Chapter 40

Early Quantum Theory:

## The Nature of the Atom

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 discrete 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 is 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 pictures 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 105 power.  Since the formula for the volume of sphere is (4/3)πR3, the volume of each atom is some 1015 (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 7.2cm3.  In such volume there exist some 6x1023 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 itto 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)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, the potential energy is Ue = -ke2/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 statistics play important roles.  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 statistics 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:

The Bohr Model 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

- ke2/r  +  (1/2) Mv2  = -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.   Knowing v2 , we can then solve for r the radius of hydrogen atom.

v2 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 -10m.  The diameter of the H-atom at its ground state becomes:  1.06x10-10m.

We define another unit for length called the "Angstrom" to be 10 -10m.   Angstrom is 10 times smaller than nanometer.

For example the red and violet wavelengths that are 700nm and 400nm  are 7000 and 4000 Angstroms.

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 r1.   The length of its circular path must be 1 as shown.   It means that the circumference of circle r1 must accommodate exactly 3 of a certain wavelength like λ1 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:  2,  3,  4,  and so on depending on the amount of energy received.  Only 4λ2 is shown here.

Figure 2

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 becomes 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 Em to a lower n-th level at energy En.  The excess energy Em- En  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 Em- En with a proportionality factor h that is named the "Planck's Constant."  The Planck's formula is therefore,

Em - En = hf .     (1)

The value of Planck's constant is h = 4.14x10-15 eV  s.   The above three paragraphs summarized the theory behind how light is generated.

Example 1: Convert the Planck's constant 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 Em- En = hf  and simply writing it as  E = hf  and also replacing  f  by c/λ  (derived from c = f λ,  we get:    E = hc/λ.   Substituting the values, we get:

E = hc/λ = 4.14x10-15eV  s (3.00x108m/s)/(105x10-9m) = 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 atom, for example.

The following equations work well for the hydrogen atom for which the number of its protons is Z = 1.

rn  =  (0.53x10-10m) n2 /Z    and

En =  - (13.6 eV) Z2 / n2

where rn is the radius of the n-th orbit or shell and En 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 En Z is the number of protons.

Solution:  For hydrogen Z =1.  Equation En = -(13.6 eV) Z2/n2  becomes:

En = {-13.6/n2}eV.   For Part (a) we need to calculate E5 - E2 and set it equal to hf according to Planck's formula and solve for frequency f .

E5 - E2 = {-13.6/52 - (-13.6/22) }eV = -13.6[1/25 -1/4] eV = 2.86 eV.

Since E5 - E2 = hf52   ;   2.86eV = hf    ;  f = 2.86eV/(4.14x10-15eV s)

f52 = 6.91x1014Hz.   Verify all calculations.

Since c = f λ; therefore, λ52 = c/f52 = 3.00x108(m/s)/6.91x1014/s = 434nm.

This transition is associated with the emission of blue light.

(b) & (c): Show that you get λ42 = 487nm (turquoise)  &  λ42 = 658nm (red).

Test Yourself 1:

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 H2 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 q1 with respect to charge q2 a distance r from it is

(a) -kq1q2/r2    (b) -kq1q2/r    (c) -kq1/r2    (d) -kq2/r2.  k is Coulomb's constant.

7) If v is the speed of electron and M its mass, its K.E. is

(a) Mv  (b)  Mv2   (c) 1/2Mv2   (d) Megh.    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)-ke2/r2  (b)-ke2/r    (c)-k/r2    (d)-k2/r2.  k is Coulomb's constant.

10) Using the Coulomb force F = ke2/r2 as the centripetal force, we may write:

(a) ke2/r2 = Mv2/r      (b) ke2/r2 = Mv/r      (c) ke2/r = Mv2/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.

Problem Set 1:

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) At the 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.

Particles and Waves:

We have so far discussed two behaviors of light: straight line motion (Geometric Optics) and the wave-like behavior (Wave Optics).  In this chapter, the particle-like behavior of light will be discussed.  In fact, the particle-like behavior is also associated with a frequency and it cannot be separated form the wave-like behavior (Quantum Optics).

Max Planck formulated this theory that as electrons orbit the nucleus of an atom, they receive energy from the surroundings in different forms.  Typical forms of receiving energy are: heat waves, light waves, and collision with other electrons and particles.  The radius at which an electron orbits is a function of electron's energy that means electron's speed as well as its distance from the nucleus.  Recall that K.E. = (1/2)Mv2.   Each electron is also under a Coulomb attraction force from the nucleus given by F = ke2/r2.  Furthermore, circular motion requires a centripetal force Fc = Mv2/r.  We know that it is the Coulomb force F that provides the necessary centripetal force Fc for the electron rotation about the nucleus.

The above discussion clarifies that, in simplest explanation, each electron takes a certain radius of rotation depending on its energy or speed.  When an electron receives extra energy, it then has to change its orbit or radius of rotation.  It has to take an orbit of greater radius.  The radius it takes is not just any radius.  The radii of rotation for any electron orbiting a nucleus are discrete values.  When such transition occurs, a vacant orbit is left behind that must be filled.  It may be filled by the same electron or any other one.  The electron that fills that vacant orbit must have the correct energy that matches the energy of that orbit. The electron that fills that orbit may have excess energy that has to be given off before being able to fill that vacant orbit.  The excess energy that an electron gives off appears as a burst of energy, a parcel of energy, a packet of energy that is called a "quantum of energy" according to Max Planck.

The excess energy is simply the energy difference between two different orbits.  If an electron returns from a greater radius orbit rm with an energy level Em to a smaller radius orbit rn with an energy level En, it releases a quantum of energy equal to the energy difference Em- En.   Max Planck  showed that this energy difference is proportional to the frequency f of that particular transition.  The proportionality constant is h with a value of  h = 6.626x10-34 J.s  called the " Planck's constant."  The parcel or quantum of energy is also called a "photon."

In electron-volts,  h = 4.14x10-15eV-sec.   The Plancks' formula is:

 Em - En = hf or,   ΔE = hf or,   E = hf .

Example 1:  Calculate (a) the energy of photons with a frequency of  3.2x1014 Hz.  (b) Find their corresponding wavelength and (c) express if they are in the visible range.

Solution: ΔE = hf  ;  ΔE = 6.626x10-34J.s)(3.2x1014/s) = 2.12x10-19J.

Note that 1eV = 1.6x10 -19J Our answer is a little more than 1eV.  In fact it is (2.12 /1.6) = 1.3 eV.

(b) c = f λ ;  λ = (3.00x108m/s)/(3.2x1014/s) = 940x10-9m = 940nm.

(c) The visible range is between 400 nm - 700 nm.  This photon is not in the visible range.  It is infrared.

Example 2:  Calculate (a) the energy (in Joules) of each photon of ultraviolet light with a wavelength of 225nm.  (b) Convert that energy to electron-volts.

Solution: (a)  E = hf  and  f = c/λ  ; therefore,  E = hc/λ.

E = 6.626x10-34J.s (3.00x108m/s)/225x10-9m = 8.83x10-19 J.

(b) Since 1eV = 1.6x10-19J ; therefore,   ΔE =  5.5eV.

The Photoelectric Effect:

The mechanism by which photoelectric effect operates may be used to verify the particle-like behavior of light.  A photoelectric cell can be made of a vacuum tube in which two metallic plates or poles are fixed.  The two plates are connected to two wires that come out of the sealed glass tube and are used for connection to other electronic components.  For time being, let us connect a photoelectric cell to just a galvanometer (sensitive ammeter) as shown in Fig. 3.  One terminal (plate) in the tube may be mounted in a slanted fashion in order for the light coming from outside to be effectively received by it.  This side forms the negative pole.  The other side that collects or receives electrons forms the positive pole.

 When photons of light are sent toward the slanted metal plate, it is  observed that the galvanometer in the circuit shows the passage of a current.  When the light is cut off, the current stops.  This shows that the collision of photons of light on the metal surface must release electrons from that surface.  The released electrons must be from the outer shells of the outermost atomic layers of the metallic plate. Each energetic photon that collides with the metal surface, releases one electron. This released electron has some speed and therefore some K.E. = 1/2Mv2.  The atoms of the outer surface that have lost electrons, replenish their electron deficiencies from the inner layer atoms of the metal oxide or the nearby free electrons.  This replenishing process transmits layer by layer through the wire and the galvanometer all the way to the pole labeled "Positive." Figure 3         The positive end pulls the released electrons from the negative end through the vacuum tube and the circuit completes itself.  This process occurs very fast.  As soon as light hits the metal plate, the circuit is on.  As soon as light is cut off, the circuit goes off.

The conclusion of the above experiment is that photons of light act as particles and kick electrons out of their orbit.  This verifies the particle-like behavior of light.

The Photoelectric Cell Formula:

The energy necessary to just detach an electron out of a metal surface is called the " Work Function" of that metal and is shown by Wo .    If the energy of each incident photon on the metal surface is hf, and the kinetic energy of the released electron is K.E., then we may write the following energy balance for a photoelectric cell:

 hf  =  Wo + K.E.

According to this equation, hf must be greater than Wo for an electron to be released.  Since h is the Planck's constant; therefore, f must be high enough for the photon to be effective.  There is a limit for frequency below which nothing happens.  That limit happens when the frequency of the incident photon is just enough to release an electron.  Such released electron has a K.E. = 0.  At the limiting frequency called the "threshold frequency", the K.E. of the released electron is zero.  With K.E. = 0 and replacing f by fth, we get:

 h fth  =  Wo or fth =  Wo / h.

The above formula gives the threshold frequency, fth .

Example 3:  The work function of the metal plate in a photoelectric cell is 1.73eV.  The wavelength of the incident photons is 366nm.  Find (a) the frequency of the photons, (b) the K.E. of the released electrons, and (c) the threshold frequency and wavelength for this photoelectric cell.

Solution:

(a) c = f λ ;  f = c/λ= (3.00x108m/s)/(366x10-9m) = 8.20x1014 Hz.

(b) hf  =  W0 + K.E.    ;    K.E. =  hf  - W0

K.E. =    (4.14x10-15eVs(8.20x1014 s-1)  -  1.73eV  = 1.66eV.

(c)  fth  Wo/h  ;   fth = 1.73eV/(4.14x10-15eVs)  =  4.18x1014 Hz.

λth = c/fth   ;      λth = (3.00x108m/s)/(4.18x1014 Hz) = 718nm.

Wave-Particle Duality:

According to de Broglie, for every moving particle of momentum Mv, we may associate an equivalent wavelength λ describing its wave motion behavior such that

 de Broglie Wavelength:

where λ is called the "de Broglie wavelength."

Example 4:  Calculate the de Broglie wavelength associated with the motion of an electron that orbits a hydrogen atom at a speed of 6.56x106 m/s.

Solution: Using λ = h/Mv, we may write:

λ = 6.626x10-34Js/[(9.108x10-31kg)(6.56x106 m/s)] = 1.11x10-10m.

The Compton Effect:

The Compton effect is another boost to the idea that energy is "quantized."   The quantization of energy means that energy coming out of an atom is in discrete parcels or packets that are called quanta (the plural of quantum).  As was mentioned in the photoelectric effect, each packet of energy or "photon" could release one electron in the form of a collision.  Each photon carries a quantum of energy or hf amount of energy.

In Compton effect, also called the "Compton Scattering", a high energy photon of wavelength λ collides with an electron causing the release of another photon that is less energetic (longer wavelength, λ' ).  Of course, the electron will be dislocated and given a higher kinetic energy as well as a different momentum because of the collision with the incident photon.  Since photons are treated as having a mass of zero and just carry a parcel or quantum of energy, they are treated as mass-less compared to electrons, and therefore scatter.  For every photon an equivalent mass may be calculated according to the Einstein mass-energy conversion formula ( E = Mc2 ).

Fig. 4 shows a photon of wavelength λ that collides with an electron (in the general case of an oblique collision) causing the electron move sidewise through angle φ while the newly emitted photon λ' moves along angle θ.  It is easy to show that the change in wavelength  λ' - λ is given by:

 Figure 4:

where Mo is the rest mass of electron.   The quantity h/(Moc) = 0.00243nm is called the Compton wavelength.

Classically, the target charge (here  electron) should oscillate at the received frequency and re-radiate at the same frequency.  Compton found that the scattered radiation had two components, one at the original wavelength of 0.071nm, and the other at a longer wavelength that depended on the scattering angle θ and not on the material of the target

This means that the collision of the incident photon λ must be with electrons and the experiment did not depend on the target material.  Also since the energy of the incident photons were about 20keV, way above the work function of any material, the electrons were treated as if they were "free electrons."

To derive this formula, both conservation of energy and conservation of momentum must be applied.

Note that the momentum of a photon may be found from the de Broglie  formula

λ = h/Mc   or,   Mc = h/λ

The momentum of a photon is p = Mc where M is the mass equivalent of the photon energy; thus,

p = h/λ.

The energy and momentum balance equations are:

Example 5: Photons of wavelength 5.00 angstrom pass through a layer of thin zinc.  Find the wavelength of the scattered photons for an scattering angle of 64.0 degrees.

Solution:   λ' - λ = [h/(Moc)](1 - cosq ) = 0.00243nm (1- cos 64.0o) = 0.00134nm = 0.0134 angstrom.

λ' = λ + 0.0134 angstrom  =  5.013 angstrom.

Line Spectra:

As was mentioned earlier, when an electron in an atom receives some energy by any means, it moves to a greater radius orbit which energy level fits that electron's energy.  Such atom is then said to be in an excited state.  The excited state is unstable however, and the electron returns to lower levels by giving off its excess energy in the form of electromagnetic radiation.  Max Planck showed that the frequency of occurrence  f  of a particular  transition between two energy levels in an atom depends on the energy difference between those two layers.

En - Em = hf

In this formula En is the energy of the n-th level, Em the energy of the m-th level (lower than the n-th level) and  h = 4.14x10-15 eV-sec is the Planck's constant and f the frequency of the released photon.

Figure 5

Possibilities for the occurrence of electron jump from one level to other levels are numerous.  It depends on the amount of energy an electron receives.  An electron can get energized when a photon hits it, or is passed by another more energetic electron that repels it, or by any other means.  The electron return can occur in one step or many steps depending on the amount of energy it loses.  In each possibility, the red arrow shows electron going to a higher energy level, and the black arrows show possible return occurrences.

Hydrogen is the simplest atom.  It has one proton and one electron.

Click on the following applet for a better understanding of the transitions:  http://www.colorado.edu/physics/2000/quantumzone/lines2.html.

In the applet, if you click on a higher orbit than where the electron is orbiting, a wave signal must be received by the electron (from outside) to give it energy to go to that higher level.  If the electron is already in a higher orbit and you click on a lower orbit, then the electron loses excess energy and gives off a wave signal before going to that lower orbit.

Also click on the following link:

http://www.walter-fendt.de/ph14e/bohrh.htm  and try both options of "Particle Mode" and "Wave Mode".

You can put the mouse on the applet near or exactly on any circle and change the orbit of the electron to anywhere you wish; however, there are only discrete orbits with circumferences each of which is an integer multiple of a certain wavelength.  It is at those special orbits that the applet shows principal quantum numbers (on the right side of the applet).

For hydrogen atom, possible transitions from the ground state E1 to 2nd state E2, 3rdstate E3, and 4th state E4 are shown in Fig. 1.   The possibilities for electron return are also shown.  The greater the energy difference between two states, the more energetic the released photon is when an excited electron returns to its lower orbit.  If the return is very energetic, the wavelength may be too short to fall in the visible range and cannot be seen in the spectroscope.  Some transitions are weak and result in larger wavelengths in the infrared region that cannot be seen either.  However some intermediate energy transitions fall in the visible range and can be seen.

Grouping of the Transitions:

Transitions made from higher levels to the first orbit form the Lyman Series.

Transitions made from higher levels to the second orbit form the Balmer Series.

Transitions made from higher levels to the third orbit form the Paschen Series.

Transitions made from higher levels to the fourth orbit form the Pfund Series.

Test Yourself 2:

1) The energy of a photon of light, according to Max Planck's formula is (a) E = 1/2Mv2  (b) E = hf   (c) E = Mgh.

2) The Planck's is (a) 6.6262x10-34 Js    (b) 4.14x10-15 eVs    (c) both a & b.   click here.

3) An electron orbiting the nucleus of an atom can be energized by (a) receiving a heat wave   (b) getting collided by another subatomic particle   (c) by getting hit by a photon   (d) both a, b, & c.

4) When an electron is energized by any means, it requires (a) a greater radius of rotation  (b) a smaller radius of rotation  (c) it stays in the same orbit but spins faster.   click here.

5) When there is a vacant orbit, it will be filled with an electron from (a) a lower orbit   (b) a higher orbit.

6) A higher orbit means (a) a greater radius   (b) a faster moving electron   (c) a greater energy   (d) a, b, and c.

7) The excess energy that an electron in a higher orbit has is released in the form of a photon (a small packet or burst of energy) as the electron fills up a lower orbit.   (a) True   (b) False   click here.

8) The excess energy is (a) the energy difference, E2 - E1, of the higher and lower orbits   (b) the energy each electron has anyway   (c) both a & b.

9) A photon has a mass of (a) zero   (b) 1/2 of the mass of an electron   (c) neither a nor b.

10) Each photon carries a certain amount of energy.  We may use the Einstein formula (E = Mc2) and calculate an equivalent mass for a photon.  (a) True  (b) False   click here.

11) The greater the energy of a photon (a) the higher its speed   (b) the higher its velocity   (c) the higher it frequency   (d) a, b, c, & d.

12) The greater the energy of a photon the lower its wavelength.  (a) True   (b) False

13) The formula for waves speed, v = f λ, takes the form of (a) c = f λ for photons of visible light only   (b) for photons of non-visible light only   (c) for the full spectrum of E&M waves that visible light is a part of.    click here.

Problem:  A student has calculated a frequency of 4.8x1016 Hz for a certain type of X-ray and a wavelength of 7.0nm.

14) Use the equation v = f λ and calculate v to see if the student's calculations is correct.  (a) Correct   (b) Wrong

15) The answer to Question 14 is (a) 3.36x108 m/s   (b) 3.36x1017m/s    (c) neither a nor b.   click here.

16) The reason why the answer to Question 14 is wrong is that v turns out to be greater than the speed of light in vacuum that is 3.0x108 m/s.   (a) True  (b) False

17) In the photoelectric effect, (a) electrons collide and release photons   (b) photons collide and release electrons  (c) neither a nor b.   click here.

18) In a photoelectric cell, the plate that receives photons, becomes (a) negative  (b) positive  (c) neutral.

19) The reason why the released (energized) electrons do no return back to their shells is that (a) their energies are more than enough for the orbits they were in  (b) the orbits (of the atoms of the metal plate) that have lost electrons, quickly replenish electrons from the inner layer atoms of the metal plate   (c) the outer shells atoms that have lost electrons will be left in loss for ever   (d) a & b.     click here.

20) When light is incident on the metal plate of a photoelectric cell, the other plate or pole of the cell becomes positive.  The reason is that (a) photons carry negative charges   (b) the other pole loses electrons to replenish the lost electrons of the metal plate through the outside wire that connects it to the metal plate  (c) both a & b.

21) In a photoelectric cell, the released electrons  (a) vanish in the vacuum of the cell  (b) accelerate toward the other pole because of the other pole is positive  (c) neither a nor b.

22) The negative current in the external wire of a photoelectric cell is (a) zero  (b) out of the metal plate  (c) toward the negative plate.   click here.

23) In a photoelectric cell, the energy of each incident photon is (a) 1/2Mv2   (b) hf  (c) Wo.

24) In a photoelectric cell, the work function of the metal plate is named (a) 1/2Mv2   (b) hf  (c) Wo.

25) In a photoelectric cell, the energy of each released electron is (a) 1/2Mv2   (b) hf  (c) Wo.   click here.

26) A 5.00eV incident photon has a frequency of (a) 1.21x10-15Hz    (b) 1.21x1015Hz    (c) 2.21x1015Hz.

27) A UV photon of f = 3.44x1015Hz has an energy of  (a)14.2eV  (b) 2.27x10-18J   (c) a & b.

28) When 3.7eV photons are incident on a 1.7eV work function metal, each released electron has an energy of (a) 2.0eV  (b) 5.4eV  (c) 6.3eV.   click here.

29) 4.7eV photons are incident on a 1.7eV work function metal. Each released electron has an energy of (a) 4.8x10-19J  (b) 3.0eV  (c) both a & b.

30) 3.7eV photons incident on a 1.7eV work function metal cause released electrons of speed (a) 8.4x10-5m/s  (b) 8.4x10 5m/s   (c) 8.4x10-15m/s.  click here.

31) A speed of 8.4x10-5m/s is not reasonable for a moving electron because (a) electrons always move at the speed of light  (b) this speed has a power of -5 that makes it very close to zero same as being stopped  (c) neither a nor b.

32) If the released electrons in a photoelectric effect have an average speed of 9.0x105 m/s and the energy of the incident photons on the average is 4.0eV, the work function of the metal is (a) 1.3eV   (b) 1.1eV   (c) 1.7eV.    click here.

33) The wavelength associated with the motion of proton at a speed of 6.2x106m/s is(a) 6.4x10-14m    (b) 9.4x10-14m   (c) 4.9x10-14m.

34) The diameter of hydrogen atom (the whereabouts of its electronic cloud) is 0.1nm or 10-10m called "Angstrom."   The diameter of the nucleus of the hydrogen atom is even 100,000 times smaller or10-15m called "Femto-meter (fm)."  The wavelength associated with the moving proton in Question 33 is (a) 6.4fm   (b) 64fm   (c) 640fm.    click here.

Problem Set 2:

1) Calculate (a) the energy of photons with a frequency of 6.40x1014Hz.  (b) Find their corresponding wavelength and (c) express if they are in the visible range.

2) Calculate (a) the energy (in Joules) of each photon of UV light with a wavelength of 107nm.  (b) Convert that energy to electron-volts.

3) The work function of the metal plate in a photoelectric cell is 2.07eV.  The wavelength of the incident photons on it is 236nm.  Find (a) the frequency of the photons, (b) the energy of each, (c) the K.E. of the released electrons, (d) their speed, and (e) the threshold frequency and wavelength for this photoelectric cell.

4) Calculate the de Broglie wavelength associated with the motion of an electron that hast a speed of (a) 1.31x107m/s.