Chapter 30
Atomic Physics
Atoms are made of three fundamental particles: electrons, protons, and neutrons. The general shape of atoms is spherical. Each atom has a nucleus that is made of a number of protons and neutrons. The nucleus is of course at the center of the sphere of the atom. The electrons of each atom spin around its nucleus at certain radii. The size of an atom is determined by the volume or the space in which its electrons require to spin. The volume they require depends on the energies of the electrons that means their temperature that translates to their speeds. The higher the temperature, the faster electronic motion and the greater the space or volume they require to spin in. If we could bring the temperature of an atom down to absolute zero to where the electronic motion seized, each negative electron would then collapse into a positive proton at the nucleus to form a neutron. But, we have not been able to do that so far!
The atomic space or volume is extremely empty:
The size of electron is much smaller than the size of the atom itself. To have an idea of the emptiness of an atom, a hydrogen atom for example, picture a sphere 6 to 7 feet in diameter. A 6-ft tall person can comfortably stand in such sphere. The majority of hydrogen atoms have just one proton at their nucleus and one electron that orbits that proton. In that 6ft-diameter sphere, the nucleus proton is like the tip of a very sharp needle at the center and the orbiting electron is like the tip of a very sharp needle moving on its surface! This shows how empty each atom is. The ratio of the radius of an atom (the radius of the sphere in which electrons spin) to the radius of electron or proton, or neutron is roughly 100,000. This means 10^{5} power. Since the formula for the volume of sphere is (4/3)πR^{3}, the volume of each atom is some 10^{15} (that means one thousand trillion) times greater than the volume of each electron or proton. That speaks of how empty the inner atomic space is.
One might ask why doesn't the electron collapse into the proton under the Coulomb force? The answer is again the electron energy, simply its speed or its temperature. It is extremely hard to shrink the volume of an atom to zero by forcing its electrons into its protons although the Coulomb attraction force is present there to help. Of course, we are not talking about a single atom. A piece of iron with a mass of 56 grams has a volume of about 7.2cm^{3}. In such volume there exist some 6x10^{23} iron atoms. We do not have any device that can compress this iron piece to a near zero volume despite knowing that each atom of it is extremely empty. It takes the gravity of a neuron star (a black hole) in the outer space to do that. A neutron star has so much accumulation of matter that creates an un-imaginable gravity in itself. Its gravity is so strong that it shrinks the atomic volume of other atoms that are easily attracted to it to zero. It is under such gravitational force that negative electrons and positive protons combine to form neutral neutrons. That's why they are also called "neutron stars." As they attract other atoms and convert them to neutrons, they become huger neutron stars. We may think of a neutron star as real solid matter. All atoms in our planet and solar system are fuzzy and extremely empty!
The motion of electrons around the nuclei of atoms is associated with K.E. If v is the average speed of an electron as it spins around its nucleus at a certain average radius r, its K.E. = (1/2)Mv^{2}. An electron, being a negative charge, is also in the electric field of the positive nucleus. The P.E. of a charge q_{1} in the field of another charge q_{2}, is U_{e} = kq_{1}q_{2} /r, where k = 8.99x10^{9} Nm^{2}/C^{2} is the Coulomb's constant. For the proton and electron of a hydrogen atom, the potential energy is U_{e} = -ke^{2}/r. One charge is +e and the other -e.
It is possible to determine the radius r of an atom by solving an energy balance equation. This way of determining the size and shape of an "orbital" (the space around the nucleus of an atom where electrons can be traced) is one basis for quantum mechanics calculations. In quantum mechanics, probability and statistic play an important role. This is because of the extremely high speeds that electrons have in that extremely small atomic space around the nucleus. We can only calculate the probability of finding the moving electron at a desired distance from the nucleus. We cannot exactly determine where it is at a given instant! A prior knowledge of probability and statistic is a must for an interested person in this field.
Here, we are going to discuss the hydrogen atom that is the simplest one.
Calculation of the Radius of Hydrogen Atom:
It can be experimentally verified that it takes 13.6eV of energy to remove the electron from a hydrogen atom when the electron is in its ground state (closest to the nucleus). If somehow the electron is already at a higher shell (farther from nucleus), then it takes less energy to remove it from its nucleus or atom. This means that if the electron of hydrogen atom is in its ground state (closest to the nucleus), it takes at least 13.6eV to detach it from its nucleus (atom). The electron energy is the sum of its K.E. and P.E. . The electron energy is negative because we have to spend energy to remove it from its nucleus. We may write:
P.E. + K.E. = -13.6 eV or
- ke^{2}/r + (1/2) Mv^{2} = -13.6 eV. (1)
In Equation (1) , we already know the values for k, e, and M. You may want to verify this by examining the terms in (1). The only quantity we don't have a value for is v or v^{2}. Knowing v^{2} , we can then solve for r the radius of hydrogen atom.
v^{2} can be found by understanding that the attraction force between the proton (+e) and the electron (-e) is the force that provides the needed centripetal force for circular motion of the electron around the proton. See Fig. 1.
Figure 1 r = 0.53x10^{-10}m. The diameter of the H-atom at its ground state becomes: 1.06x10^{-10}m. |
We define another unit of length called "Angstrom (Å)" to be 10^{ -10}m. Å is 10 times smaller than nanometer. For example the red and violet wavelengths that are 700nm and 400nm are 7000 and 4000 Angstroms, written as 7000Å and 4000Å.
As electrons spin around the nuclei of atoms, they receive energy by many means. If an electron receives energy, its K.E. increases and therefore has to change its orbit and jump to an orbit of a greater radius. A higher energy level of electron corresponds to a greater radius of rotation.
The possible radii are discrete and not continuous. This means that the radius at which electron rotates can not have just any arbitrary value. The reason for discreteness is the electron's wavy motion. Electron has also a wavy motion along any circular path that it is in. At each discrete radius that electron moves, the circumference of that orbit must be an integer multiple of a certain wavelength.
Fig. 2 shows two of such radii. Suppose an electron is orbiting at radius r_{1}. The length of its circular path must be 3λ_{1} as shown. It means that the circumference of circle r_{1} must accommodate exactly 3 of a certain wavelength like λ_{1} in it. When this electron receives a certain amount of energy, it jumps to a bigger radius orbit. The new circumference must again be such that it accommodates an integer multiple of another certain wavelength in it. The possible integer multiples are: 4λ_{2}, 5λ_{3},_{ } 6λ_{4}, and so on depending on the amount of energy received. Only 4λ_{2} is shown here.
Figure 2
The energy levels are discrete the same way that the radii are. Any radius corresponds to a certain energy level particular to that radius. Since there are many many possible radii, there are many many energy levels accordingly. When, for whatever reason, an electron jumps from a certain level to a higher energy level, its original energy level will be vacant. It will be quickly filled by another electron. The electron that fills the vacant position, loses some energy because it fills a lower energy level.
In general, in filling a vacant position, an electron goes from the m-th level at energy E_{m} to a lower n-th level at energy E_{n}. The excess energy E_{m}- E_{n} will be released as a pulse of electromagnetic radiation or simply a photon of light.
Max Planck showed that the frequency f of the emitted light is proportional to this energy difference E_{m}- E_{n} with a proportionality factor h that is named the "Planck's Constant." The Planck's formula is therefore,
E_{m} - E_{n} = hf . (1)
The value of Planck's constant is h = 4.14x10^{-15} eV ∙ s. The above three paragraphs summarizes the theory behind light generation.
Example 1: Convert the Planck's constant to to its SI unit of J ∙ s .
Solution: Verify that you get h = 6.626x10^{-34} J ∙ s by replacing eV by its Joule equivalent.
Example 2: Calculate the energy of a UV ray that has a wavelength of 105nm.
Solution: Using E_{m}- E_{n} = hf and writing it as simply E = hf and also replacing f by c/λ from c = f λ , we get: E = hc/λ. Substituting the values, we get:
E = hc/λ = 4.14x10^{-15}J ∙ s (3.00x10^{8}m/s)/(105x10^{-9}m) = 11.8eV
Emission and Absorption Spectra:
When a gas like hydrogen is under a high enough voltage, its electrons separate from the nuclei of its atoms and are pulled toward the positive pole of the external voltage source while the positive (ionized) nuclei move toward the negative pole of the source. During this avalanche-like motion in opposite directions, many different atomic formations and separations (de-ionizations and ionizations) between the electrons and nuclei occur. As long as the external voltage is kept on, this process continues. The transitions of electrons from many different levels to other many different levels generate many different frequencies. See Equation (1). Different frequencies, of course, means different colors of light. Not all the generated frequencies are in the visible range. In fact a very high percentage of them fall either in the ultraviolet range or in the infrared range that we can not see.
The gas becomes hot due to such transitions. A hot gas emits light. The spectrum of a hot (excited) gas is called the "emission spectrum." The same gas, when cold, absorbs all such emissions. A cold gas has "absorption spectrum." You will observe the emission spectrum of hydrogen and helium gases in a physics experiment. A high voltage source for gas excitation, a low pressure hydrogen tube, and a spectrometer are enough to see the emission spectrum of hydrogen atoms, for example.
The following equations work well for the hydrogen atom for which the number of its protons is Z = 1.
r_{n} = (0.53x10^{-10}m) n^{2} /Z and
E_{n} = - (13.6 eV) Z^{2 }/ n^{2}
where r_{n} is the radius of the n-th orbit or shell and E_{n} is its energy. For atoms that have more protons and consequently more electrons, the calculations are more involved, and require higher levels of mathematics. In such calculations, the repulsion of electrons and their interactions with each other must also be taken into consideration.
Example 3: Calculate the frequency and wavelength associated with electron transition (a) from 5th shell to 2nd shell, (b) from 4th shell to 2nd shell, and (c) from 3rd shell to the 2nd shell of the hydrogen atom. Note that in the formula for energy levels E_{n}, Z is the number of protons.
Solution: For hydrogen Z =1. Equation E_{n} = -(13.6 eV) Z^{2}/n^{2} becomes:
E_{n} = {-13.6/n^{2}}eV. For Part (a) we need to calculate E_{5} - E_{2} and set it equal to hf according to Planck's formula and solve for frequency f .
E_{5} - E_{2 }= {-13.6/5^{2} - (-13.6/2^{2}) }eV = -13.6[1/25 -1/4] eV = 2.86 eV.
Since E_{5} - E_{2 } = hf_{52} ; 2.86eV = hf ; f = 2.86eV/(4.14x10^{-15}eV∙ s)
f_{52} = 6.91x10^{14}Hz. Verify all calculations.
Since c = f λ; therefore, λ_{52} = c/f_{52} = 3.00x10^{8}(m/s)/6.91x10^{14}/s = 434nm.
This transition is associated with the emission of blue light.
(b) & (c): Show that you get λ_{42 }= 487nm (turquoise) & λ_{42 } = 658nm (red).
Questions:
1) In most atoms, the ratio of atomic radius to electron radius is close to
(a) 100 (b) 10,000 (c) 100,000 (d) 10 click here.
2) The atomic size is the size of
(a) nucleus (b) neutron (c) space determined by the electronic cloud.
3) The electronic cloud in a hydrogen atom is caused by
(a) a large number of electrons randomly flowing around the nucleus
(b) a single electron spinning so fast that it appears everywhere
(c) a fuzz of dust click here.
(d) two electrons orbiting opposite to each other's direction.
4) The electronic cloud in a H_{2} molecule is caused by
(a) a large number of electrons randomly flowing around the nuclei
(b) two fast-moving electrons spinning in opposite directions repelling each other
(c) a fuzz of dust
(d) an electron orbiting opposite to proton's motion. click here.
5) The energy of an atom is the energy of its
(a) protons (b) neutrons (c) electrons with respect to its protons
(d) the K.E. of its electrons plus the P.E. of its electrons with respect to its nucleus.
6) The P.E. of charge q_{1} with respect to charge q_{2} a distance r from it is
(a) -kq_{1}q_{2}/r^{2} (b) -kq_{1}q_{2}/r (c) -kq_{1}/r^{2} (d) -kq_{2}/r^{2}. k is Coulomb's constant.
7) If v is the speed of electron and M its mass, its K.E. is
(a) Mv (b) Mv^{2} (c) 1/2Mv^{2} (d) M_{e}gh. click here.
8) 13.6 eV is the
(a) min. energy for ionization of a hydrogen atom from ground state
(b) max. energy for ionization of a hydrogen atom from ground state
(c) average energy for ionization of a hydrogen atom click here.
(d) average energy for ionization of all hydrogen atoms in a tube.
9) The P.E. of the electron in a hydrogen atom is
(a)-ke^{2}/r^{2} (b)-ke^{2}/r (c)-k/r^{2 }(d)-k^{2}/r^{2}. k is Coulomb's constant.
10) Using the Coulomb force F = ke^{2}/r^{2} as the centripetal force, we may write:
(a) ke^{2}/r^{2} = Mv^{2}/r (b) ke^{2}/r^{2} = Mv/r (c) ke^{2}/r = Mv^{2}/r.
11) The diameter of hydrogen atom is approximately
(a) 1nm (b)10nm (c)0.1nm (d) 0.2nm. click here.
12) The radius of hydrogen atom is approximately
(a)1 (b)10 (c) 0.1 (d) 0.53 Angstroms.
13) The wavelength of red light is
(a) 40nm (b)400nm (c) 4000nm (d)700nm.
14) The wavelength of violet light is
(a)70 (b)700 (c)7000 (d)4000 Angstrom.
15) An electron around the nucleus of an atom orbits at
(a) a fixed radius that never changes click here.
(b) a variable radius that can have any value
(c) a variable radius that can have certain discrete values.
16) The radius of rotation of electron about the nucleus is such that
(a) the P.E. of the electron equals its K.E.
(b) the P.E. of the electron equals (1/2) its K.E.
(c) (1/2)the P.E. of the electron equals its K.E.
(d) an integer number of a wavelength fits in its wavy path of motion.
17) The discreteness of orbit positions (radii) in atoms is for the fact that
(a) electrons have wavy motion as they orbit the nucleus and the wavelength of their wavy motion must fit a whole number of times in their circular path
(b) electron position is fixed
(c) electrons repel each other
(d) P.E. is discrete by itself.
18) The discreteness of electron energy (energy levels) is because of
(a) the discreteness of the atomic radii
(b) the discreteness of P.E.
(c) the discreteness of K.E.
(d) equalization of P.E. and K.E. around atoms.
19) Light is generated when
(a) a higher energy electron fills a vacant energy level and loses some energy
(b) an electron is energized and moves to a higher possible orbit
(c) an electron stops moving
(d) none of the above. click here.
20) The color of an emitted photon of light depends on
(a) the energy difference between the energy of an excited electron and the energy of the level it fills up
(b) its frequency of occurrence
(c) Its wavelength
(d)all of the above. click here.
Problems:
1) Find the wavelength of the emitted light associated with the electron transition from shells 4, 3, and 2 to Shell 1 of hydrogen atom. In each case state whether the emitted radiation is visible, uv, or infrared.
2) An instants that hydrogen atom at its ground state, calculate the attraction force between its electron and proton.
3) Knowing the mass of electron and the centripetal force on it in hydrogen atom as found in Problem 2, find the speed of electron at the instant it is in ground state.
4) Knowing the speed of electron at the instant it is in ground state, find (a) its K.E., and (b) convert it to eV.
5) At the ground state, calculate (a) the P.E. of the electron with respect to the proton in hydrogen atom in both Joule and eV. (b) Use the result of Problem 4 to calculate the total energy of the hydrogen atom at ground state. By total energy, it is of course meant the total energy of the electron.
Answers:
1) 97.4nm, 103nm, & 122nm (All UV) 2) 8.2x10^{-8}N 3) 2.18x10^{6}m/s
4) 2.17x10^{-18}J, 13.6eV 5) (a) -4.34x10^{-18}J, -27.2eV (b) -13.6eV