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The Discovery of the Neutron
Abstract
During the first half of the 20th century, there were many important
discoveries regarded the structure of atoms. One such discovery was James
Chadwick’s discovery of the neutron in 1932. His discovery of its existence
as well as its nature helped to provide a clearer picture of the atom and helped
build the framework for the modern use of nuclear technology.
Background
James Chadwick
James Chadwick was born in Bollington, England, in 1891. He
accidentally signed up for physics at Victoria University of Manchester
where he met Rutherford. During WWI he was interned in Germany, but he
still continued his education. After the war, Chadwick went back to England
where he became Rutherford’s assistant at the Cavendish Laboratory in
Cambridge, where he did much of his work including his discovery of the
neutron. In 1935 he won the Nobel Prize in Physics for his discovery of the
neutron.
Prior Nuclear Theory
Many advancements were made in the field of nuclear physics during
the early 20th century. After Rutherford’s gold foil experiment, Rutherford
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developed the nuclear atomic theory, in which a dense nucleus contains the
positive charge of an atom and is surrounded by orbiting negatively charged
electrons. When Rutherford discovered that hydrogen nuclei (protons) could
be creating by disintegrating larger nuclei, he combined this observation with
the known emission of electrons by nuclei (beta rays) in order to confirm his
theory that the nucleus was composed of protons and electrons. In order to
account for the difference between the atomic number (the positive charge of
the nucleus) of an element and its atomic mass, the model placed extra
electrons and protons inside of the nucleus. For instance, prior to Chadwick’s
discovery, nitrogen-14’s nucleus was believed to have 14 protons and 7
electrons and the alpha particle (a helium nucleus) was believed to have 4
protons and 2 electrons. The number of the extra proton-electron pairs in the
nucleus was equal to the difference between the atomic number and the
atomic mass, thus giving atoms their observed mass. Rutherford even
suggested in a lecture that there might be a type of nucleus containing one
proton and one electron which would be very hard to detect, which he dubbed
a “neutron.”
Discovering the neutron
Prior Experiments
One experiment that helped to lead to Chadwick’s discovery was done
by W. Bothe and H. Becker in which they found that beryllium, when
exposed to alpha particles from polonium, emitted a penetrating radiation
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(more penetrating than any other observed electromagnetic radiation) that
they assumed was a kind of electromagnetic radiation. A later experiment
done by Irène and Frédéric Joliot-Curie further explored this “beryllium
radiation” and found that it ejected protons from paraffin wax (which
contains hydrogen). They calculated that the energy of the electromagnetic
radiation would have to be 55 MeV, far higher than any other observed
radiation. The highest possible energy acquired from the collision between
beryllium and alpha particles would be from the formation of C13 or:
𝐵𝑒 9 + 𝛼 4 → 𝐶 13 + 𝛾
At the time the mass defect of beryllium-9 was not known. Instead,
knowing the mass defect of carbon-13 to be 10 MeV and assuming the mass
defect of beryllium-9 to be 0 (to give a maximum value of the energy
released) yields an energy of 10 MeV. Together with the kinetic energy of the
alpha particle, the max energy that the quantum could have would only be 14
MeV, far less than the required 55 MeV.
Chadwick’s Experiments
Chadwick started out his search for the neutron by exposing other
elements to the “beryllium radiation”. He found that the radiation also ejected
particles from all other light elements which he examined.
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The apparatus used to observe the effects of the beryllium radiation
Chadwick observed that the velocity of the protons released from the
paraffin wax (placed in front of the counter) was 3.3 × 109 𝑐𝑚/𝑠,
corresponding to an energy of 5.7 × 106 electron volts. He then observed the
radiation’s effects on other elements. He did this by affixing the element
(such as lithium, beryllium, boron, carbon, and nitrogen) on a clean brass
plate placed close to the counter opening. He observed that the emitted
particles had very small ranges but produced many more deflections
(collisions) than protons. He thus concluded that the particles were recoil
atoms from these elements. Knowing the velocity of the protons produced by
the paraffin wax, the required energy of the quantum can be calculated by
𝐸
𝐸𝑚𝑐 2
2
2
using the equation ℎ𝑣 = ( ) + √(
)+
𝐸2
2
, (see next page for derivation
of this equation) where E is the kinetic energy of the particle, and this yields
an energy of 55 MeV.
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𝐸
𝐸𝑚𝑐 2
2
2
Derivation of ℎ𝑣 = ( ) + √(
We start out with 𝐸 = 2ℎ𝑣/(2 +
𝑚𝑐 2
ℎ𝑣
)+
𝐸2
2
) (the max kinetic energy given to a
particle by a photon of energy hv during a Compton recoil collision)
𝑚𝑐 2
𝐸 = 2ℎ𝑣/(2 +
)
ℎ𝑣
𝐸 = 2(ℎ𝑣 )2 /(2ℎ𝑣 + 𝑚𝑐 2 )
𝐸(2ℎ𝑣 + 𝑚𝑐 2 ) = 2(ℎ𝑣 )2
2𝐸ℎ𝑣 + 𝐸𝑚𝑐 2 = 2(ℎ𝑣 )2
2𝐸ℎ𝑣 + 𝐸𝑚𝑐 2 = 2(ℎ𝑣 )2
𝐸𝑚𝑐 2 = 2(ℎ𝑣 )2 − 2𝐸ℎ𝑣
𝐸𝑚𝑐 2
= (ℎ𝑣 )2 − 𝐸ℎ𝑣
2
𝐸𝑚𝑐 2 𝐸 2
𝐸2
2
+
= (ℎ𝑣 ) − 𝐸ℎ𝑣 +
2
4
4
𝐸𝑚𝑐 2 𝐸 2
𝐸 2
+
= (ℎ𝑣 − )
2
4
2
𝐸𝑚𝑐 2 𝐸 2
𝐸
√
+
= ℎ𝑣 −
2
4
2
ℎ𝑣 = √
𝐸𝑚𝑐 2 𝐸 2 𝐸
+
+
2
4
2
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Based on this photon energy the energy of recoil nitrogen atoms can be
calculated using the formula 𝐸 = 2ℎ𝑣/(2 +
𝑚𝑐 2
ℎ𝑣
) (the max kinetic energy
given to a particle of mass m by a photon of energy hv in a Compton recoil
collision), which comes out to 450,000 eV. Assuming that it takes 35 eV to
produce a pair of ions in air, there should be no more than 13,000
(450,000/35) ion pairs produced. However, Chadwick’s experiment detected
30,000 to 40,000 ion pairs. This would require over 90 MeV of energy from
𝐸
𝐸𝑚𝑐 2
2
2
the photon (using ℎ𝑣 = ( ) + √(
)+
𝐸2
2
) in order to generate the
required KE for the nitrogen. Similar discrepancies were detected with the
other elements that were tested. Since assuming that the beryllium radiation
is electromagnetic makes the required radiation for different collisions, either
conservation of energy or conservation mass is violated or it is not
electromagnetic.
The Mass of the Neutron and the Nitrogen Paradox
Initial Mass
Chadwick’s initial calculation of the mass of the neutron placed it in
between 1.005 and 1.008 AMU. This placed its mass beneath that of a proton
and electron combined, which supported the theory that it was a composite
between those two particles. The slightly smaller mass was assumed to be the
mass defect (binding energy) which was at a reasonable value of between 1 to
2 MeV.
The Nitrogen Paradox
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Spin had been discovered prior to Chadwick’s discovery of the neutron.
Each particle would have either +1/2 or -1/2 spin. Therefore the spin of a
nucleus with an odd number of particles should be non-integer while the
particle of a nucleus with an even number of particles should be odd. The
nitrogen nucleus, according to the combined proton and electron model
would have an odd number of nuclear particles (14 protons 7 electrons).
However, when the spin of the nucleus was measured it was an integer
number, meaning that it should have an even number of particles in the
nucleus. One possible solution to this problem would be making the neutron
an elementary particle, instead of a composite between a proton and an
electron, as this would make the number of particles in a nitrogen nucleus
even (7 protons and neutrons), and therefore would solve the nitrogen
paradox.
Calculating a more Precise Mass of the Neutron
In order to calculate the mass of the neutron more precisely, Chadwick
decided to use the splitting of deuterium (𝐻12 , or a hydrogen with a neutron).
In order to do this, he utilized a 𝛾 from thorium-C, which was the strongest
source of radiation at the time, producing radiation with an energy of 2.6
MeV. A previous radium source that he tried to use with an energy of 1.8
MeV failed to split deuterium. The equation for the splitting of deuterium is
as follows:
𝐻12 + 𝛾 = 𝐻11 + 𝑛10
Knowing that the difference in mass between deuterium and hydrogen is
1.0058 amu, the possible range of mass of the neutron can be calculated.
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Since the 1.8 MeV quantum failed to overcome the binding energy and the
2.6 MeV did, the binding energy must range from 1.8-2.6 MeV. Assuming
that the binding energy is about 1.8 MeV yields a neutron mass of 1.0077
amu while assuming a binding energy of 2.6 MeV yields a neutron mass of
1.0086 amu. Since the mass of a free proton and electron together is 1.0078
amu, the mass of the neutron is greater than that of a proton and electron and
therefore it is not composed of them and is an elementary particle.
Conclusion
Through his experiments, Chadwick was able to prove that the
beryllium radiation was a neutron and not electromagnetic, and he later
proved its elementary nature by making a precise measurement of his mass.
His discoveries would be vital for the development of future nuclear
technology and models of the atom.
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Works Cited
Brown, Andrew. The Neutron and the Bomb: A Biography of Sir James
Chadwick. Oxford: Oxford UP, 1997. Print.
Weinberg, Steven. The Discovery of Subatomic Particles. New York:
Scientific American Library, 1983. Print.
Wright, Stephen. Classical Scientific Papers: Physics. London: Mills and
Boon, 1964. Print.