- 3080 - ATOMS - Stability with Uncertainty? The stability of all atoms, in all the elements, in all of matter, depends on the “Principle of Uncertainty” and the “wave-particle duality” of all matter. How can stability depend on uncertainty?
--------------- 3080 - ATOMS - Stability with Uncertainty
- This review will use the Uncertainty Principle to calculate the ground energy state of the atom. The “rest state” of all atoms can not be zero. If it were zero all atoms would be unstable. Instead, the rest state of all atoms is some specific energy level we can express in electron-volts.
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- All atoms have a positively charged nucleus of protons and negatively charged electrons orbiting around the nucleus. The attractive electric charge is pulling the electron into the nucleus, much like the force of gravity pulling on an orbiting satellite.
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- The electron does not collapse into the center due to its angular momentum in orbit. Even at rest, the ground state of an atom’s energy level is not zero. The energy of an electron’s angular momentum is always positive.
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- This review will calculate the energy ground state of a hydrogen atom, one proton and one electron, using centripetal force, using “Conservation of Angular Momentum“, using the energy of the Electric Charge, using the “Conservation of Energy“, and using the Uncertainty Principle.
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- In the first case lets treat the electron as a particle at the end of a spring. The spring force is pulling it in and the angular momentum is pushing it out. The two are in balance in the stable atom.
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-------------------- The Kinetic Energy of the orbiting electron can be expressed as:
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-------------------- KE = ½ mass * velocity squared.
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------------------- KE = ½ m*v^2
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-------------------- Momentum = mass * velocity, p = m*v.
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-------------------- Expressing Kinetic Energy as a function of momentum.
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------------------- KE = ½ * p^2 / m
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---------- The elastic energy of the spring, which is potential energy, must be equal to the centripetal energy which is centripetal force * distance, in this case the distance is the radius of the electron’s orbit.
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------------------ Potential Energy = ½ mass * radian velocity^2 * radius^2
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------------------ PE = ½ m*w^2 * r^2
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------------------ Total Energy = Kinetic Energy + Potential Energy
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----------------- Energy = KE + PE
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----------------- Energy = ½ * p^2 / m + ½ m*w^2 * r^2
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- The Uncertainty Principle states that the uncertainty of the electron’s momentum,(dp), times the uncertainty of the electron’s location,(dr), is greater than Planck’s Constant of Action ,(h). In this case the electron’s location is its radius of orbit.
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--------------- dp * dr = h
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-------------- h = 1.05 *10^-34 joule-seconds
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-------------- dp = h / dr
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------------- p = h / r, substituting this into the energy equation in order to get energy as a function of momentum and location which the terms in the Uncertainty Principle:
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-------------- Energy = ½ * h^2 / m *r^2 + ½ m*(w*r)^2
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- This equation for total energy of the atom is Energy as a function of radius. If we study this equation we can see that as the Kinetic Energy decreases the Spring Energy must increase because the total energy is constant according to the Conservation of Energy.
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- Further, we can see that if the radius increases the Kinetic Energy decreases and the Spring Energy increases. Therefore the shape of the curve plotting energy versus radius must be an upside down “ U” because as the radius increases the Kinetic Energy decreases, it goes through a minimum where the slope is zero, and then increases as the Spring Energy increases with increasing radius.
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- From calculus we know that the first derivative of a function is the slope of that function, and the slope of the minimum point is zero.
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-------------- Energy = ½ * h^2 / m *r^2 + ½ m*w^2* r^2
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-------------- 1st derivative = dE = - h^2 / m * r^3 + m*w^2^r
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-------------- setting dE = 0, to find the radius where energy is minimum
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-------------- h^2 / m * r^3 = m*w^2^r
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-------------- r^2 = (h/m*w) at the minimum energy point.
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- Substituting the radius at minimum back into the energy formula to determine the Energy level at minimum we get:
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-------------- Energy at minimum = ½ * h^2 / m *(h/m*w) + ½ m*w^2*(h/m*w)
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-------------- Energy at minimum = w*h
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-------------- Energy at minimum = w*10^-34 joule * seconds
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- “w” is the radian velocity of the electron’s orbit in radians per second.
This times 10^-34 joule* seconds is very small but always a positive number of energy in joules.
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- Therefore, we can conclude that the minimum energy level can not be zero. It is some positive number depending on the radian velocity. But, we don’t know the radian velocity, so, let’s try another approach using the electric force in our calculations.
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- We can also assume that the stability of the atom is due to the balance of the centripetal force of the spinning electron and the electric force.
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--------- Centripetal Force = Electric Force
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----------- Centripetal Force = mass * velocity^2 / radius.
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------------ Centripetal Force = m*v^2/ r
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----------- We can calculate this formula from Force = mass * acceleration, F=m*a.
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----- ----- Angular acceleration = radius * radial velocity^2.
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----------- Radial velocity = tangential velocity / radius.
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----------- Therefore, acceleration = tangential velocity^2 / radius. a = v^2 /r.
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------------ where: m = mass of the electron = 9.11*10^-31 kilograms
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---------- Electric Force = e^2 / 4*Pi* Eo*r^2
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---------- where: e = charge of the electron = 1.6*10^-19 coulombs
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---------- Eo = permittivity of free space = 8.85*10^-12 coulombs^2*sec^2 / kg*m^3
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--------- Centripetal Force = Electric Force
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------------- m*v^2/ r = e^2 / 4*Pi* Eo*r^2
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------------ v^2 = (e^2 / 4*Pi*Eo*r*m)
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- Solve for v^2 and calculate the Angular Momentum as a function of radius. Angular Momentum = mass * velocity * radius.
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------------ L - m*v*r
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------------ L^2 = m^2 * v^2 * r^2
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------------ L^2 = m^2 * r^2 * (e^2 / 4*Pi*Eo*r*m)
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------------ L^2 = m * r * e^2 / 4*Pi*Eo
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- Substituting the values in for the constants from the above paragraphs:
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----------- L^2 / r = 2.1*10^-58 kg^2*m^3/sec^2
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- The ratio of angular momentum to radius is a constant. When radius increases momentum must also increase and vice versa. It also says that momentum varies continuously with the radius.
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- Here is where we encounter another problem. The electron is a charged particle and an electric charge in motion radiates energy . That is how all the electromagnetic energy surrounding us gets generated.
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- But, if this were true then the electron’s angular momentum would lose energy through radiation and the electron would spiral down and crash into the proton. Other calculations determine that this crash into the proton would take 10^-10 seconds. All the atoms would become neutral neutrons. All matter would be in the form of a neutron star. We must be missing something?
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- We can not treat the electron just as a particle. Remember, the wave-particle duality says that the electron is also an orbiting wave. If the length of orbit is exactly equal to the integer half-wavelengths of the electron than the electron is a standing wave in a stationary orbit.
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- An electron in a “stationary state” does not radiate energy. Therefore, the energy in these stationary states must take on discrete values. They can not be continuous.
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- Electrons orbiting the nucleus in a stationary state are standing waves. You can visualize a standing wave as a rope loosely tied to a wall. You shake the rope up and down. If you time your shaking up and down just right you can get the rope to oscillate at exactly ½ wavelengths.
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- If you double your rate of shaking up and down you can get the rope to oscillate at a full wavelength. And, so on. In order for the electrons to have stationary states in orbit the length of orbit must be in integer multiples of half-wavelengths.
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- The energy of the stationary states occur in values of integer multiples of Planck’s Constant of Action. In order to calculate the energy level of an atom with its stationary state at “ground level”, an integer multiple of 1, the angular momentum must be equal to 1.05*10^-34 joule-seconds.
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- From our formula for angular momentum, substituting Planck’s Constant for L:
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------------ r = (1.05*10^-34 joule-seconds)^2 / 2.1*10^-58 kg^2*m^3/sec^2
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----------- r = 0.5*10^-10 meters
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- This is then the radius of a stable orbit 0.5*10^-10 meters. This is also the diameter of an atom, 10^-10 meters, or 0.1 nanometers. We can substitute this value for the radius back into the Energy formula to determine the energy of an electron in the ground state of every atom:
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---------- Electric Force = e^2 / 4*Pi* Eo*r^2
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---------- Electric Energy = Force * distance = e^2 / 4*Pi* Eo*r
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---------- Electric Energy = - 13.6 electron volts
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- It is 13.6 eV that is holding an electron in orbit. That is also how much energy it takes to knock an electron out of its ground state and to separate the electron form the nucleus.
The electron is said to be a free electron and the nucleus is said to be ionized with its positive charge.
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- Energy of radiation, of photons, is equal to Planck’s Constant * frequency, E=h*f.
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- 13.6 eV of energy has a frequency of 3,313,000,000,000,000 cycles per second.
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- Wavelength = speed of light / frequency, which is 90 nanometers wavelength. When astronomers find photons of light emitted at this specific wavelength they know it is coming from ionized hydrogen.
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- Heavier elements have different energy levels for each orbiting electron. When an element is ionized and its radiation is detected, the wavelength will identify which element that is. Hundreds of elements and molecules have been identified in outer space by astronomers.
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March 6, 2021 ATOMS - Stability with Uncertainty? 1032 3080
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--------------------- --- Saturday, March 6, 2021 ---------------------------
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