probability density (1/nm) x (nm)

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1 14 Spring 99 Problem Set 1 Solutions Physics pril Handout Beginning pril 6: Serway Wee 4.5 particle of mass m is conned to a region of x-axis extending x to x. potential V in this region. properly from solution to time-independent Schrodinger equation for normalized has form particle r 1 cos nx; which n (n1) and n 1;;3;::: in We consider system at t so that we can ignore timedependent a) part of wavefunction e i Ent principle of superposition any linear combination of wavefunctions is a solution to Schrodinger equation. Consider quantum states of also smaller than n so that n (and n (n1) Construct superposition of all wavefunctions with running variable. to by doing integral: from 1 total(x) Z n (x) 1 d p Z Z cos sin x (x) d x sin x total(x) p 1 x Square resulting wavefunction to get probability density and b) with equation for intensity in a single slit diraction compare sin I ; a () sin max I is equivalent of in terms of x? You now that rst What minimum occurs at. Find x min value of x this diraction specic case of 1 nm and 8 plot probability For density as a function of x. What is numerical value of x min for SOUTION: probability density is tot(x) sin x P 1 total(x)j j. probability density is plotted below: x 1. Reading ssignment Reading ssignment: Wee Beginning pril 19: Serway so total wavefunction is. Particle in a Box and Uncertainty Principle pattern n(x). Since wavefunctions obey corresponds to. very large n for which change in n from one state to next is much ) is essentially a continuous this case? d cos (x) equivalent ofis (x). Setting this equal to gives x min SOUTION: We have 1

2 this case x min 8 4 :5 nm. In From plot in part (b) and from similarity in form of c) you can see that probability density P total has same equation as a diraction pattern with rst minimum at x min.thus form is localized to within x x min. Show that this is consistent particle SOUTION: result from part (b) we have Using But x : p h so p h : Setch wavefunctions for three lowest standing wave states a) energies E 1 ;E ;E 3. with probability density (1/nm).1 Particle in a Box 3. an electron of mass m trapped in following potential: Consider x (nm) with Heisenberg Uncertainty Principle. SOUTION: n x h p ; which is consistent with uncertainty relation xp h:

3 For rst excited state of energy E where is it most probable to b) electron? nd n integral number of half-wavelengths must t in box in SOUTION: to satisfy boundary conditions. Thus n n so n4n order x. solutions are n n(x) Bcos nx for n odd and at sin nx for n even. We are concerned in this section with n(x) solutions for n even. What is? It comes from normalization condition: nx sin p 1. probability of nding particle between and so integral of probability density over that range: is 1 n sin n nx sin 1 : If electron is in its ground state (with energy E 1 ) what is e) that it will be found in interval between x and x x probability Z x j n(x)j 1 x cos Z x x cos ' x 1 Z j n(x)j 1; 1 so Z j Z n(x)j 1; Z j Z n(x)j for n even. where x <<? SOUTION: It is most probable to nd electron at. form of wavefunction for n1 is n(x) Bcos x SOUTION: B is given from normalization condition as where How will energy of lowest energy state change if width of c) box is reduced by a factor of? Z j B Z n(x)j B 1; probability that electron will be found in SOUTION: energy varies lie 1 so it will increase a factor of 4 so B 1 p. interval between x and x xis What is probability that you will nd electron between d) + when electron is in energy state E n and n is even? and for x <<. and n n n(). In addition wavefunction must go to zero 4. Serway Chapter 41 pg 147 Problem 8 3

4 r x sin j 1 j Z 3 sin x x x 3 sin x 1 sin p 3 : (b) Note j 1(x) j is symmetric about x by denition P ( and P ( 3 x 3 )1P( x 3 )1( :196) :68 Innite Well with rounded egdes 5. particle of mass m moves in a potential well of width (from m x 1 x> >: addition particle is in a stationary state described by wave In function x >< x 1 x >: a) Determine energy of particle in terms of and m. m (E U) m 4 m (E U) E + m x x + Em + x b) Determine numerical value of. normalization condition is given by SOUTION: 1 Z 1 Z 1 j (x)j x 5 4 SOUTION: wavefunction function is given 8 1 (x) (a) probability is determined from integral <x< Z3 x P ) ( 3 Z 3 Time-independent Schrodinger equation is given by SOUTION: d 1 1 cos for given (x) wehave Hence d 3 x for E yields E Solving. m x ) 1. Hence P ( x 3 )P( 3 x) and refore Hence (1 x ) Z x x4 (1 + 4 ) x to x ) and in this well potential is given by x x3 + x x< >< x U(x) x : 4

5 r 15 Thus. 16 c) Determine most probable location(s) of particle. SOUTION: probability density P (x) j (x)j (x) dp x 1 (1 x ). happens at x and x. But P () so most probable This for particle is x. location By substitution show that for any non-zero values of! and a) wave function (x; t) sin (x!t) does not satisfy time- dependent Schrodinger equation with V (x; t) : SOUTION: time-dependent Schrodinger equation with V (x; t) is sines and cosines are linearly independent functions and since purely Since numbers cannot equal purely imaginary numbers Schrodinger real must be added to above to obtain a superposition which function describe a bound state? does SOUTION: wave function e i(x!t) corresponds to a traveling wave moving +x direction. This is not a standing wave solution. To obtain a in wave we must add a traveling wave moving in x direction standing total(x; t) e i(x!t) + e i(x!t) cos xe i!t sinusoidal one-dimensional traveling wavefunction for an electron is by given a) What is momentum of electron? We now that general form of a sinusoidal one-dimensional SOUTION: wavefunction is traveling p 1:511 m 1 b) What is total energy of electron (in ev)? wave function (x; t) e i(x!t) satises time-dependent b) equation. Explain why it does not describe a particle in a Schrodinger What wave bound (standing wave) state of an innite potential well. Maxima and minima ocuur when 4 x that is e i(x!t). resulting superposition is a standing wave and describes a bound state. 6. Time-dependent Schrodinger Equation 7. Wavefunction of an Electron (x; t) e i(1:511 m 1 x31 16 s 1 t) (x; t) m space derivative is sin(x!t) (x; t) e i (pxet) : time derivative is So by comparison with above relation we have!cos(x!t) we have Substituting which gives p 1: g-m/s. m sin(x!t) 6 i!cos(x!t) equation is not satised for arbitrary x and t. 5

6 SOUTION: by comparison with relation in part (a) we have gain E 3116 s 1 c) What is potential energy of electron (in ev)? SOUTION: part (a) we now momentum. inetic energy is given From m 9: g. Using E from part (b) we havev EK using ev. 11: which gives E 3: J 19.8 ev. by p K 1: J8:6eV m 6

d)p () = A = Z x x p b Particle in a Box 3. electron is in a -dimensional well with innitely high sides and width An Which of following statements mus

d)p () = A = Z x x p b Particle in a Box 3. electron is in a -dimensional well with innitely high sides and width An Which of following statements mus 4 Spring 99 Problem Set Optional Problems Physics April, 999 Handout a) Show that (x; t) =Ae i(kx,!t) satises wave equation for a string: (x; t) @ = v @ (x; t) @t @x Show that same wave function (x; t)