Reason for alpha particles

[AntonioLao]

In nuclear disintegration, three types of radiation were discovered. These are, not in any order of importance: (1) the alpha radiation, (2) the beta radiation and (3) the gamma radiation. These radiations could have personified three key famous and so famous scientists in the name of Ralph Alpher, Hans Bethe, and George Gamow. But only Hans Bethe was awarded the 1967 Nobel Prize for Physics crediting his works in nuclear processes of the stars. As often is the case, the middleman always gets the cream of the crop.

But it was George Gamow with possibly the help of Ralph Alpher (his student) who did come up with the most convincing hypothesis as to why alpha particles can escape the strong nuclear force of the decaying atoms. His understanding of alpha particles led him to hypothesize the big bang model of the universe and he calculated the CMBR at 5 degrees Kelvin in 1948. In 1964, not aware of Gamow’s calculation, Arno Allan Penzias and Robert Wilson discovered serendipitously the CMBR at a temperature of 3.5 degrees Kelvin. Penzias and Wilson shared the 1978 Nobel Prize for Physics without so much as mentioning Gamow’s name since no Nobel Prize was ever awarded posthumously. Gamow died August 19, 1968 in Boulder, Colorado.

[Guille]

What does the strong nuclear force do to the quarks of the protons and neutrons to convert them? I am studying beta radiation in physics class at school this week, and I have a question: how can electrons be "created", if mass can't be created or destroyed? are the electrons inside the protons and neutrons and come out of the. Or are they just formed? If so, how can they be "formed"?

[AntonioLao]

how can electrons be "created", if mass can't be created or destroyed?

From a quantum mechanical point of view, everything can be described by wave functions. These are dependent on Cartesian position coordinates and time coordinate as
[math] \psi (x,y,x,t) [/math]
So the electron was never created. Its wave functions were superposed with the wave function of the protons forming the wave function of the neutrons.
The superposition of wave functions can be used to explain the H-effect at the other thread by subversion.

[Guille]

Can two or more types of radiaton happen at the same time? Can the atomic decay be stopped by humans? Is it a possible source of energy? Are there any investigations for these sources of energy?

[AntonioLao]

The splittings of heavy nuclei, e.g. uranium isotopes, produce energy by the process of fission. The joinings of light nuclei, e.g. hydrogen isotopes, produce energy by the process of fusion. Both of these processes require very narrow initial conditions to succeed such as extreme temperatures and extreme densities as well as the purities of the materials.

[mkirkpatrick]

The splittings of heavy nuclei, e.g. uranium isotopes, produce energy by the process of fission. The joinings of light nuclei, e.g. hydrogen isotopes, produce energy by the process of fusion. Both of these processes require very narrow initial conditions to succeed such as extreme temperatures and extreme densities as well as the purities of the materials.

What level of purity are we talking about Antonio,To produce fusion!
And can we in the coming few years actually bringthisabout!I tendto think we will.What do you think Antonio?

kind regards michael

[AntonioLao]

What level of purity are we talking about

Purity as in nothing at all. Nothing in this case means there are no neutrinos in the fusion reactor. My suggestion is that these neutrinos must be artificially introduced into the reactor in order for the fusion process to start.

[mkirkpatrick]

Purity as in nothing at all. Nothing in this case means there are no neutrinos in the fusion reactor. My suggestion is that these neutrinos must be artificially introduced into the reactor in order for the fusion process to start.

If what you say is right Antonio,how in Gods name,would you do that!Can one "artifically" introduce neutrinos!

kind regards michael.

[baudrunner]

Hi, Guys

Guille, when the electron beam of a cathode ray tube strikes the phosphor dots on your tv screen, alpha and beta particles and gamma rays are emitted. Your tv program involves the application of radioactive decay to flouresce the phosphors! Remember radioactive phosphorus?

I think I left the following discussion of fusion somewhere else on the TOEquest site but I'm too lazy to dig it up so I'll just repost it.

In the process known as atomic fusion, ideally two hydrogen atoms fuse to form one Helium atom. As we shall see further in our discussion this is not actually the case in practical fusion applications. If we were to create a Helium atom from two hydrogen atoms and two neutrons, the new nucleus is formed by the combination of two protons and two neutrons. If this were accomplished without a change of mass, the mass of the helium atom would be 4.0333 amu; however, it is only 4.002 amu. This difference represents a loss in mass of 0.031 amu and represents a strongly exothermic reaction. To calculate the energy released this way using Einstein's equation...

m (grams) = 0.031 amu x (1.66 x 10^-24 g/amu)

m = 5.146 x 10^-26 g

c2 = (3.0 x 10^10)² = 9 x 10^20 cm/sec

E = ( 5.146 x 10^-26 ) x ( 9 x 10^20) ...(E=mc²)

E = 4.63 x 10^-5 erg per atom

since 1 erg = 6.24 x 10^11 electron volts...
E in electron volts is = (4.63 x 10^-5) x (6.24 x 10^11 electron volts/erg)

Therefore, E = 28,891,200 electron volts. This is the energy released in our formation of a single Helium atom.

Continuing our hypothetical experiment, it has been shown that there are the same number of atoms of any element in the gram-atomic weight of that element. This number is known as Avogadro's number and is..

6.023 x 10^23 atoms

Since the gram-atomic weight of Helium is 4.0026 the number of atoms in one gram of Helium atoms is..

(6.023 x 10^23 ) / 4.0026 = 1.505 x 10^23 atoms

The energy released in the formation of 1 gram of Helium is therefore..

28,891,200 x (1.505 x 10^23 ) = 4.348 x 10^30 electron volts

Since the force of one dyne is equivalent to 6.24 x 10^11 electron volts then the formation of 1 gram of Helium produces a force of ..

(4.348 x 10^30 ) / (6.24 x 10^11 ) = 6.968 x 10^18 dynes

Or in other words, the work done in ergs..

(1.6 x 10^-12 (eV)) x (4.348 x 10^30 ) = 6.957 x 10^18 ergs

It can be seen that this particular fusion reaction releases a tremendous amount of energy considering that 1 dyne of force accelerates a mass of 1 gram at a rate of 1 cm per sec². A fusion reaction of this type using enough Hydrogen and enough neutrons to yield just 1 gram of Helium would generate enough force to accelerate a one million kilogram mass at the rate of ~7 million kilometres /sec² !

This, however is not typical of the fusion reaction in a hydrogen bomb. In this case, a deuterium ion and a tritium ion, the positive isotopes of the heavy isotopes of hydrogen, undergo fusion to form a Helium nucleus and a neutron. This releases energy which is ideally about 40% less by calculation than that of our example - still considerable, but in reality it is actually less than that since the conversion is far from perfect. The sun's energy is primarily generated by the fusion of four Hydrogen nuclei into one Helium nucleus. The energy released in this process is only about 7% less than that of our hypothetical experiment.

[AntonioLao]

Can one "artifically" introduce neutrinos!

Yes. But we would need a neutrino factory of sort.