Chapter 40:
Early Quantum Theory:
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 needletip 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 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 U_{e} = ke^{2} / r.
It is possible to determine the whereabout of the electrons around the nucleus of atoms and also the way they are oriented by solving different caseproblems 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,
The Bohr Model
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 e^{2} / r + ½ Mv^{2} = 13.6 eV, (1)
where v^{2} 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: ke^{2} / r^{2} = Mv^{2} / r. Solving for v^{2} yields: v^{2} = (ke^{2}) / (rM) (2) Substituting (2) in (1) changes (1) to :  k e^{2} / r + ½ M( ke^{2} / rM) = 13.6 eV (3) M cancels and we get:  ( ½) k e^{2} / 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 ^{–10}m. The diameter of Hatom is therefore 1.06 X 10 ^{–10}m.

Note that 1x10^{10}m is defined as 1 angstrom shown as 1Å 
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 3λ_{1} and 4λ_{2} are fit into orbits of radii r_{1} and r_{2}.
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 mth level that has an energy of E_{m} goes to a lower energy level E_{n}, its excess energy (Em –En ) appears as a photon of electromagnetic radiation that has an energy of (hf).
E_{m} – E_{n} = hf
Where h = 4.14x10^{15} eVsec 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 avalanchelike 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 (Htube) and (Hetube), 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.
r_{n} = (0.53x10^{10}m) n^{2} / Z
E_{n} = (13.6 eV) Z^{2 }/ n^{2}
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.
Chapter 40 Test Yourself 1:
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 H_{2} molecule is caused by
(a) a mixture of a large number of electrons randomly flowing around the nuclei.
(b) two extremelyfastmoving 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)  kq_{1}q_{2}/r^{2} (k is the Coulomb’s constant.) (b)  kq_{1}q_{2}/r. (c)  kq_{1}/r^{2}. (d)  kq_{2}/r^{2}.
7) If v is the average speed of electron, then its K.E. is
(a) M_{e}v where M_{e} is the electron mass. (b) M_{e}v^{2}. (c) ½ M_{e}v^{2}. (d) M_{e}gh. 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) ke^{2}/r^{2} (k is the Coulomb’s constant.). (b) ke^{2}/r. (c)  k/r^{2.}. (d)  k^{2}/r^{2}.
10) Using the Coulomb force (F = ke^{2}/r^{2}) between the electron and the proton of a hydrogen atom as the centripetal force for its electron rotation, 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. (d) ke^{2}/r^{2} = 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.
Particles and Waves:
We have so far discussed two behaviors of light: straight line motion ( Geometric Optics) and the wavelike behavior and interference (Wave Optics). In this chapter, the particlelike behavior of light will be discussed. In fact, the particlelike behavior is also associated with a frequency and it cannot be separated form the wavelike behavior.
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 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 K.E. and therefore electron's speed. Recall K.E. = (1/2)Mv^{2}. Each electron is also under a Coulomb attraction force from the nucleus given by F = ke^{2} / r^{2}. Furthermore, circular motion requires a centripetal force F_{c} = Mv^{2}/r. We know that it is the Coulomb force F that provides the necessary centripetal force F_{c} for electron spin.
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. 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 or 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 R_{m} with an energy level E_{m} to a smaller radius orbit R_{n} with an energy level E_{n}, it releases a quantum of energy equal to the energy difference E_{m}  E_{n}. Planck showed that this energy difference is proportional to the frequency of occurrence ( f ) of the released quantum or the packet of energy. The proportionality constant is h with a value of h = 6.626x10^{34} J.s called the " Planck's constant." The packet or quantum of energy is also called a "photon."
In electronvolts, ( h ) has a value of h = 4.14x10^{15}eVsec. The Plancks' formula is:
E_{m}  E_{n} = hf or, ΔE = hf
Example 1: Calculate (a) the energy of photons whose frequency of occurrence is 3.2x10^{14} Hz. (b) Find their corresponding wavelength and (c) express if they are in the visible range.
Solution: (a) ΔE = hf ; ΔE = ( 6.626x10^{34} J.s )( 3.2x10^{14} /s) = 2.12x10^{19} J
Note that 1eV = 1.6x10^{ 19} J . Our answer is a little more than 1eV. In fact it is (2.12 /1.6) = 1.3 eV.
(b) c = f λ ; λ = c / f ; λ = (3.00x10^{8}m/s)/ (3.2x10^{14}/s) = 9.4x10^{7}m = 940x10^{9}m = 940nm
(c) The visible range is between 400 nm  700 nm; this is not in the range. It is infrared.
Example 2: Calculate (a) the energy ( in Joules) of each photon of ultraviolet light whose wavelength is 225nm. (b) Convert that energy to electronvolts.
Solution: (a) ΔE = hf = hc /λ ; ΔE = ( 6.626x10^{34} J.s )(3.00x10^{8}m/s) / 225x10^{9}m = 8.83x10^{19} J.
(b) Since 1eV = 1.6x10^{19}J; therefore, ΔE = 5.5eV.
Photoelectric Effect:
The mechanism by which photoelectric effect operates may be used to verify the particlelike behavior of light. A photoelectric cell is 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 the figure below. One terminal (plate) in the tube may be mounted in a slanted way in order for the light coming from outside to effectively shine on it. This side forms the negative pole. The other side collects or receives electrons and 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 the outer shells of the outermost atomic layers of the metal oxide coating. Each energetic photon that collides with the metal surface, releases one electron. This released electron has some speed and therefore some K.E. = 1/2Mv^{2}. The atoms of the outer surface that have lost electrons, replenish their electron deficiencies from the inner layer atoms of the metal oxide. This replenishing process transmits layer by layer through the wire and the galvanometer all the way to the other pole that makes it "Positive." The positive end then 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.

Fig. 1 
The conclusion of the above experiment is that photons of light act as particles and kick electrons out of their orbit. This explains the particlelike behavior of light and a verification that energy is "quantized."
Photoelectric Effect 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 W_{o} . Since, according to Planck's formula. the energy of each incident photon on the metal surface is hf, and the kinetic energy of the released electron is K.E., we may write the following energy balance for a photoelectric cell.
hf = W_{o} + K.E..
According to this equation, hf must be greater than W_{o} for each electron to be released. Since h is a 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 kinetic energy of the released electron is zero. Setting KE = 0 and replacing f by f_{th}, we get:
h f_{th} = W_{o } or f_{th} = W_{o}_{ /} h.
The above formula gives the threshold frequency, f_{th} .
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 = fλ ; f = c/λ= (3.00x10^{8}m/s) / (366x10^{9}m) = 8.20x10^{14 }Hz
(b) hf = W_{0} + K.E. ; K.E. = hf  W_{0}
K.E. = ( 4.14x10^{15}eVs )( 8.20x10^{14 }/s )  1.73eV = 1.66eV
(c) f_{th} = W_{o} / h ; f_{th} = 1.73eV / (4.14x10^{15}eVs) = 4.18x10^{14 } Hz
λ_{th} = c / f_{th} ; λ_{th} = (3.00x10^{8}m/s ) / (4.18x10^{14 }Hz) = 718nm
WaveParticle 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 (a) the DeBroglie wavelength associated with the motion of an electron that orbits a hydrogen atom at a speed of 6.56x10^{6} m/s, and (b) compare the calculated wavelength with the diameter of hydrogen atom.
Solution: (a) Using λ = h/Mv, we may write: λ = (6.626x10^{34} Js) / [(9.108x10^{31}kg)(6.56x10^{6} m/s)] = 1.11x10^{10}m.
(b) The diameter of hydrogen atom is 1.06x10^{10}m, almost equal to the calculated deBroglie wavelength in Part (a). This shows why the electron of each hydrogen atom has a wavy motion around its nucleus. Note that for electron speeds higher than 6.56x10^{6} m/s, given in this example, λ will be smaller, and a greater number of such λ's will fit in each circular path that electron may have around nucleus.
The Compton Effect:
The Compton effect is another boost to the idea that energy of a system 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 "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 and 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 massless compared to electrons, and therefore scatter. For every photon an equivalent mass may be calculated according to the Einstein massenergy conversion formula ( E = Mc^{2 }).
The figure on the right 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.
The figure on the right shows a photon of wavelength λ that collides with an electron (in the general case of 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:
where m_{o} is the rest mass of electron. The quantity h_{/}(m_{o}c) = 0.00243 nm is called the Compton wavelength. Classically, the target charge (here the electron) should oscillate at the received frequency and reradiate 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 that 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 Å 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/m_{o}c) (1  cosq ) = (0.00243 nm) ( 1 cos 64.0 ) = 0.00134 nm = 0.0134 Å
λ' = λ + 0.0134 Å = 5.013Å
Line Spectra:
As was mentioned earlier, when an electron in an atom receives some energy by any means, it moves to a greaterer radius orbit whose 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 ( visible light is a small part of the E&M waves spectrum). 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.
E_{n}  E_{m} = hf
In this formula E_{n} is the energy of the nth level, E_{m} the energy of the mth level (lower than the nth level) and h = 4.14x10^{15} eVsec is the Planck’s constant. f is the frequency of the released photon.
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 (s) 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 this 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.walterfendt.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 whose each circumference is an integer multiple of a certain wavelength. It is at those special orbits that the applet shows principal quantum numbers for the electron on the right side.
For hydrogen atom, possible transitions from the ground state (E_{1}) to 2^{nd} state (E_{2}), 3^{rd} state (E_{3}), and 4^{th} 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 lower orbits. 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.
Emission and Absorption Spectra
A hot gas emits light because of the energy it receives by any means to stay hot. As was mentioned earlier, the received energy by an atom sends its electrons to higher levels, and in their returns, the electrons emit light at different wavelengths. The emitted wavelengths can be observed in a prism spectrometer in the form of a few lines of different colors. Each element has its own unique spectral lines that can be used as an ID for that element. Such spectrum coming from a hot gas is called emission spectrum. For a host gas spectral lines are discrete.
For white light entering a spectrometer the spectrum is a continuous band of rainbow colors. This continuous band of colors in a spectrometer ranges from violet to red and gives the following colors: violet, blue, green, yellow, orange, and red. Light emitted from the Sun contains so many different colors (or electronic transitions) that its spectrum gives variety of colors changing gradually from violet to red. It contains so many different violets, blues, greens, yellows, oranges, and reds that it appears continuous.
Chapter 40 Test Yourself 2:
1) The energy of a photon of light, according to Max Planck's formula is (a) E = 1/2Mv^{2}. (b) E = hf. (c) E = Mgh.
2) The Planck's constant, h, is (a) 6.6262x10^{34} J.sec. (b) 4.14x10^{15} eV.sec. (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 an electron in a higher orbit has is released in the form of a photon (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 = Mc^{2}) 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 = fλ, takes the form of (a) c = fλ for photons of visible light only. (b) for photons of nonvisible light only. (c) for the full spectrum of E&M waves which visible light is a part of. click here.
Problem: A student has calculated a frequency of 4.8x10^{16} Hz for a certain type of Xray and a wavelength of 7.0nm.
14) Use the equation v = fλ and calculate v to see if the student's calculations is correct. (a) Correct (b) Wrong
15) The answer to Question 14 is (a) 3.36x10^{8} m/s. (b) 3.36x10^{17} m/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 3.0x10^{8} m/s. (a) True (b) False
17) In the photoelectric effect, (a) electrons collide and release photons. (b) photons collide and release electrons. 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 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 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 (from the metal plate as a result of incident photons), (a) vanish in the vacuum of the cell. (b) accelerate toward the other pole because of the other pole being positive. (c) neither a nor b.
22) The negative current in the external wire of a photoelectric cell is (a) zero. (b) from the metal plate. (c) toward the negative plate. click here.
23) In a photoelectric cell, the energy of an incident photon is (a) 1/2Mv^{2}. (b) hf. (c) W_{o}.
24) In a photoelectric cell, the work function of the metal plate is named (a) 1/2Mv^{2}. (b) hf. (c) W_{o}.
25) In a photoelectric cell, the energy of each released electron is (a) 1/2Mv^{2}. (b) hf. (c) W_{o}. click here.
26) A 5.00eV incident photon has a frequency of (a) 1.21x10^{15}Hz. (b) 1.21x10^{15}Hz. (c) 2.21x10^{15}Hz.
27) An ultraviolet photon of frequency 3.44x10^{15}Hz has an energy, hf, of (a) 14.2 eV. (b) 2.27x10^{18}J. (c) a & b.
28) When 3.7eV photons are incident on a 1.7eV work function metal, each released electron has a K.E. 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^{19}J. (b) 3.0eV. (c) both a & b.
30) 3.7eV photons are incident on a 1.7eV work function metal. Each released electron has a speed of (a) 8.4x10^{5}m/s. (b) 8.4x10^{ 5}m/s. (c) 8.4x10^{15}m/s. click here.
31) A speed of 8.4x10^{5}m/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 9.0x10^{5} 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 6.2x10^{6} m/s is (a) 6.4x10^{14}m. (b) 9.4x10^{14}m. (c) 4.9x10^{14}m.
34) The diameter of hydrogen atom (the where about of its electronic cloud) is 0.1nm or 10^{10}m called "Angstrom." The diameter of the nucleus of the hydrogen atom is even 100,000 times smaller or 10^{15}m called "Femtometer (fm)." The wavelength associated with the moving proton in Question 33 is (a) 6.4fm. (b) 64fm. (c) 640fm. click here.