constant A. Hints: You may nee tanh 2 θ + sech 2 θ =, coshθ = sinhθ, sinhθ = coshθ θ θ, an sech 2 (ax)x = 2 a. The Schröinger equation is!2 2m ψ x = A sech ( ax ) x 2 ψ x 2 = A ( ( x cosh ax) ) = Aacosh 2 ax + Vψ = Eψ. We nee to etermine 2 ψ x 2. sinh ax = a tanh ax ψ 2 ψ x = ψ 2 x x = ( a tanh ( ax x )ψ ) = a ψ tanh ( ax ) x 2 ψ x = a { a ( ( 2 tanh2 ax) ) a tanh 2 ( ax) }ψ = ( 2 tanh 2 (ax) )a 2 ψ Now the Schröinger wave equation says!2 2 ψ + Vψ = Eψ 2m x 2!2 a 2 2m ( 2 tanh2 (ax) )ψ!2 a 2 ( ) +!2 a 2 2!2 a 2 2m sech2 ax!2 a 2 m +!2 a 2 2m = E E =!2 a 2 2m ( m sech2 ax)ψ = Eψ 2m!2 a 2 ( m sech2 ax) = E + tanh ax x ψ An we see that ψ is a solution to the Schröinger wave equation with energy E =!2 a 2 2m.

3 . Suppose that you have the wave function given as ik a tanh(ax) ψ k β eikx where k = 2mE an β = ik + a. Does this wave! 2 function correspon to a state of efinite energy? If so, what is the energy? If not, explain why it oes not. The Schröinger equation is Ĥ ψ =!2 2 2m x + V 2 ψ = E ψ. We nee to etermine 2 ψ x 2. ψ k = A ik a tanh(ax) β eikx = ika e ikx + ia β k eikx tanhax x ψ k = ika ike ikx ae ikx tanhax + ia2 β k eikx sech 2 ax where we evaluate ( x tanhax) = sinh ( ax ( x )cosh ax) x tanhax = asech2 ax The secon erivative: 2 x 2 ψ k = ika β 2 x 2 ψ k = x x ψ k k 2 e ikx + ia2 k = ika β eikx = a ( sinh 2 ( ax)cosh 2 ( ax) ) = ika β x ikeikx ae ikx tanhax + ia2 k eikx sech 2 ax ( ike ikx sech 2 ax) + ia2 k eikx k 2 2a 2 sech 2 ax 2ia k = a tanh2 ax ( 2a tanhaxsech 2 ax) ikae ikx tanhax a 2 e ikx sech 2 ax tanhaxsech2 ax ika tanhax Now the wave equation looks like, factoring out all of the common terms:!2 ika βm eika =!2 ik 2 A 2βm eika k 2 2 a2 sech 2 ax ia k tanhaxsech2 ax ika 2 tanhax + a2 sech 2 ax + ia k tanhaxsech2 ax [ k + ia tanhax] =!2 k 2 ik a tanhax A 2m β Therefore, Ĥ ψ k =!2 k 2 energy!2 k 2 2m. eika 2m ψ k = E ψ k an is a state of efinite energy with

4 The graph elow shows the hyperolic tangent function as a function of theta. Using the graph, show that for large values of negative x that ik a tanh(ax) ψ k ik + a e ikx Ae ikx an also etermine the asymptotic form for ψ k (x)for large positive x. From your results, etermine the transmission an reflection coefficients an comment on the result. Tanh( ) ik a tanh(ax) For large negative or positive x, we have ψ k ik + a e ikx. From the graph, we see that as x, the hyperolic tangent function goes to ik a tanh(ax) negative one. Thus, ψ k ik + a e ikx ~ A ik + a ik + a e ikx ~ Ae ikx. This represents a wave coming into the arrier from the left with no accompanying reflecte wave. Further, from the graph, we see that as x +, the hyperolic tangent function ik a tanh(ax) goes to positive one. Thus, ψ k ik + a e ikx ~ A ik a ik + a e ikx. This represents a transmitte wave on the right of the arrier. The transmission coefficient: T = Amplitue of outgoing wave Amplitue of incoming wave 2 = The reflection coefficient: R = T =. A * ik a ik + a e ikx A ik a ik + a e ikx A * A =. This particular potential prouces no reflections (a reflectionless potential) an every particle regarless of its energy passes.

5 2. Suppose that an electron in a hyrogen atom is in the state ψ 2. a. What is the proaility of fining the electron insie of the nucleus of hyrogen of raius? Express your answer in terms of a an. Hints:. The raial portion of the wave function is R 2 (r) = 2. x n e x x = ex xn nxn + n( n )xn ( )n n! n 24a 5 re r 2 a.. The Taylor expansion of the exponential function is e x = x + x2 2! x! Assume that << a. We must first etermine ψ 2 = R 2 Y. m Y l ( 2l +) ( l m l ( θ,φ) l )! = ε 4 ( l + )! eimφ m l ( cosθ ); ε = ( 2( ) +) ( )! Y ( θ,φ) = ( ) 4 ( + )! eiφ P ( cosθ ) = where, P ( cosθ ) = ( cos 2 θ ) ml ( cosθ ) = cos 2 θ cosθ 2 2 cosθ cosθ P ( cosθ ) an where, ( cosθ ) = l 2 l l! cosθ cos 2 θ P ( cosθ ) = 2! cosθ cos2 θ Thus, P 2 ( cosθ ) = cos 2 θ Therefore, Y ( θ,φ) = l = cos(θ) = sinθ cosθ P ( cosθ ) 8 eiφ P ( cosθ ) = Now we can form the state ψ 2. ( ) 8 eiφ P ( cosθ ) cosθ cosθ 8 eiφ sinθ. = sinθ.

6 ψ 2 = R 2 (r)y (θ,φ) = 24a 5 re r 2 a r ψ 2 = 24 8a 5 re 2 a eiφ ψ 2 = r r a 8a 2 e 2 a sinθeiφ The proaility is given in the usual way: ψ 2 = r r a 8a 2 e 2 a sinθe iφ P = ψ 2 ψ 2 = ψ * ψ r = 2 φ = 2 sin θ θ = cos[θ] 4 64a 5 + cos[θ] 2 = cos cos cos eiφ sinθ φ sin θ θ r 4 e r a r + cos 2 = = = r 4 e a r r = ae a r r 4 + 4ar +2a 2 r a r + 24a 4 = 2 24 = 4 = e a a 4 + 4a 2 +2a a a 5 e 24a 5 4 = 24a 5 a 5 a a a 2 a a 5 a 5 a + 24 e a P = a 5 24a 5 a 5 a + 24 P = 2 a + 2 P = 2 a 2 e a e a = a + a

7 . What is the expectation value of r for the electron in the state ψ 2? Express your answer in terms of a. Hint: You ll nee a hint from part a. r = 2 ( ψ * rψ )r 2 r sinθ θ ϕ = r = 64a 5 r 5 e r a r sin θ θ 4 r = 64a 5 2a6 2 r = 5a [ ] The raial integral evaluates to: 2 φ 2 r a 64a 5 e sin 2 θ r 5 r sinθ θ φ r 5 e r a r = ae a r r 5 5ar 4 + 2a 2 r 6a r 2 +2a 4 2a 5 r 5 e r a r = 2a 6 = 2a 6 an the theta integral was evaluate in part a. c. Determine L z ψ 2 an the associate eigenvalue. L z ψ 2 = i! φ r a 8a 2 re 2a sinθe iφ =! ψ 2 an the eigenvalue is!.

8 Physics 22 Equations Useful Integrals: x n x = xn+ n + sin x x = cos x cos x x = sin x cos 2 ( qx)x sin 2 ( qx)x cos ( qx)x sin ( qx)x a 2 x cos 2 qx x = a 2 a 2 xsin 2 qx x = a 2 a 2 = x 2 + sin[2qx] 4q = x 2 sin[2qx] 4q sin( qx)cos(qx) x = a 2 e ax x = eax a e ax2 x = a xe ax2 x = x 2 e ax2 x = = sin[qx] + sin[qx] 4q 2q = cos[qx] + cos[qx] 4q 2q 2 2a 2hc 2 λ 5 hc λ = 97 W m λkt e 2 7nm 4nm Constants: g = 9.8 m s 2 G = 6.67 Nm 2 c = 8 m s σ = k B =.8 2 J K kg 2 ev =.6 9 J e =.6 9 C h = Js; m e = 9. kg =.5 MeV c 2 m p = kg = 98 MeV c 2 m n = kg = 99 MeV c 2 m E = 6 24 kg R E = m a =.5 m Formulas : c = υλ E = hυ = hc λ S λ = 2hc2 λ 5 hc e λkt S υ = 2hυ c 2 hυ e S λ = 2ckT λ 4 kt λ max = 2.9 m K T S = σt 4 ev stop = hf φ λ ' = λ + h ( cosφ ) mc! = h 2 ; k = 2 λ ; ω = 2 f!2 2m 2 ψ + Vψ = i! ψ t Ê = i! t ˆp = i! x T ˆ =!2 2 2m x 2 Ĥ =!2 2 2m x + V 2 ˆx = x O = ψ * Ôψ r P = ψ * ψ x 2! 2 E n = n 2 2ma 2 = Eψ Ψ n (x,t) = 2 a sin ( k nx)e i E n! t

9 H n (q) = Y l ( θ,φ) = ε n e q2 q n e q2 ; q = ( 2l +) ( l )! 4 ( l + )! eimφ ( cosθ ) = ( cos 2 θ ) ml ( cosθ ) = 2 l l! cosθ 2l+ 2r L n l na L n+2l cosθ 2 l mω 2! x ( cos 2 θ ) l ( cosθ ); ε = cosθ = ( ) 2l+ na ( 2 ) 2l+ ( r ) 2l+ 2r L n+2l ( na ) 2r ( na ) = e 2 na r na ( 2 ) n+2l ( r ) n+2l e 2 na r 2r n+2l na ψ nlml = 2 na! 2 r 2n ( n + l)! e na n l a a ± = ip + mω x 2m!ω H = ( a ± a ± 2 )!ω L ± = L x ± il y 2r ( na ) l 2l+ 2r L n l na L 2 =! 2 sinθ θ sinθ 2 θ + sin 2 θ φ 2 L z = i! φ P = ψ * ψ r = 2 ψ * ψ r 2 r sinθ θφ Y l θ,φ ( )