Topic – Physics 2a
Mass defect and binding
energy
•
•
•
•
Prior learning
Atomic structure
Electrical forces
Key words
– Atomic nucleus,mass difference, mass of
products, nuclear reactions, total mass
Topic – Physics 2a
By the end of this lesson we should be able to:
• Know how binding energy is linked to
Einstein's equation – E=mc2
• Calculate mass difference
• Complete problem solving questions on
binding energy
Topic – Physics 2a
What is binding energy?
• Nuclei are made up of protons and neutron, but
the mass of a nucleus is always less than the sum
of the individual masses of the protons and
neutrons which constitute it.
• The difference (missing mass) is a measure of the
nuclear binding energy which holds the nucleus
together.
• This binding energy can be calculated from the
Einstein relationship:
• Nuclear binding energy = Δmc2
Topic – Physics 2a
Mass defect
• The measured mass of a nucleus is always
less than the sum of the individual masses of
its nucleons.
• This difference is called MASS DEFECT
Topic – Physics 2a
Binding energy and mass
defect
• The mass defect is a measure of the atoms
binding energy
• The more binding energy per nucleon the
greater the atoms stability – held together
stronger!!
Topic – Physics 2a
To calculate binding energy
of a nucleus…
• Add the mass of the individual
nucleons (protons and neutrons)
• Then subtract the mass of the actual
atom itself
• The mass left over – Mass defect
Topic – Physics 2a
Values and Equation
•
•
•
•
Mass of proton 1.67262 x 10-27kg or 1.00727u
Mass of neutron 1.67493 x 10-27 kg or 1.00867u
Mass of electron 9.11 x 10-31kg or 5.49 x 10-4u
See tables/data sheet for other values of atoms
• Mass difference = mass of particles– mass of atom
Topic – Physics 2a
Example: When a neutron combines
with a proton  Deuteron nucleus
• Neutron mass = 1.00867u
• Proton mass = 1.00728u
• = 2.01595u
• Deuterium nuclear mass = 2.01350u
• 2.01595 – 2.01350 =
• Difference (loss) = 0.00245u
Topic – Physics 2a
Converting into energy
•
•
•
•
•
Constants:
1eV = 1.6 x 10-19J
1 MeV = 1.6 x 10-13 J
1 u = 1.6606 x 10-27 kg
1u = 931 MeV
• speed of light, c = 3.0 x 108 ms-1
Topic – Physics 2a
Example: converting to energy
• Mass defect is then converted into an
energy equivalent
• x 931
• 0.00245u x 931 = 2.3MeV of energy
• This is the energy that would be
required to pull the nucleus apart
Topic – Physics 2a
Worksheet
• Eb = md c2
where:
• Eb = the binding energy in joules (J)
• md = mass defect in kg.
• c = speed of light, 3.0 x 108 ms-1
Topic – Physics 2a
Example- Calculate the binding energy
of a lithium-7 nucleus, in joules and
MeV.
• mass defect = (mass of neutrons and protons) - (mass of nucleus)
= [(3 x mass proton) + (4 x mass neutron)] - (mass of lithium )
In kg
= [(3 x 1.67 x 10-27) + (4 x 1.68 x 10-27)] - 1.165 x 10-26
= 5.01 x 10-27 + 6.72 x 10-27 - 1.165 x 10-26
= 8.0 x 10-29 kg
E = Mc2
E = 8.0 x 10-29 x (3.0 x 108)2 = 7.12 x 10-12 J
In u units
1 u = 1.6606 x 10-27 kg therefore
8.0 x 10-29 / 1.66 x 10-27 = 0.4818 u
0.4818 x 931 = 44.86MeV
(check that’s correct by x 1.6 x 10-13 to get back into J)
Topic – Physics 2a
In the sun, hydrogen atoms combine to make helium
in the process of fusion. In this process, energy is
released. One of the reactions is shown.
Determine the binding energy release for this
reaction in joules and MeV.
1
1
2
1
H = 1.67 x 10-27 kg
H = 3.34 x 10-27 kg
md = (mass 11H + 21H ) - (mass
3
2
He = 5.00 x 10-27 kg
= 1.00 x 10-29 kg
3
2
He )
= (1.67 x 10-27 + 3.34 x 10-27) – 5.00 x 10 -27
In joules:
In MeV
Eb = md x c2
= 1.00 x 10-29 x (3.0 x 108)2
Eb = 9.00 x 10-13 J
mass in u =
1.00 x10 -29
1.66 x10 - 27
= 6.024 x 10-3 u
MeV = 6.024 x 10-3 x 931
= 5.61 MeV
Topic – Physics 2a
A common reaction in nuclear fission power
235
1
141
92
1
U
+
n
®
Ba
+
Kr
+
3
plants
is
,
92
0
56
36
0n
calculate the energy released from this reaction
in joules and MeV.
25
235
92 U = 3.90 x 10 kg
-25
141
=
2.34
x
10
kg
Ba
56
-25
92
=
1.53
x
10
kg
Kr
36
92
1
235
1
md = 141
56 Ba + 36 Kr + 3 0 n 92 U + 0 n
= (2.34 x 10-25 + 1.53 x 10-25 + [3 x 1.68 x 10-27]) - (3.90 x 10-25 +1.68 x 10 -27 kg)
= 3.9204 x 10-25 - 3.9168 x 10 -25
= 3.6 x 10-28 J
md in u =
3.6 x10 -28
1.66 x10 - 27
MeV = 0.2169 x 931
= 202 MeV
= 0.2169 u
1 u = 1.6606 x 10-27 kg therefore
Topic – Physics 2a
Extension Questions
1. What is the binding energy per nucleon in MeV for
the following atoms involved in nuclear energy:
•
a. U-238 nucleus
b. He-4 nucleus
c. Fe-56 (one of the most stable atoms)
(Fe = 55.934939 u)
2. The oxygen atom O has an isotope, O. Find the
binding energy of each nucleus and thus
•
determine which is more stable.
3. Calculate the energy released from one decay of
U-238.
Topic – Physics 2a
1. What is the binding energy per nucleon in MeV for the following atoms involved in
nuclear energy:
a. U-238 nucleus
U-238 = 92p + 146n
md = ([92 x 1.00783) + (146 x 1.00867)] - U-238
= 92.72036 + 147.26582 – 238.05079
= 1.93539 m
= 1.93539 x 1.6606 x 10-27
= 3.2130 x 10-27 J
E = 3.2130 x 10-27 x (3 x 108)2
= 2.8925 x 10-10 J per atom
as there are 238 nucleons, then
binding energy per nucleon =
2.8925x10 -10
= 1.21 x 10-12 J
238
b. He-4 nucleus
He-4 = 2p + 2n
= (2 x 1.00783) + (2 x 1.00867) – 4.00260
= 0.0304 m
= 0.0304 x 1.6606 x 10-27
= 5.04822 x 10-29 J
E = 5.04822 x 10-29 x (3 x 108)2
= 4.5434 x 10-12 J per atom
binding energy per nucleon =
4.5434x10 - 12
= 1.136 x 10-12 J
4
Topic – Physics 2a
c. Fe-56 (one of the most stable atoms) (Fe = 55.934939 u)
Fe-56 = 26p + 30n
= (26 x 1.00783) + (30 x 1.00867) – 55.934939
= 0.528741 m
= 8.78027 x 1028 kg
E = 8.78027 x 10-28 x (3 x 108)2
= 7.90225 x 10-11
7.90225 x10 -11
binding energy per nucleon =
= 1.41 x 10-12 J
56
You will notice that the binding energy per nucleon is greater than the
previous two examples.
Topic – Physics 2a
2. The oxygen atom 168 O has an isotope,
and thus determine which is more stable.
17
8 O.
Find the binding energy of each nucleus
O-16 = 8p + 8n
= (8 x 1.00783) + (8 x 1.00867) – 15.99943
= 0.13257 m
= 2.20146 x 1028 kg
E = 2.20146 x 10-28 x (3 x 108)2
= 1.9813 x 10-11
1.9813x10 - 11
binding energy per nucleon =
= 1.238 x 10 -12 J
16
O-17 = 8p + 9n
= (8 x 1.00783) + (9 x 1.00867) – 16.99913
= 0.14154 m
= 2.3504 x 1028 kg
E = 2.3504 x 10-28 x (3 x 108)2
= 1.1154 x 10-11
2.1154x10 -11
binding energy per nucleon =
= 1.244 x 10 -12 J
17
Oxygen 17 has the greater binding energy so is more stable.
Topic – Physics 2a
3. Calculate the energy released from one decay of U-238.
238
92
U® 42 He+ 23490Th
md = 238.05079 – (4.00260 + 234.04360)
= 4.59 x 10-3 u
= 7.622154 x 10-30 kg
E = mc2
= 7.622154 x 10-30 x (3 x 108)2
= 6.86 x 10
-13
J
and
6.86 x10 - 13
1.6 x10
- 13
= 4.29 MeV
Topic – Physics 2a
links
• http://www.s-cool.co.uk/alevel/physics/nuclear-energy/reviseit/mass-defect
• http://www.colorado.edu/physics/200
0/periodic_table/amu.html
• http://www.alevelphysicstutor.com/we-nucbinding-energy.php
Topic – Physics 2a
By the end of this lesson we should be able to:
• Know how binding energy is linked to
Einstein's equation – E=mc2
• Calculate mass difference
• Complete problem solving questions on
binding energy