Download HW Wk9 Solutions

HW9 Solutions (HW9 due Tues, Mar 31)
1. T&M 36.P.38
For ground state of the hydrogen atom, calculate the probability
of finding the electron in the region between r and !r = 0.04a0 at
the positions r = a0 and r = 1.5a0 .
Solution:
In the ground state, ! (r) =
1
3
0
"a
e# r a0 , and hence the radial
4r 2 #2 r a0
probability density is P(r) = 4! r " (r) = 3 e
. The probability of
a0
2
2
finding the electron in the small interval between r and
r + !r = r + 0.04a0 is approximately P(r)!r . For r = a0 , this results in
P(r = a0 )!r =
4 "2
e (0.04a0 ) = 0.0217 , while for r = 1.5a0 , the probability is
a0
4(1.5)2 "3
P(r = a0 )!r =
e (0.04a0 ) = 0.0179 .
a0
2. T&M 36.P.47
Compute the probability that the electron in the ground state of a
hydrogen atom is in the region 0 < r < 3.50a0 .
Solution:
From the previous problem, we know that the probability density
4r 2 #2 r a0
is P(r) = 4! r " (r) = 3 e
. The probability that the electron is in
a0
2
2
the region 0 < r < 3.50a0 is
3.5a0
!
0
3.5a0
P(r)dr =
!
0
4r 2 "2 r a0
e
dr = 0.97 .
a03
3. T&M 36.P.48
The potential energy of a magnetic
moment in an external
r r
magnetic field is given by U = ! µ " B .
(a) Calculate the difference in energy between the two
possible orientations of an electron in the external field
r
B = 0.75Tk̂ .
(b) If these electrons are bombarded with photons of energy
equal to that of the energy difference, “spin-flip”
transitions can be induced. Find the wavelength of the
photons needed for such transitions.
Solution:
(a) The difference in energy levels is !E = 2 µ B . The magnetic
moment due to spin for the electron is µ = 5.79 ! 10 "5 eV T
(Bohr magneton). Hence,
!E = 2(5.79 " 10 #5 eV T)(0.75T) = 8.68 " 10 #5 eV .
(b)
hc
. Putting in
"
hc 1240eV #10 $9 m
=
= 0.0143m .
the numbers, one obtains ! =
"E
8.68 % 10 $5 eV
The photon wavelength is given by !E =
4. T&M 36.P.49
The total angular momentum of a hydrogen atom in a certain
excited state has the quantum number j=3/2. What can you say
about the possible values of the orbital angular momentum
quantum number l?
Solution:
The total angular momentum quantum number j is given by
j = l ± 12 , and hence l = 2 or l = 1 .
5. T&M 36.P.32
Using the fact that in an atom there are two quantum states for
each combination of l and m l , because of the electron spin,
find the total number of electron states that have the following
values of n.
!
!
!
(a) n = 4
(b) n = 2
Solution:
6. T&M 36.P.033
Find the minimum value of the angle θ between and the +z
direction for an electron in an atom that has the following
values of l .
(a) l = 1
(b) l = 4
(c) l = 50
Solution:
7. T&M 36.P.054
What is the ground-state electron configuration of an atom of
each of the following:
(a) carbon
(b) oxygen.
Solution:
8. T&M 40.P.015
Calculate the binding energy and the binding energy per
nucleon from the masses given in the table for the following.
(a) 12C
(b) 56Fe
(c) 238U.
Solution:
9. T&M 36.P.025
The counting rate from a radioactive source is 8000 counts/s
at time t = 0, and 10 min later the rate is 1000 counts/s.
(a) What is the half-life?
(b) What is the decay constant?
(c) What is the counting rate after 20 min?
Solution:
10. T&M 36.P.039
The rubidium isotope 87Rb is a β emitter that has a half-life of
4.9 x 1010 y that decays into 87Sr. It is used to determine the age
of rocks and fossils. Rocks containing the fossils of early
animal have a ratio of 87Sr to 87Rb of 0.0100. Assuming that
there was no 87Sr present when the rocks were formed,
calculate the age of the fossils.
Solution:
11. T&M 36.P.028
Use the table below to calculate the energy release in MeV for
the α decay of the following:
(a) 226Ra
(b) 242Pu
Solution: