LECTURE 2: ATOMIC AND NUCLEAR STRUCTURE
Problem Solutions
2.1 Calculate the energy required for transition of an electron from the K shell to the L shell in
tungsten. Compare the result to the energy necessary for a similar transition in hydrogen.
Explain the difference.
For tungsten, the binding energies are
K shell
-69,500 eV
L shell
-11,280 eV
so the transition energy is 58,220 eV.
For hydrogen, the binding energies are
K shell
-13.5 eV
L shell
-3.4 eV
so the transition energy is 10.1 eV.
It takes more energy to move a tungsten electron from the K to the L
shell
than for hydrogen, since the electron experiences a greater
attractive force
from tungsten. This is due to the greater number of
protons in the tungsten
nucleus.
2.2 What is the energy equivalent to the mass of the electron? Since the mass of a particle
increases with velocity, assume the electron is at rest.
E
=
= 9.1x10
=
mc
−31
2
kg × (3.00 × 10 m / sec)
819
. × 10 −14 J
8
2
2.3 What are the atomic and mass numbers of the oxygen isotope with 16 nucleons? Calculate
the nuclear mass defect, binding energy, and binding energy per nucleon for this nucleon
for this nuclide. The mass of the nucleus is 15.9949 amu.
The oxygen isotope with 16 nucleons has 8 protons and 8 neutrons, so Z=8 and
A=16.
proton mass: 1.00727 amu x 8
neutron mass: 1.00866 amu x 8
=
=
8.05816 amu
8.06928 amu
16.12744 amu
-15.9949 amu
mass defect
=
binding energy =
x
931 MeV/amu
0.1325 amu
123.4 MeV
binding energy/nucleon =
123.4 MeV/amu
= 7.71 MeV/nucleon
2.4 Natural oxygen contains three isotopes with nuclear masses of 15.9949, 16.9991, and
17.992, and relative abundances of 2500:1:5, respectively. Determine to three decimal
places the average nuclear mass of oxygen.
15.9949 × 2500 + 16.9991 × 1 + 17.992 × 5
2506
= 15.999
avg nuclear mass =
2.5 The energy released during the atomic explosion at Hiroshima was estimated to be equal to
that released by 20,000 tons of TNT. Assume that a total energy of 200 MeV is
released during a fission of a 235U nucleus and that a total energy of 3.8 x 109 J is
released during detonation of 1 ton of TNT. Find the number of fissions which
occurred in the Hiroshima explosion and the total decrease in mass.
energy released in explosion = 20,000 × 3.8 × 109 J
7.6 × 1013 J
=
1.6 × 10 −19 J / eV
=
4.75 × 10 32 eV
energy released per fission = 200 MeV / fission
= 2.00 × 108 eV / fission
4.75 × 10 32 eV
2.00 × 108 eV / fission
= 2.38 × 1024 fissions
number of fissions =
Assuming the 7.6 × 1013 J comes completely from a decrease in mass,
E = mc 2 → m =
E
c2
Then
m =
7.6 × 1013 J
=
(3.00 × 108 m / sec) 2
8.44 × 10 -4 kg
=
0.844 g
2.6 A “4-megaton thermonuclear shot” means that a nuclear explosion releases as much energy
as that liberated during detonation of 4 million tons of TNT. Using of 3.8 x 109 J/ton as
the heat of detonation of TNT, calculate the total energy in joules and in kilocalories
released during the nuclear explosion (1 kilocalorie = 4186 joules).
energy released = 4 × 106 ton × 3.8 × 109 J / ton
15.2 × 1015 J
=
4186 J / kcal
=
3.63 × 1012 kcal