Assigned problems for Chapter 10 on the particle in a box - Cobalt

2. Calculate the probability that a particlewill be found between 0.49 L and 0.51L in a 1 -Dbox of length Lwhen it has (a) n =1, (b) n=2.The the wavefunction to be a constant in this rangethe wavefunctions aren = 2L sin ⎛ n x ⎞⎜ ⎟⎝ L ⎠The propability is0.51LP = ∫ n 2 dx ≈ 2 n x0.49L(a)P = 2 L sin 2 ⎛⎜⎝n xLFor x =.5L n=1⎞⎟ x⎠P= 2 L sin 2⎛⎞⎜ ⎟ x0.02L = 0.04⎝ 2⎠

3.(a) Write the Schrödinger equation <strong>for</strong> a particle in a 1-D square wellof length L(b) Calculate the expectation value of p and p 2 <strong>for</strong> a particle in the state n=1Answ:(a)− h22md 2dx 2= Ewhich has the solutionn =2L sin ⎛ n x ⎞⎜ ⎟⎝ L ⎠ˆ p = − h iddx< ˆ p >=L∫0n * p ˆn dx = 2hiL 2L ⎛sin n x ⎞∫ ⎜ ⎟⎝ L ⎠0ddx sin ⎛ n x ⎞⎜ ⎟ dx⎝ L ⎠= 2hiL 2L ⎛sin n x ⎞ ⎛∫ ⎜ ⎟ cos n x ⎞⎜ ⎟ dx = 0⎝ L ⎠ ⎝ L ⎠0

4. What are the most likely locations of a particle in abox of length L in the state n = 3 ?3 = 2L sin ⎛ 3 x ⎞⎜ ⎟⎝ L ⎠P(x)∝ 3 3 ∝sin 2 ⎛ 3 x⎞⎜ ⎟⎝ L ⎠The maxima and minima in P(x) corresponds to dP(x)dx = 0dP(x)dx∝ sin ⎛ 3 x ⎞ ⎛⎜ ⎟ cos 3 x ⎞ ⎛⎜ ⎟ ∝ sin 6 x ⎞⎜ ⎟⎝ L ⎠ ⎝ L ⎠ ⎝ L ⎠⎛sin = 0 when = 6 x ⎞⎜ ⎟ = n' , n = 0,1,2.....⎝ L ⎠which corresponds to x = n' L6, n'< 6.n' = 0,2,4, and 6 corresponds to minima inn' = 1, 3, and 5 to maxima

5. Set up the Schrödinger equation <strong>for</strong> a particle of mass m in a threedimensionalsquare well withsides L x ;L y ;L z . Show that the wavefunction is defined by three quantum numbersand that the Schrödinger equantion is separable . Find the energy levels, and specializethe results to a cubic box of side LPlease see handout on particle in 3-D box (Lecture-<strong>10</strong>)6. To a crude first approximation, a electron in a linear polyenemy be considered to be a particle in a one dimensional box.The polyene - carotene contains 22 conjugated C atoms, and theaverage internuclear distance is 140 pm. Each state up to n =11 isoccupied by two electrons.Calculate(a) the separation in energy between the ground state andthe first excited state in which one electron occupies the state with n=12.(b) The frequency of the radiation required to produce a transition betweenthese two states(c) The total probability of finding an electron between C atoms 11 and 12 in thegroundsate of the 22-electron moleculeAnsw:(a): L =(21)x(1.40x<strong>10</strong> -<strong>10</strong> m) = 2.94x<strong>10</strong> −9 m=E = E 12 − E 11 = (2n + 1)x h 22h2= (2x11 + 1)x8mL 8mL 2(23)x(6.626x<strong>10</strong> −34 Js)(8)x(9.11x<strong>10</strong> −31 Kg)x(2.94x<strong>10</strong> −9 ) 2 = 1.603x<strong>10</strong> −19 J = 1.60x<strong>10</strong> −19 J

(b) : = Eh = 1.603x<strong>10</strong>−19 J6.626x<strong>10</strong> −34 Js = 2.42x<strong>10</strong>14 s −1 = 2.42x<strong>10</strong> 14 Hz7. Consider a particle in a cubic box. What is the degeneracy of the levelsthat has an energy three times that of the lowest level.