Sun
Nuclear
Reactions
Sun
Nuclear Physics
The work of Einstein – A new view of mass:
E = m c2
Conceptual Implication: Mass is a
highly concentrated form of energy.
Therefore mass can be converted to other forms
of energy, as long as Conservation of Energy is
satisfied
Sun
Nuclear Physics
How much mass would be required to operate a 200 Megawatt power plant
for one year? Note: The total amount of energy required to operate the
facility for a year is 6.4 x 1015 Joules.
Solution:
m c2 = 6.4 x 1015 Joules
6.4 x 1015
m=
(3 x
108)2
6.4 x 1015
=
= .071 kg = 71 g
9x
1016
Note: the mass of a paperclip is about 1 g
Sun
Nuclear Physics – The Sun’s Core
The nuclear reaction that produces the energy in the core of the
sun is
4 ( 1H )  4He + energy + 2 neutrinos
Given the following information:
Mass of 1H = 1.6736 x 10-27 Kg
Mass of 4He = 6.6466 x 10-27 Kg
What is the mass on the left side of this reaction?
Total MassLeft Side
= 4 times the mass of 1H
= (4) (1.6736 x 10-27 kg)
= 6.6943 x 10-27 kg
Sun
Nuclear Physics – The Sun’s Core
The nuclear reaction that produces the energy in the core of the
sun is
4 ( 1H )  4He + energy + 2 neutrinos
Given the following information:
Mass of 1H = 1.6736 x 10-27 Kg
Mass of 4He = 6.6466 x 10-27 Kg
What is the mass on the right side of this reaction?
Since there is no mass associated with the energy or
the neutrinos on the right side,
Total MassRight Side = 6.6466 x 10-27 kg
Sun
Nuclear Physics – The Sun’s Core
The nuclear reaction that produces the energy in the core of the
sun is
4 ( 1H )  4He + energy + 2 neutrinos
Given the following information:
Mass of 1H = 1.6736 x 10-27 Kg
Mass of 4He = 6.6466 x 10-27 Kg
What is the difference in mass in this reaction?
m = 4.77 x 10-29 kg
Sun
Nuclear Physics – The Sun’s Core
The nuclear reaction that produces the energy in the core of the
sun is
4 ( 1H )  4He + energy + 2 neutrinos
Given the following information:
Mass of 1H = 1.6736 x 10-27 Kg
Mass of 4He = 6.6466 x 10-27 Kg
How much energy is released in this reaction?
Energy = m c2 = 4.3 x 10-12 Joules per reaction
Sun
Nuclear Physics – The Sun’s Core
How many nuclear reactions per second are required to account for the
luminosity of the sun?
Recall that he luminosity of the sun is 4 x 1026 J/sec. To determine the
number of reactions per second, we must know the total energy
released per second by the sun, and divide by the amount of energy per
reaction, that is,
Number of reactions per second = total energy per second/energy per
reaction
Therefore,
Reactions per second = (4 x 1026 J/sec) / 4.3 x 10-12 Joules/reaction
= 9.3 x 1037 reactions per second
Sun
Nuclear Physics – The Sun’s Core
A kilogram of mass is equivalent to how much every (assume a perfect
conversion of mass into energy)? Another way to think about it – how many
joules of energy are contained in 1 kg of mass?
Recall
E = m c2
Therefore
E/m = c2 = (3 x 108 m/sec)2 = 9 x 1016 J/kg
Sun
Nuclear Physics – The Sun’s Core
A The units in the calculation of the previous slide should be m2 / sec2 . How
did they magically become J/kg?
The fundamental MKS units of the Joule is
Joule → kg m2 / sec2
If you don’t remember this, think about he MKS units on Kinetic Energy
It should be clear, therefore, that
Joule / kg → m2 / sec2
And that the units on the previous slide are correct
Sun
Nuclear Physics – The Sun’s Core
Given the luminosity of the sun, how many kilograms of mass are being
converted to energy in 1 second?
The mass converted per second can be determine by knowing the J/sec
produced by the sun (luminosity) and dividing that number the by the
J/kg calculated in the previous calculation. Therefore
Mass / second
= (Luminosity of the Sun) / (Energy – Mass Equivalent)
= (4 x 1026 J/sec) / (9 x 1016 J/Kg)
= 4.4 x 109 kg/sec !!!
On Earth, this mass is conversion is equivalent to converting
over 4 millions tons of material PER SECOND.
Sun
Nuclear Physics – The Sun’s Core
How many hydrogen nuclei are converted into Helium every second in the
core of the sun?
Our discussions in class show that 4 hydrogen nuclei participate in the
creation of each helium nucleus. As a result, the number of hydrogen
being converted each second can be determine by multiplying the
number of reactions per second (see previous slide) by 4, since 4
hydrogen nuclei participate in each reaction. Therefore
# hydrogen nuclei converted into Helium every second is:
( 9.3 x 1037 reactions/second ) / ( 4 Hydrogen/reaction )
= 3.7 x 1038 Hydrogen/second
Sun
Nuclear Physics – The Sun’s Core
How many kilograms of hydrogen are being converted to helium each second in
the core of the sun?
The mass of hydrogen being converted is determined by multiplying the
number of hydrogen being converted each second and multiplying by the
mass of each hydrogen nucleus.
Since the mass of hydrogen is 1.67 x 10-27 kg,
The mass of hydrogen being converted into Helium every second is
( 3.7 x 1038 Hydrogen/second ) (1.67 x 10-27 Kg/Hydrogen)
= 6.18 x 1011 Kg/second
That is, about 6 x 1011 Kg of hydrogen is being converted
to Helium every second in the core of the sun.
Sun
Nuclear Physics – The Sun’s Core
About 6 x 1011 Kg of hydrogen is being converted to
Helium every second in the core of the sun.
On the earth, this would be equivalent to
Over 600,000 million tons of Hydrogen
every second