Download Physics 249 Homework 5 Due Oct 19th 1) Consider two potential

Physics 249 Homework 5
Due Oct 19th
1) Consider two potential steps. One from V(x<0)=0 to V(x>0)=V0 and the other from
V(x<0)=0 to V(x>0)=-V0. Consider a particle incident on this step potential from
–x with total energy E=2V0. Note this is the total energy including the potential
energy. For each calculate.
a) The wave numbers in regions two (x>0) in terms of the wave number in region 1 (x<0)
and V0.
b) The reflection coefficients.
c) The transmission coefficients.
Note since the potential energy is zero in region one the total energy is the same as the
kinetic energy.
a) The wave number for both ases i (V0) and ii (-V0) in region one is.:
!! =
4!V!
ℏ
2!"
=
ℏ
in region two the wave number are
!!! =
!!! =
2! 2V! − V!
=
ℏ
2! 2V! − −V!
ℏ
2!V!
ℏ
6!V!
ℏ
=
b) The reflection coefficients.
!! − !!
!! =
!! + !!
!
=
!
!! − !!
!!! =
!! + !!
=
!
2− 2
= 0.0294
2+ 2
2− 6
!
= 0.0102
2+ 6
c) The transmission coefficients.
!! =
4!! !!
!! + !!
!
=
8 2
2+ 2
!
= 0.9706
!!! =
4!! !!
!! + !!
!
=
8 6
2+ 6
= 0.9898
!
Note that T+R=1 in both cases!
2) Consider potential barrier 0.5nm wide and 10eV high. Electrons accelerated through
5V approach the barrier. What percentage of electrons will tunnel through the
barrier?
!!
! = !"
!
!
!
!
= 1+
!
!!"ℎ !"
!
!
4! 1 − !
!
!
Note that there was an error in the formula in the book and in my notes. The sinh should
have been squared. The formula has been corrected in my notes from lecture 16.
Solutions with either sinh2 or sinh were accepted.
2! !! − !
!!"ℎ
ℏ
! = 1+
!
!
4! 1 − !
!
!
!
!!"ℎ
! = 1+
!
!!
!
2!! ! !! − !
!!"ℎ
ℏ!
= 1+
!
!
4! 1 − !
!
!
2 ∗ 0.511!10! 10 − 5
197.3!"#$
4 ∗ 0.5 1 − 0.5
!
!!
!
!!
0.5
= 4.23!10!! = 4.23!10!! %
3) Consider a infinite 3D box potential with L2=2L1 and L3=4L1. What are the quantum
numbers of the lowest degenerate energy levels? List, ordered by energy, the
quantum numbers and energies of all the levels up two the lowest energy set of
degenerate energy levels.
!=
ℏ! !
ℏ! ! ! !!! !!! !!!
!! + !!! + !!! =
+ +
2!
2!!! 1
4 16
1, 1, 4 and 1, 2, 2
Systematically you increment up the third eigenstate comparing it to incrementing up the
second (or first) eigenstate at each step. 1, 2, 1 is smaller then 1, 1, 4 and 1, 2, 2 is
the same as 1, 1, 4. Both give fractions of ¼ and 1.
Define
ℏ! ! !
!! =
2!!!
for the constant term in front of the terms
1, 1, 1: ! = 1.313!!
1, 1, 2 ! = 1.5!!
1, 1, 3 ! = 1.813!!
1, 2, 1 ! = 2.063!!
1, 1, 4 ! = 2.25!!
1, 2, 2 ! = 2.25!!
4) The radial probability distribution function for the hydrogen atom ground state can be
writer as ! ! = !! ! ! !!!/!! , where C is a normalization constant.
a) Why is the factor of ! ! included?
b) Show that P(r) has it’s maximum value at r=a0.
c) If you considered equal infinitesimal volumes, what at what radius is the maximum
probability to find an electron?
d) Calculate the expectation value of r.
a) The probability at a give radius is integrated over a spherical shell of area 4!! ! .
b)
!"(!)
2 !!!
= 2!"! !!!/!! − !! ! ! !! = 0
!"
!!
!
! − ! !! = 0
!
! = !!
!!!
! ! !(!)
2 !!!!
2 !!!!
2
!! − !2!
! − !2!
=
2!!
!
! ! + !! !
!
!!
!!
!!
!!
!
! !(!! )
= 2!! !! − 8!! !! + 4!! !! = −2!! !!
!! !
indicating a local maxima
c)
!(!) ! = !" !!!/!!
Which has a maxima at r=0.
d)
!
!
!
!!
!!
4
!!!
!
! ! ! !!!/!!
!
using
4
!!!
!
!
! ! ! !!!/!! =
4 3!
!!! 2
!!
!
=
4 6 ! 3
! = !
!!! 16 ! 2 !
As expected, larger than the maximum.