As we know, an atom has protons and neutrons at its nucleus and electrons spinning around it. The atomic space is extremely empty. In other words, the sizes of electrons, protons, and neutrons are much smaller than the size of the atom itself. If the surface of a basketball is the place where electrons of needle-tip size are spinning over, protons and neutrons of almost the same size are at the nucleus. The ratio of the radius of the atom (the radius of the basketball) to the radius of electron, proton, or neutron (a needle tip) is of the order of (10,000 to 100,000) closer to 100,000.
The motion of electrons around the atom is associated with K.E. If (v) is the average speed of an electron as it spins around the nucleus at a certain average radius (r), its K.E. = ½ Mv2. An electron, being a negative charge, is also in the electric field of the positive nucleus. The P.E. of a charge q1 in the field of another charge q2, is Ue = -kq1q2 / r, where k = 8.99X109 Nm2/C2 is the Coulomb’s constant. For the proton and electron of a hydrogen atom Ue = -ke2 / r.
It is possible to determine the where-about of the electrons around the nucleus of atoms and also the way they are oriented by solving different case-problems in an energy balancing way. This way of determining the size and shape of an “orbital” (space around the nucleus of an atom where electrons can be traced) is the basis for quantum mechanics calculations. Here, we are going to discuss the hydrogen atom, the simplest one.
Calculation of the Radius of the Hydrogen Atom
It can be experimentally verified that it takes 13.6 eV of energy to remove the electron from a hydrogen atom when it is in its ground state (closest to the nucleus). If the electron is somehow moved to higher shells, then it takes less energy to remove it from its atom. This means that if the electron of hydrogen atom is in its ground state, it takes at least 13.6eV to detach it from its atom. The electron energy is the sum of its K.E. and P.E.. We may write:
P.E. + K.E. = -13.6 eV, or
- k e2 / r + ½ Mv2 = -13.6 eV, (1)
where v2 can be found by understanding that the Coulombic force between the proton and electron of the hydrogen atom provides the necessary centripetal force for the electron spin around the proton.
Equating the expression for these two forces yields: ke2 / r2 = Mv2 / r.
Solving for v2 yields: v2 = (ke2) / (rM) (2)
Substituting (2) in (1) changes (1) to :
- k e2 / r + ½ M( ke2 / rM) = -13.6 eV (3)
M cancels and we get:
- ( ½) k e2 / r = -13.6 ( 1.6 X 10 –19 ) Joules.
Substituting for (k) and (e), yields the value for r of
r = 0.53 X 10 –10m. The diameter is therefore 1.06 X 10 –10m.
Here, we define another unit of length called “Angstrom” to be 10 -10 meter. This is 10 times smaller than one nanometer. For example the red and violet wavelengths that are 700nm and 400nm or, 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 for its rotation. The possible orbits are discrete positions and not continuous. The reason for this discreteness is the fact that an electron’s motion must fit into a path that is an integer multiple of a certain wavelength that is proportional to its energy. Look at the following figure where 3λ1 and 4λ2 are fit into orbits of radii r1 and r2.
The energy levels are discrete as well as the radii. When, for whatever reason, an electron jumps from a certain level to a higher energy level, its original energy level (orbit) is left vacant. That must be filled, not necessarily with the same electron, but with any other electron that can lose the correct amount of energy. When a more energized electron of the m-th level that has an energy of Em goes to a lower energy level En, its excess energy (Em –En ) appears as a photon of electromagnetic radiation that has an energy of (hf).
Em – En = hf
Where h = 4.14x10-15 eV-sec is the Planck’s constant. That is how light is generated.
When hydrogen, helium, or any other gas 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 source, while the positive (ionized) nuclei move toward the negative pole. During this avalanche-like motion in opposite directions, many different recombinings and separations between the electrons and nuclei occur. These transitions from many different levels to other many different levels generate so many different (hf)s and (λ)s of different colors some of which fall in the visible range. The gas becomes hot because of these transitions. An enough hot gas emits light. The spectrum of a hot (excited) gas is called an “emission spectrum”. The same gas, when cold, absorbs all such emissions. A cold gas has “absorption spectrum”. As a demo, we are going to observe the emission spectrum of (H) and (He) gases in our physics lab. A high voltage source for gas excitation, low pressure (H-tube) and (He-tube), as well as a spectrometer are needed. The following equations give the radius and the energy corresponding to different layers of atoms, of course atoms that are singly ionized only.
rn = (0.53x10-10m) n2 / Z
En = -(13.6 eV) Z2 / n2
For atoms that have more than one of their electrons around them, the calculations are more difficult and involved, and require higher levels of mathematics. In such calculations, the repulsion of electrons and their interactions must be taken into consideration.
1) In most atoms, the ratio of the atomic radius to the 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. (d) electron itself.
3) The electronic cloud in a hydrogen atom is caused by
(a) a mixture of a large number of electrons randomly flowing around the nucleus.
(b) a single electron spinning around the nucleus so fast that it appears everywhere.
(c) a fuzz of dust. (d) two electrons orbiting opposite to each other’s direction. click here.
4) The electronic cloud in a H2 molecule is caused by
(a) a mixture of a large number of electrons randomly flowing around the nuclei.
(b) two extremely-fast-moving electrons spinning around the nuclei 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 formula for the P.E. of charge q1 with respect to charge q2 a distance r from it is
(a) - kq1q2/r2 (k is the Coulomb’s constant.) (b) - kq1q2/r. (c) - kq1/r2. (d) - kq2/r2.
7) If v is the average speed of electron, then its K.E. is
(a) Mev where Me is the electron mass. (b) Mev2. (c) ½ Mev2. (d) Megh. click here.
8) 13.6 eV is the
(a) minimum energy for ionization of a hydrogen atom from its ground state.
(b) maximum energy for ionization of a hydrogen atom from its ground state.
(c) average energy for ionization of a hydrogen atom.
(d) average energy for ionization of all hydrogen atoms in a narrow tube. click here.
9) The P.E. of the electron in a hydrogen atom is
(a) -ke2/r2 (k is the Coulomb’s constant.). (b) -ke2/r. (c) - k/r2.. (d) - k2/r2.
10) Using the Coulomb force (F = ke2/r2) between the electron and the proton of a hydrogen atom as the centripetal force for its electron rotation, we may write:
(a) ke2/r2 = Mv2/r. (b) ke2/r2 = Mv/r. (c) ke2/r = Mv2/r. (d) ke2/r2 = Mgh.
11) The diameter of hydrogen atom is approximately
(a) 1 nm. (b) 10nm. (c) 0.1nm. (d) 0.2nm. click here.
12) The radius of hydrogen atom is approximately
(a) 1 Å. (b) 10 Å. (c) 0.1 Å. (d) 0.53 Å.
13) The wavelength of red light is
(a) 40 nm. (b) 400nm (c) 4000nm.. (d) 700nm
14) The wavelength of violet light is
(a) 70 Å. (b) 700 Å. (c) 7000 Å. (d) 4000 Å. click here.
15) An electron around the nucleus of an atom orbits at
(a) a fixed radius that never changes.
(b) a variable radius that can have any value.
(c) a variable radius that can have certain discrete values.
16) The radius at which an electron orbits the nucleus of an atom must be such that
(a) the P.E. of the electron equals its K.E.
(b) the P.E. of the electron equals ½ its K.E.
(c) ½ the P.E. of the electron equals its K.E.
(d) an integer number of a certain wavelength fits in its wavy path of motion. click here.
17) The reason for discreteness of possible positions (radii) for the electron orbit around atoms is 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 electronic levels (radii of orbitting) click here.
(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.