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>>
View the other Key Equations and Concepts in this
chapter
Energy Calculations, Equation 1.5
>> Parts of this equation/concept include:
E = mc2 (Equation 1.5)
where E is the binding energy in Joules, m is the mass defect (kg),
and c is the speed of light (2.998 x 108 m/s).
| A. Binding Energy Calculations, Equation 1.5 |
Mass defect is used as m in equation 1.5 to determine the
binding energy of the nucleus. The mass defect is the difference
between the sum of the weight of each particle in the nucleus and
the actual mass of the nucleus. Alternatively, if the actual mass
is that of the entire atom, then electrons need to be included in
the summation. The actual mass is a measured value that must be
given as part of the problem.
Because of the relatively small differences, it is important to
keep all numbers without rounding until the very last step.
(Don't round for significant figures until the last step.) Keep
in mind that the numbers of protons, neutrons, and electrons are
exact numbers and therefore have infinite significant figures.
>> Example 1
What is the binding energy and the binding energy per nucleon
for 118Sn? Actual mass of atom = 1.957827 x 10–22
g.
Solution:
118Sn has 50 protons, 68 neutrons, and 50 electrons.
Since the actual mass is the mass of the atom, the mass of electrons
must be included in the calculation of the theoretical mass.
| protons |
= |
50 (1.67263 x 10–24 g) |
| |
= |
8.3632 x 10–23 g |
| neutrons |
= |
68(1.67494 x 10–24 g) |
| |
= |
1.139 x 10–22 g |
| electrons |
= |
50(9.100 x 10–28 g) |
| |
= |
4.55 x 10–26 g |
| sum |
= |
1.975775 x 10–22 g |
| mass defect |
= |
sum – actual mass |
| |
= |
1.975775 x 10–22 g – 1.957827 x 10–22 g |
| |
= |
1.7947975 x 10–24 g |
| mass defect |
= |
 m |
= |
1.7947975 x 10–24 g |
|
|
= 1.7947975 x 10–27 kg |
| binding energy |
= |
E |
| |
= |
(m)c2 |
| |
= |
(1.7947975 x 10–27 g)(2.998 x 108 m/s)2 |
| |
= |
1.613 x 10–10 J |
The use of four significant figures is based on the speed of
light used.
Since nucleons are both protons and neutrons, 118Sn
has 118 of them. (mass number = protons + neutrons)
| BE/nucleon |
= |
|
| |
= |
|
| |
= |
1.367 x 10–12 J |
>> Example 2
What is the binding energy of 55Mn? (Actual mass of
nucleus = 9.120517 x 10–26 kg.)
Solution:
55Mn has 25 protons and 30 neutrons. Because the actual
mass is that of the nucleus, the electrons don't need to be included.
| protons |
= |
25(1.67263 x 10–24 g) |
| |
= |
4.1816 x 10–23 g |
| neutrons |
= |
30(1.67494 x 10–24 g) |
| sum |
= |
9.2064 x 10–23 g |
|
|
= 9.2064 x 10–26 kg |
Recall that in doing addition and subtraction, each number must
have the same units.
mass defect = m = 9.2064
x 10–26 kg – 9.120517 x 10–26
kg = 8.5883 x 10–28 kg
BE = (8.5883 x 10–28 kg)(2.998 x 108
m/s)2 = 7.719 x 10–11 J
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| B. Energies of Nuclear Reactions |
The same equation used for binding energies can be used to calculate
the energy associated with nuclear reaction. Instead of a mass defect,
the mass difference between the products and reactants is used for
m. This requires that the actual masses of the products and
reactants be given. (These cannot be calculated from theory.)
For matter-antimatter annihilation, the mass of the products is
zero. After all, that is what annihilation means.
>> Example 3
How much energy is created when 1000 atoms of 239Pu
decay in the following reaction?
239Pu  4He
+ 235U
Actual masses: 239Pu = 239.05218 amu, 4He
= 4.0026033 amu, 235U = 235.04394 amu
1 amu = 1.66056 x 10– 24 g
Solution:
mass difference = 239.05218 – (4.0026033 + 235.04394) =
0.0056367 amu
(With correct significant figures, the answer is 0.00564 amu.
Because it was addition, decimal places are used to determine
the significant figures of the answer. This answer has three significant
figures. The entire number will be used for the rest of the calculation,
but the final answer can't have more than three significant figures
because of this first calculation.)
For use in the equation, atomic mass units must be converted
to kilograms.
| 0.0056367 amu |
|
|
|
|
= 9.360079 x 10–30 kg |
An alternate way to obtain this answer is to convert each mass
in atomic mass units to grams or kilograms before doing the subtraction.
E = mc2
= (9.360079 x 10–30 kg)(2.998 x 108
m/s)2 = 8.4128 x 10–13 J
This value was calculated for one atom. Since the problem says
there are 1000 atoms
energy generated = 1000(8.4128 x 10–13
J) = 8.41 x 10–10 J
Since the mass difference only had three significant figures,
the final answer only has three. We have assumed that 1000 atoms
can be counted and that this value is exact (infinite significant
figures).
>> Example 4
How much energy is created from the annihilation of an electron
(9.100 x 10–28 g) by a positron?
Solution:
Since a positron is the antimatter of an electron, it will have
the same mass as an electron. Since matter and antimatter annihilate
each other, the mass of the product is zero. In other words, energy
is the only product.
m = 2(9.100 x 10–28 g) = 1.820 x 10–27 g(1 kg/1000 g) = 1.820 x 10–30 kg
E = mc2
= (1.820 x 10–30)(2.998 x 108)2
= 1.636 x 10–13 J
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