Q. What is the de Broglie wavelength of an electron that... kinetic energy of 100 eV?

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Q. What is the de Broglie wavelength of an electron that has a
kinetic energy of 100 eV?
After an electron is accelerated in 100 V potential difference, its
kinetic energy is 100 eV.
eV unit has to be converted into SI unit, Joule.
1 eV = 1.6 x 10-19 J
Ek = (1/2)mov2 = 1.6 x 10-17 J
v2 = 2Ek/mo = 2(1.6 x 10-17 J)/(9.1 x 10-31 kg)
= 3.52 x 1013 m2/s2
v = 5.93 x 106 m/s
low speed: no need to use relativistic
l = h/p
= h/mov = (6.6 x 10-34 Js)/(9.1 x 10-31 kg x 5.93 x 106 m/s)
= 1.23 x 10-10 m = 0.123 nm
The Atomic Nucleus
Composed of nucleons: proton (+e) and neutrons (neutral)
mp = 1.673 x 10-27 kg = 938.28 MeV
E = mc2
mn = 1.675 x 10-27 kg = 939.4 MeV
me = 9.1 x 10-31 kg = 0.511 MeV
Positively charged (# of protons = # of electrons)
Heavy object: most of the atomic mass is concentrated in the nucleus
Size ≈ 10-14 m = 10 femtometer (10-15 = femto)
Considering the size of an atom ≈ 10-10 m, an atom is mostly
empty.
The force binding nucleons together should overcome Coulomb repulsion!
Nuclear force is very strong and attractive.
Each element has different # of protons  periodic table is arranged
by the # of protons.
atomic number: Z
Z = # of electrons: chemical property
8 neutrons
Missing information about the # of neutrons?
146 neutrons
Dmitri Ivanovich Mendeleev (1834-1907)
“I began to look about and write down the elements with their
atomic weights and typical properties, analogous elements and
like atomic weights on separate cards, and this soon convinced
me that the properties of elements are in periodic dependence
upon their atomic weights.”
--Principles of Chemistry, 1905, Vol. II
atomic mass number (A): total # of protons and neutrons
A = Z + N (integer)
A
X
Z
Element
Hydrogen
Helium
Lithium
Uranium
Symbol
Z
N
Mass #
H
1
0
1
He
2
2
4
3
92
4
146
7
238
1
1
4
2
element
7
3
238
92
Li
U
Approximate radius, R, for the nucleus
R = (1.2 x 10-15)A1/3 m
The actual mass of atom or nucleus is expressed in atomic mass units (U).
1 u = 1.6607 x 10-27 kg
= 931 MeV
Exactly equal to one-twelfth of the mass of
the most abundant form of the carbon atom
mp = 1.673 x 10-27 kg = 1.007276 u
mn = 1.675 x 10-27 kg = 1.008665 u
me = 9.1 x 10-31 kg = 0.00055 u
include everything (p, n, and e)
The average atomic mass of mercury (Hg) is 200.59 u. What is
the average nuclear mass for Hg atom?
Z = 80 for Hg  there are 80 electrons.
Mass of electrons in Hg atom = 80 x 0.00055 u
= 0.044 u
Nuclear mass = atomic mass – electron mass
= 200.59 u – 0.044 u
= 200.55 u
Isotopes
Atoms that have the same atomic number (Z) but
different mass numbers (A).
A
X
Z
7
Li
3
Natural abundance:
6
Li
3
92%
58
Co
27
8%
60
Co
27
Helium Nucleus
4
He
2
atomic mass = 4.002602 u
2 electrons: electron mass = 2 x 0.00055 u
= 0.0011 u
nucleus mass = 4.002602 u – 0.0011 u
= 4.001502 u
We know He nucleus is composed of 2 protons and 2 neutrons:
2 proton mass = 2 x 1.007276 u
2 neutron mass = 2 x 1.008665 u
total mass = 4.031882 u > He nucleus mass
Dm = 4.031882 u – 4.001502 u = 0.030376 u
All nuclei have a mass that is smaller than the sum of the
masses of its separated nucleons. This difference in mass
is called the mass defect.
22
E = mc
DE
Dmc
binding energy of the nucleons in the nucleus
Dm for He = 0.030376 u
= 0.030376 x (1.66 x 10-27 kg)
= 5.0 x 10-29 kg
1 u = 1.6607 x 10-27 kg
= 0.030376 x 931 MeV
= 931 MeV
= 28.2 MeV
DE = Dmc2 = (5.0 x 10-29 kg)(3 x 108 m/s)2
= 4.5 x 10-12 J
= 28 MeV
binding energy per nucleon in He = 7 MeV
28 MeV
p n
He nucleus
Ex. 31.4 How much energy is released as 1 kg of neutrons and 1 kg
of protons are changed into 2 kg of helium 4 nuclei?
# of He nuclei in 2 kg = 2 kg/(4 x 1.66 x 10-27 kg)
= 3.0 x 1026
28 MeV is released per nucleus:
total energy released = 28 x 3.0 x 1026 MeV
= 8.4 x 1027 MeV
= 1.5 x 1015 J
New York city needs 1010 W of electricity.
This amount can run NY city for 2 days!!
Radioactivity
Sometimes particles or photons are emitted
from the nuclei of atoms: unstable nuclei
radioactive
unstable nuclei  stable nuclei
all naturally occurring elements with Z > 83
Three major forms of radioactive emission:
Alpha particle (a): nucleus of a He atom 4a
2
Beta particles (b): beta minus particle (electron)
beta plus particle (positron)
0
1
b
0
1
b
Gamma rays (g): high energy photons, short wavelength x-ray
A
X
Z
A
A
X
Z
A-4
X
+
Z-2
4
2
A
A
+
X
Z+1
0
1
X
Z
+
X
Z
g
a
b
A and Z are conserved in nuclear reaction!!
g
T1/2: half-life
Initial # of unstable nuclei at time t = 0
N(t) = Noe-pt
# of unstable nuclei at time t
decay constant
At T1/2, N(T1/2) = (0.5)No
N (T1/ 2 )
1
 pT1 / 2
e

No
2
ln e
 pT1 / 2
pT1/ 2
  ln  12 
 
 ln 2  0.692
If you know T1/2, you can get p from this equation.
Then, at any time you can calculate # of unstable nuclei.
Or, with the knowledge of # of unstable nuclei, we can calculate
how much time has passed.  Carbon dating
Carbon Dating
should be dead!
By measuring the ratio of 14C isotope to 12C in a biological sample
one can calculate the age of the object from which the sample came
with the knowledge of the initial ratio when the biological activity
ceased.
half-life: 5730 yrs
?
A mummy found in Egypt
bbc.co.uk
14
7N +
1 n
0
 146C + 1?1p
Ratio of 14C to 12C?
Before 1900 (atomic bomb testing)
1.5 x 10-12
14 C
6
 147N + ?0-1b
beta decay
A mummy found in Egypt
Measured the ratio of 14C to 12C
r(t) = 1.05 x 10-12
ro = 1.5 x 10-12
T1/2 = 5730 yrs
bbc.co.uk
Total # of 12C = N12
# of 14C = N14 = r(t)N12
Let’s calculate decay const, p
p = 0.693/T1/2
= 0.693/5730
= 1.21 x 10-4 (yr-1)
r(t)N12 = roN12e-pt
r(T)/ro = (1.05/1.5) = e-pT
ln(0.7) = -pT
T = -ln(0.7)/(1.21 x 10-4)
= 2848 yr
Ex. 31.10 Cobalt-60 has a half-life of 5.25 yrs (1.66 x 108 s).
Given 1 mg of the substance, find (a) the number of atoms in the
sample, (b) the decay constant, (c) the time taken for 0.01 mg
to decay.
mass of Co-60 = 59.93381 u = 60 u
= 60 x 1.66 x 10-27 kg
= 1.0 x 10-25 kg
No = (1 x 10-6 kg)/(1 x 10-25 kg)
= 1 x 1019 atoms
p = 0.693/T1/2 = 0.693/(1.66 x 108 s)
= 4.2 x 10-9 s-1
How long does it take to reach N/No = 0.99?
ln(N/No) = -pt
ln(0.99) = -(4.2 x 10-9)t
about a month
t = 2.4 x 106 s
7
1 yr = p x 10 s
Which of the following nuclear reaction is(are) impossible?
(a) 23191Pa  a + 22789Ac
(b) 146C  147N + +1b
(c) 24092U  -1b + 24093Np
20%
20%
20%
N
on
e
ll
A
)
(c
5.
)
4.
(b
3.
)
2.
20%
(a)
(b)
(c)
All
None
(a
1.
20%
45
Nuclear Fission
In a chemical reaction, changes in atoms are in their electron
configuration. However, in a nuclear reaction, the configuration of
nucleus changes by emitting or absorbing typically high energy
photons, nucleons, alpha particles, and/or light charged particles.
In this process, charge, nucleons, and mass-energy are conserved.
Nuclear fission:
Unstable heavy nucleus  multiple intermediate mass nuclei
+ large energy
In 1938, Otto Hahn, Lise Meitner, and Fritz Strassmann discovered:
stable isotope
T1/2 ≈ 109 yrs
unstable
235
236
92
92
U
U
n
200 MeV per fission
138
Ba
n
56
n
95
n
Kr
36
235U
236U
141Ba
n
Chain Reaction
92Kr
"When you sit with a nice girl for two hours, it seems like
two minutes. When you sit on a hot stove for two minutes,
it seems like two hours that's relativity."
"Gravitation is not responsible for people falling in love."
Through the release of atomic energy, our generation has brought
into the world the most revolutionary force since prehistoric man's
discovery of fire. This basic force of the universe cannot be fitted
into the outmoded concept of narrow nationalisms. For there is no
secret and there is no defense; there is no possibility of control
except through the aroused understanding and insistence of the
peoples of the world. We scientists recognise our inescapable
responsibility to carry to our fellow citizens an understanding of
atomic energy and its implication for society. In this lies our only
security and our only hope - we believe that an informed citizenry will
act for life and not for death.
A. Einstein, 1947
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