-
---------------------- 2709 - ATOMS - measuring how atoms work?
-
---------------------------- The energy level at each orbit for hydrogen is:
-
-------------------- n = 1 = -13.6 electron volts
-
-------------------- n = 2 = -3.4 electron volts
-
-------------------- n = 3 = -1.51 electron volts
-
-------------------- n = 4 = -0.85 electron volts
-
-------------------- n = 5 = -0.54 electron volts
-
- Note that for each orbit of the electron the energy level is higher. The ground level for the electron is -13.6 volts. For an electron to jump from the second orbit down to the ground level orbit it would have to emit +10.2 electron volts of energy.
-
- The frequency of the photon radiation being emitted would be 118 nanometers wavelength, far into the ultraviolet. When astronomers see this wavelength in the spectrum of light they are observing they know that they are looking at hydrogen gas.
-
----------------------------- 10.2 eV, E = h*f , E/f = h , E/f = 4 * 10^-15
-
----------------------------- h = 4.1356692*10^-15 eV*seconds.
-
- E/f is equal to a constant. Planck’s Constant of Action. Action is the product of Energy and Time, therefore, units eV * seconds. Note that the constant is a very small number (10^-15). Note that if Energy increases frequency increases in order to keep the ratio, E/f, constant.
-
---------------------------- f = 2.55 * 10^15 cycles per second.
-
---------------------------- f*w = c = constant speed of light = 3*10^8 meters per second.
Note that as frequency increase wavelength decrease in order to keep the product constant.
-
------------------------- Wavelength = 118 nanometers, which is in the far ultraviolet.
- So, what determines these exact energy levels and why does the electron only have these specific orbits? And, why does each atom only emit specific frequencies of radiation?
-
- For the simplest atom, hydrogen, we can use some classical physics to determine the answers to these questions. The force that is pulling the electron toward the nucleus is the electric force of opposite charges. The force that is pushing the electron away from its orbit is the centripetal force of angular momentum.
-
------------ Electric Force = (Q1 * Q2 / r^2) * 9,000,000,000
-
------------ Electric Force = (e^2 / r^2 )* 9,000,000,000
-
- The Electric Force is the product of the charges divided by the square of the distance between them. Q1 and Q2 are the charges of the proton and the electron which are the same = “e”.
-
- In this case the charge is the electron charge squared = e^2 = 2.6*10^-38 coulombs^2.
-
------------------------ Electron charge is 1.60217733 * 10^-19 coulombs.
-
------------------- Electron charge squared = 2.566972197*10^-38 coulombs squared
-
- The distance is the radius of the orbit. 9*10^9 is the Coulomb’s Constant of proportionality, “K“.
-
- K = ¼*pi*Eo = 8.987552*10^9 newtons * meters^2 / coulombs^2. Eo is the electric permittivity in a vacuum.
-
------------ Electric Force = 9*10^9 * 2.6*10^-38 / r^2
-
------------ Electric Force = 23*10^-29 / r^2
-
------------ If we knew the radius we could calculate the Electric Force. The Electric Force * the radius^2 is equal to a constant, very small number.
-
------------- Centripetal Force = m*v^2 / r
-
------------- Centripetal Force = 9*10^-31 * v^2 / r
-
- Centripetal Force is the force of angular momentum. Force is the product of mass and velocity^2 per distance of the radius of orbit. “m” is the mass of the electron = 9*10^-31 kilograms.
-
------------------------ m = Mass of the electron = 500,000 electron volts /c^2 = 9.1093897*10^-31 kilograms.
-
-------------------------- m^2 = 83 * 10^-62 kg^2
-
- The electron is in a constant circular orbit therefore the two forces, electric and centripetal, must be equal and opposite:
-
----------------------- Centripetal Force = Electric Force
-
----------------------- 9*10^-31 * v^2 / r = 23*10^-29 / r^2
-
----------------------- r * v^2 = 254
-
- The product of the radius and the velocity squared is a constant. If the radius increases the velocity decreases in order to keep the product constant, 254.
-
- Next we use the laws of Conservation of Angular Momentum and Conservation of Energy.
-
------------ Angular Momentum = m*v*r
-
------------ Angular Momentum = mass of electron * velocity* radius.
-
------------- Angular Momentum^2 = m^2 * v^2 * r^2
-
------------- Angular Momentum^2 = 83*10^-62 * 254/r * r^2
-
------------- Angular Momentum^2 = 2*10^-58 * r
--------------- Angular Momentum^2 / r = 2*10^-58.
-
- The ratio of angular momentum squared to radius is a constant. As the radius increases the angular momentum, which is the velocity of the orbiting electron must also increase in order to keep the ratio constant, 2*10^-58.
-
- So, the electric force equation and the angular momentum equation have the opposite effect on the electron’s velocity. That is how the two forces can become in balance.
-
------------ Total Energy = Angular Kinetic Energy + Electric Force Energy.
-
- Since the Centripetal Force and Electric Force are equal and opposite, and since Energy is the product of change in force causing a change in distance, dE = dF * ds.
-
--------------------Angular Kinetic Energy = ½ m*v^2.
-
- Instead of using calculus and intergrating the Energy function over distance we will assume the function to have an average of ½.
-
----------- Total Energy = Kinetic Energy - Electric Energy
-
----------- Total Energy = + 1/2m*v^2 - Electric Force * Radius
-
------------ E = -½ * (23*10^-29 / r^2)* r
-
----------- E = 12*10^-29 / r
-
- We should be able to use this equation to calculate the energy level for every radius of orbit. However, these equations suffer one major flaw. The electron is a charged particle and in a circular orbit it is constantly accelerating. An accelerating charged particle radiates energy.
-
- Consequently an orbiting electron would steadily lose energy by radiation, the radius would continually decrease and the electron would spiral into the nucleus. This would happen in 10^-10 seconds.
-
- Just like the Blackbody radiation problem that Max Planck and Albert Einstein solved, we need quantum mechanics to solve this problem too. Energy is not continuous. Energy comes in quantum’s. The radius orbits for the electrons are not continuous. Angular momentum is not continuous. They all come in quantum steps.
-
- At the atomic level everything is digital and it just seems continuous at the macro-level that we see. In order for electrons to not radiate when they circle the nucleus they have to reside in integer wavelengths to create a complete standing wave in their orbits.
-
- This is what creates the discrete orbits for every atom. Each orbit has a circumference that is multiples of ½ wavelength. In half wavelengths a wave starts and zero, goes through a maximum, and ends at zero. This allows the electron to accelerate around this particular orbit without radiating away any energy.
-
- All the allowable orbits have ½ wavelength standing waves in their circumference. A wavelength is Planck’s Constant of Action / momentum
-
- Angular momentum = an integer of Planck’s constant per 2*pi wavelengths.
-
--------------------------- h/2*pi = 1.0545727 *10^-34 joule * seconds
-
---------------------------- h/2*pi = 6.582121*10^-16 eV * seconds
-
---------- 2*pi radians is one wavelength. “n” is an integer = 1, 2, 3, 4, etc.
--------- Angular Momentum = n*h / (2*pi)
-
--------- Angular Momentum = n * 10^-34
-
--------- Angular Momentum^2 = n^2 * 10^-68
-
- From our previous calculation:
-
--------- Angular Momentum^2 = 2*10^-58 * r
-
- In the case of the hydrogen atom and the ground level orbit where n = 1
-
--------------------------10^-68 = 2*10^-58 * r
-
-------------------------- r = .5 * 10^-10 meters.
-
- Double the radius and get the diameter of an atom 10^-10 meters, or 0.1 nanometers.
-
- From our previous calculation:
-
----------- E = 11.5*10^-29 / r
-
----------- E = 11.5*10^-29 / .46 * 10^-10
-
----------- E = 24 * 10^-19 joules
-
---------- 1 eV = 1.6022*10^-19 joules
-
------------E = 15 eV
-
- The energy level of orbit 1 in the atom is -15 electron volts. These equations used “Classical” physics. “Newton’s” physics. Our result is 12% in error, because ground level for an atom is - 13.6 electron volts.
-
- In order to calculate a more accurate answer we must use calculus and quantum mechanics equations. Also, our calculations only work on the simplest atom, hydrogen. As soon as we add more electrons and more protons the multiple interactions become too difficult to calculate.
-
- However, we did discover and interesting result, that the quantization of angular momentum is the result of the circumference of the electron orbit accommodating and integral number of electron wavelengths. Only in these orbits is the atom stable and the electron not radiating energy.
-
------------ Angular Momentum = Integer * h / 2*pi
-
------------ Angular Momentum = Integer * 6.6*10^-16 eV*seconds.
-
- There is an observational way to measure the energy levels inside an atom. And, all atoms are different, that is why we have different elements in the Periodic Table. From hydrogen to iron we can determine the energy levels between orbiting electrons by measuring the frequency of the photons that are emitted or absorbed when an electron jumps between orbits.
-
- Frequencies are one of the most precise measurements we know how to make. Frequencies are the basis for atomic clocks, lasers, interferometers, the most accurate measuring instruments.
-
- The photons are always at specific frequencies unique to the change in energy level, dE.
-
-------------------- dE = h*f
-
-------------------- dE = Planck’s Constant of Action * frequency
-
------------------- dE = 6.6*10^-34 * f
-
------------------ dE / f = 6.6*10^-34
-
- The ratio of the change in energy to the frequency is a constant, 6.6*10^-34.
-
- Planck’s Constant of Action = 6.6260755*10^-34 (kilogram * meters^2 / second^2) * seconds, or joule*seconds Action is the product of Energy and Time.
- If the energy increase the frequency increases in order to keep the ratio constant at 6.6*10^-34. Note that the product is a very, very small number.
- If a bottle of hydrogen gas is at room temperature virtually all the atoms in the bottle would be at their lowest energy level, called the ground level state, -13.6 electron volts.
-
- At room temperatures ordinary thermal collisions between atoms do not supply enough kinetic energy to excite the atom to the next higher energy level. The excited state for the atom would be for the electron to have jumped from the ground level state to the second orbit.
- If the gas is heated up, thermal energy, which is the kinetic energy of moving atoms, could kick the electron to its higher energy level. To do this the electron has to absorb a discrete amount of energy.
-
- When the excited electron falls back to the lower energy level it will emit a discrete amount of radiation energy that is exactly the same.
-
- We calculated the energy levels for the hydrogen atom:
-
------------------- Orbit 1 ----------- -13.6 electron volts.
-
------------------- Orbit 2 ----------- - 3.4electron volts.
-
------------------- Orbit 3 ----------- - 1.51 electron volts.
-
------------------- Orbit 4 ----------- - .85 electron volts.
-
- When the electron drops from the second orbit to the first orbit it releases +10.2 electron volts in the photon that is emitted.
-
----------------------- dE = - 3.4 - (-13.6) = + 10.2 electron volts
-
----------------------- dE = 6.6 *10^-34 * f = 10.2 electron volts
-
------------------------ 1 electron volt = 1.6 *10^-19 joules
-
---------------------- f = 2.5*10^15 wavelengths / second
-
--------------------- wavelength * frequency = speed of light
-
-------------------- wavelength = 3.10^8 meters/second / 2.5*10^15 wavelengths / second
-
------------------- wavelength = 83 nanometers, which is in the ultraviolet
-
------------------- wavelength * Temperature = 0.0029 m*Kelvin
-
- The product of wavelength and temperature is a constant 0.0029. If wavelength increases than temperature must decrease. If wavelength decreases than temperature must increase in order to keep their product constant at -0.0029. (See Review 2563, “The Temperature of Light, for an explanation of Boltzman’s Constant“)
-
----------------------- Temperature = 35,000 Kelvin
=
- So when astronomers measure this wavelength in the light from the heavens they know they are looking at hydrogen gas and that its temperature is above 35,000 Kelvin. This calculation also confirms that all of the hydrogen gas in the glass bottle is at the ground state energy level since room temperature is 300 Kelvin.
-
- This type of observation and measurement works for all electromagnetic radiation coming from atoms. For example let’s take a laser that is specified to have a coherent light wavelength of 339.1 nanometers.
-
----------------------- dE = h*c / wavelength
-
----------------------- dE = 6.6*10^-34 * 3*10^8 / 3.391*10^-7
-
----------------------- dE = 3.6 electron volts
-
- The energy gap between the excited states for the laser material is 3.6 eV. If the physicists makes enough of these type measurements on a particular material the entire spectrum of emissions and absorptions can be determined for the atoms of that particular element.
-
- This is how astronomers find the elements in the distant stars and gas nebulae. My math is very rudimentary, but, it serves as an illustration of how the measurements are done to understand the workings of the ubiquitous atoms. Hope you enjoyed the journey.
-
- April 17, 2020 983 985 2709
----------------------------------------------------------------------------------------
----- Comments appreciated and Pass it on to whomever is interested. ----
--- Some reviews are at: -------------- http://jdetrick.blogspot.com -----
-- email feedback, corrections, request for copies or Index of all reviews
--- to: ------ jamesdetrick@comcast.net ------ “Jim Detrick” -----------
- https://plus.google.com/u/0/ -- www.facebook.com -- www.twitter.com
--------------------- Friday, April 17, 2020 -------------------------
-----------------------------------------------------------------------------------------
No comments:
Post a Comment