for Solutions Hand-In Set #9 PHYS 212 Spring 2007
A62 -- Uncertainty
(c) For the second trial solution we have 2 (x) = B sin kx d2 = Bk cos kx dx d 2 1 = -Bk 2 sin kx. dx2 Plugging these into the rearranged Schrdinger equation o gives -Bk 2 sin kx = -
?
The uncertainty in the x position is given in the problem statement: x = 10-12 m. This implies a minimum uncertainty in the x component of the momentum given by the Heisenberg Uncertainty Principle: xp h h p . 2 2x
The mass is certain, so this can be rewritten as h mv , 2x or v h 2mx 1.06 10-34 Js 2 0.04 kg 10-12 m 1.3 10-21 m/s
2m (E - U0 )B sin kx. h2
To see if 2 is a soluton, first cancel all common factors, leading to -k 2 = -
?
2m (E - U0 ). h2
A65 -- Schrdinger equation for classically o forbidden situation
In a classically forbidden region, U0 > E, so (E-U0 ) < 0. This means that the righthand side is a positive constant. But the lefthand side is a negative constant (for real k), and the two sides can't be equal. Therefore the function 2 is not a solution. (You could argue that 2 is a solution if k is an imaginary number. This actually leads to solutions like 3 . If you're interested in pursuing this further, talk to your instructor.) (d) For the third trial solution we have 3 (x) = Ce-x - d3 d 2 1 = -Cke-x - = +Ck 2 e-x . dx dx2
(a) Rearranging the Schrdinger equation gives o 2m d2 (x) = - 2 (E - U0 )(x). dx2 h (b) For the first trial solution we have 1 (x) = Ax d1 = 2Ax dx d 2 1 = 2A. dx2 Plugging these into the rearranged Schrdinger equation o gives 2A = -
? 2
Plugging these into the rearranged Schrdinger equation o gives Ck 2 e-x = -
?
2m (E - U0 )Ce-x . h2
To see if 2 is a soluton, first cancel all common factors, leading to k2 = -
?
2m (E - U0 ). h2
2m (E - U0 )Ax2 . h2
?
In a classically forbidden region, U0 > E, so (E-U0 ) < 0. This means that the righthand side is a positive constant. The lefthand side is also a positive constant, and if we choose k= 2m (U0 - E), h2
where I have used the symbol = to emphasize that were are tying to see if the function 1 "works," i.e., makes the equality true for all values of x. In this case 1 is not a soluton. The lefthand side is a positive constant, and the righthand side is quadratic in x; these clearly can't be equal for all x. The function 1 (x) is not a solution.
then the left and right sides are equal, and 3 (x) is a solution.
A72 -- Barrier tunneling
Tipler 36-24
a) The tunneling probability is given by Tipler's Eq. (35-29): where a = 0.05 nm is the barrier thickness and = 2m(U0 - E) , h2 T e-2a ,
and U0 is the constant potential energy (which is greater than energy E) and m is the mass of the particle. (Note that in Eq. (35-29) there is a "proportional to" that looks like an "alpha.") In calculating the quantity under the radical in , I make the substitution h h/2, and for "units I convenience" multiply the numerator and denominator by c2 , giving 2mc2 (U0 - E) (2)2 2m(U0 - E) = 2 (hc)2 h 2 511 103 eV(22 eV - 15 eV)(2)2 = (1240 eVnm)2 = 184 nm-2. (You could, of course, convert everything to SI units.) Substituting this into Tipler's equation gives T e-2
1840.05
The transitions in the Balmer series in the hydrogen spectrum are illustrated in energy-level diagram of Fig. 36-4 of your text on p. 1177. All of the transitions in the Balmer series have the n = 2 level as their final state. The three transitions giving the longest wavelengths are those involving the smallest energy differences between states, i.e., the n = 3 n = 2, n = 4 n = 2, and n = 5 n = 2 transitions. The relevant energies are E5 = E4 E3 E2 13.6 eV 52 13.6 eV = 42 13.6 eV = 32 13.6 eV = 22 = -0.54 eV = -0.85 eV = -1.51 eV = -3.40 eV
The energies of the photons are E52 = E5 - E2 = -0.54 + 3.40 = 2.86 eV, E42 = E4 - E2 = -0.85 + 3.40 = 2.55 eV, E32 = E3 - E2 = -1.51 + 3.40 = 1.89 eV. The wavelengths are 52 = hc E52 1240 eVnm = 2.86 eV = 434 nm, hc E42 1240 eVnm = 2.55 eV = 486 nm, hc E32 1240 eVnm = 1.89 eV = 656 nm.
= e-1.36 = 0.26.
b) From part a) we know that approximately 26% make it through, or about 260,000. c) If we increase the barrier width by a factor of 10 to 0.5 nm, the tunneling probablity becomes T e-2
1840.5
= e-13.6 = 1.3 10-6 .
42 =
d) Now on average one electron gets through. Increasing the width by a factor of 10 makes a huge difference!
32 =
Tipler 31-22
The energy of the photon is Eph = hf = hc 1240 eVnm = 368 nm = 3.37 eV.
The photon is absorbed, and all of its energy goes to raising the energy of the atom to a state 3.37 eV above the ground state. (The other given information is irrelevant.)
Tipler 36-38
Tipler 36-47
In the ground state of hydrogen, the wavefunction for the electron is 1 (r) = the probability density is |(r)|2 = 1 1 a0
3
1 a0
3/2
e-r/a0 ,
The radial probability distribution function for the ground state of hydrogen (a one-electron atom with Z = 1) is illustrated in Fig. 36-10 of your text on p. 1184. The maximum of this function is near the value 1 on the horizontal axis, which corresponds to a radius of r = a0 (because the variable on the horizontal axis r/a0 . To see exactly where this maximum occurs, take the derivative of P (r) with respect to r, and set this equal to 1: dP d Cr2 e-2Zr/a0 = dr dr = 2Cre-2Zr/a0 - Cr2 2Z a0 . e-2Zr/a0
e-2r/a0 ,
and the radial probability density is P (r) = |(r)|2 4r2 = 4r2 1 a0
3
e-2r/a0 .
= 2Cre-2Zr/a0 1 -
Zr a0
At r = a0 these expresssions become (a0 ) = 1 e a3 0 ,
This expression will be zero when (1 - Zra0 ) = 0, or when a0 . Z
1 |(a0 )| = , a3 e2 0
2
r=
and P (a0 ) = |(a0 )2 |4a2 = 0 4 . a0 e2
For hydrogen, Z = 1, so the maximum of the radial probability distribution is exactly a0 , the radius of the electron orbit given by the simple Bohr model.