Physics SS 3 Week 6
Topic: QUANTUM
INTRODUCTION
Some of the important postulates of the Bohr’s model of the atom were that the electrons moved around the nucleus in specified orbits. In these orbits they can move without radiating energy.
They can acquire or lose energy only in discrete units called quanta. Thus Bohr suggests that electrons in the atom exist in discrete (or quantized) energy states.
In 1902, Max Planck was able to show that experimental observations in the radiation emitted by substances could be explained on the basis that the energy from such bodies emitted in separate or discrete packets of energy known as energy quanta of value hf, where f is the frequency of radiation and h is a constant known as Planck’s Constant. Thus the energy E of the quantum of radiation or photon is given by E = hf.
This is known as Planck’s theory of radiation. The term quantum means amount fixed amount, or discrete or separate amount as distinguished from a continuous quantity.
Planck’s quantum hypothesis thus suggests (and this is accepted today) that the energy of radiation can be E = hf, or 2hf or 3hf, etc. but there cannot be vibrations or radiations whose energy lies between these values. That is energy radiated is not a continuous quantity but is rather quantized – i.e., it exists only in discrete amounts. This is the meaning of the concept of energy quantization.
The electrons in an atom can only have certain, specific amounts of energy. This is an unusual concept—if an electron is whirling around an atom with a given amount of energy.
Energy Levels in an Atom
In Figure 25.1, we’ve drawn a few of the energy levels of a hypothetical atom. Let’s start by looking at E1. This is the lowest energy level, and it is called the ground state energy of the atom. When an electron is sitting as close to the nucleus as possible, and when it’s completely unexcited, it will have the energy of E1, -10 eV.
To increase the energy of the electron—to move the electron into a higher energy level—energy must somehow be transferred to the electron. This energy transfer is done by a photon. To jump up in energy, an electron absorbs a photon, and to drop down in energy, an electron emits a photon.
The energy diagram in Figure 25.1 tells us that to get to E2 from the ground state, an electron must absorb a photon carrying 3 eV—and only 3 eV!—of energy. If a photon with an energy of 2.9 eV comes along and knocks into the electron, nothing will happen. Similarly, if a photon with an energy of 3.1 eV comes along, nothing will happen. It’s all or nothing; either the electron gets just the right amount of energy to go from one energy level to the next, or it doesn’t.
How does the electron get from the ground state to E3? There are two ways. The electron could first absorb a photon with an energy of 3 eV, taking it from E1 to E2, and then it could absorb another photon with an energy of 1 eV, taking it from E2 to E3. Or, the electron could start in E1 and simply absorb a photon with an energy of 4 eV.
We’ve been talking about photons having certain energies, but we haven’t yet told you how to figure out the energy of a photon. Here’s the formula:
E = hf
This formula tells us that the energy of a photon is equal to Planck’s constant, h, which is 6.63 × 10-34 J·s (this value is given to you on the constants sheet), multiplied by the frequency of the photon. You should remember from Chapter 23 that the frequency of a wave is related to the wavelength by the formula
v = λf
For light, the velocity is c, or 3 . 108 m/s, so we can instead write
c = λf
This means that we can rewrite the equation for the energy of a photon to read
E = hc/λ
These formulas tell us that a photon with a high frequency, and therefore with a small wavelength, is higher in energy than a photon with a low frequency and long wavelength. So gamma rays, for example, are a lot higher energy than radio waves because gamma rays have a higher frequency.
This is a simple plug-and-chug problem. Figure 25.1 gives us the energy levels, and we just need to use our formula to find the wavelength of the absorbed photon. But what value do we use for the energy? -10 eV? -7 eV? Wrong… the value we use is the energy of the jump, not the energy of one of the states. To get from E1 to E2, an electron must absorb 3 eV of energy, so that’s the value we use in the formula:
At this point, the problem is simply a matter of converting units and using your calculator. However, solving it is a lot easier if you know that the quantity hc equals 1240 eV nm. This value is on the constants sheet, and it is very important to know! by knowing this value, we can solve for the photon’s wavelength very quickly.
In other words, for an electron in Figure 25.1 to go from the ground state to E2, it must absorb light with a wavelength of exactly 410 nm, which happens to be a lovely shade of violet.
If you look back at Figure 25.1, you might wonder what’s going on in the gap between E3 and E∞. In fact, this gap is likely filled with lots more energy levels—E4, E5, E6… However, as you go up in energy, the energy levels get squeezed closer and closer together (notice, for example, how the energy gap between E1 and E2 is greater than the gap between E2 and E3). However, we didn’t draw all these energy levels, because our diagram would have become too crowded.
The presence of all these other energy levels, though, raises an interesting question. Clearly, an electron can keep moving from one energy level to the next if it absorbs the appropriate photons, getting closer and closer to E∞. But can it ever get beyond E∞? That is, if an electron in the ground state were to absorb a photon with energy greater than 10 eV, what would happen?
The answer is that the electron would be ejected from the atom. It takes exactly 10 eV for our electron to escape the electric pull of the atom’s nucleus—this minimum amount of energy needed for an electron to escape an atom is called the ionization energy2—so if the electron absorbed a photon with an energy of, say, 11 eV, then it would use 10 of those eV to escape the atom, and it would have 1 eV of leftover energy. That leftover energy takes the form of kinetic energy.
As we said above, it takes 10 eV for our electron to escape the atom, which leaves 1 eV to be converted to kinetic energy. The formula for kinetic energy requires that we use values in standard units, so we need to convert the energy to joules.
1eV(1.6 x 10-19J/eV) = ½ mv2
If we plug in the mass of an electron for m, we find that v = 5.9 × 105 m/s. Is that a reasonable answer? It’s certainly quite fast, but electrons generally travel very quickly. And it’s several orders of magnitude slower than the speed of light, which is the fastest anything can travel. So our answer seems to make sense.
The observation that electrons, when given enough energy, can be ejected from atoms was the basis of one of the most important discoveries of twentieth century physics: the photoelectric effect.
Photoelectric Effect: Energy in the form of light can cause an atom to eject one of its electrons, but only if the frequency of the light is above a certain value.
This discovery was surprising, because physicists in the early twentieth century thought that the brightness of light, and not its frequency, was related to the light’s energy. But no matter how bright the light was that they used in experiments, they couldn’t make atoms eject their electrons unless the light was above a certain frequency. This frequency became known, for obvious reasons, as the cutoff frequency. The cutoff frequency is different for very type of atom.
A metal surface has a work has a work function of 10eV. What is the cutoff frequency for this metal?
Remembering that hc equals 1240 eV·nm, we can easily find the wavelength of a photon with energy of 10 eV:
λ = 1240eV.nm/10eV
= 124nm
Using the equation c = λf, we find that f = 2.42 × 1015 Hz. So any photon with a frequency equal to or greater than 2.42 × 1015 Hz carries enough energy to eject a photon from the metal surface.
Franck-Hertz Experiment
The discreteness of atomic energy levels was first shown directly in the Franck-Hertz experiment. This experiment is one of the classic demonstrations of the quantization of atomic energy levels and was first performed by J. Frank and G. Hertz in 1914. Their goal was to verify the quantum theory assumptions about the existence of discrete energy levels in atoms and that quantized amounts of energy are transferred in emission and absorption. By accelerating a beam of electrons through a mercury vapor, they found that when the kinetic energy of the electrons reached about 4.9 eV, the vapor emitted a spectrum line at 254 nm. This experiment led to the detailed investigation of the atomic structure of many elements. When an electron encounters an atom and bounces off without losing any of its energy then such an event is called an “elastic scattering”. The electron will be elastically scattered unless it has sufficient energy to cause a change in the internal energy of the atom. Since atomic energy levels are quantized, this means that electrons flowing through a gas of atoms with less energy than the first excited state of the atoms will not lose any energy as they travel. On the other hand, an electron with enough energy to cause a transition to an excited state of the atom may induce such a transition with subsequent loss of kinetic energy. Such an event is called an “inelastic scattering” of the electron by the atom. In the Franck Hertz experiment electrons are emitted from a hot cathode into a tube filled with mercury vapor. The experimental arrangement is shown in figure below.
The driving potential Vg1 reduces the space charge and causes a large current to flow in the tube. The electrons are then accelerated through a positive potential by the accelerating potential Vg2. After being accelerated, the electrons are slowed by a potential drop in the opposite direction by the braking potential. Then the electrons are collected at a far end of the tube and the current is measured. In a vacuum tube that contains no gas the current would rise steadily as the accelerating voltage Vg2, is increased. The presence of the gas changes this behavior because of collisions of the electrons with the gas atoms. At first the current does rise with the potential, but when the electrons get enough energy they inelastically collide with the gas atoms and excite higher energy levels in the gas. After these collisions the electrons will have lower energy and due to the opposing potential, they will not make it to the end of the tube. This will cause the current to decrease to a minimum. After this minimum, as the potential increases the current will again increase until the electrons get enough energy to excite the gas twice. This process continues with the electrons repeatedly exciting the gas atoms. The potential difference between the minima (or maxima) is equivalent to the energy of the excited level. The graph plotted between the accelerating potential and anode current is shown below.
At each of the critical potentials (V1, V2, or V3 etc.) the electrons have just sufficient kinetic energy to raise the internal energy of the sodium atom by collision.
Spectra
Embedded in an atom which consists of a nucleus that is positively charged and which revolves round the orbit is an electron which is negatively charged.
When the atom or metal is heated and it absorbs energy to a particular intensity, the electrons starts to collide with one another and with the electron shell until the electrons are able to liberate themselves from the orbit. They create an atom with high energy level which teams up with the low energy level. When this occurs, radiation is emitted. Some of the particles and energies in the radiation are equal to the difference in energy of the atom between initial and final stage.
E = E0 – E1
The study of spectra is referred to as spectroscopy.
Types of Spectra
(i) Emission spectra:
These are spectra observed due to light emitted when the temperature of an atom, gas or metal is raised, e.g. incandescent solid or vapour , a spark discharge, arc discharge, the discharge of electricity through a gas or vapour contained in a discharged tube, etc.
(ii) Line Spectra
Line spectrum is obtained from atoms in gases such as hydrogen or neon at low pressure in a discharge tube. This spectrum mainly occurs in sodium, mercury vapours and hydrogen gas. When a light is incidented on a vapour-like mercury or hydrogen, a particular temperature is reached when or vapour starts to emit light of different wavelengths which depends on the nature of gas or metal. The emission can be viewed with the aid of a spectrometer in a slit. Differences are observed to be discontinuous but not overlapping, with equal distance between them, i.e. they are regularly spaced but do not form a series.
Line spectra occur when there is transmission of electron from one region to another usually from higher to lower energy level or vice-versa. Line spectra observed can be represented by:
∆E = hf = En – Eo
Where hf is the change of energy level (eV)
En = higher energy level or excitation energy level (eV)
Eo = lower energy level or ground state (eV)
Where h Planck’s constant with value at 6.6 x 10-34J
(iii) Continuous Spectrum
The spectrum formed from white light contains all colors, or frequencies, and is known as a continuous spectrum. Continuous spectra are produced by all incandescent solids and liquids and by gases under high pressure. A gas under low pressure does not produce a continuous spectrum but instead produces a line spectrum, i.e., one composed of individual lines at specific frequencies characteristic of the gas, rather than a continuous band of all frequencies. If the gas is made incandescent by heat or an electric discharge, the resulting spectrum is a bright-line, or emission, spectrum, consisting of a series of bright lines against a dark background. A dark-line, or absorption, spectrum is the reverse of a bright-line spectrum; it is produced when white light containing all frequencies passes through a gas not hot enough to be incandescent. It consists of a series of dark lines superimposed on a continuous spectrum, each line corresponding to a frequency where a bright line would appear if the gas were incandescent. The Fraunhofer lines appearing in the spectrum of the sun are an example of a dark-line spectrum; they are caused by the absorption of certain frequencies of light by the cooler, outer layers of the solar atmosphere. Line spectra of either type are useful in chemical analysis, since they reveal the presence of particular elements.
(iv) Band Spectrum
It consists of a number of bright bands with a sharp edge at one end but fading out at the other end. Band spectra are obtained from molecules. It is the characteristic of the molecule. Calcium or barium salts in a bunsen flame and gases like carbon-di-oxide, ammonia, and nitrogen in molecular state in the discharge tube give band spectra. When the bands are examined with high resolving power spectrometer, each band is found to be made of a large number of fine lines, very close to each other at the sharp edge but shaped out at the other end. Using band spectra the molecular structure of the substance can be studied.
(v) Absorption Spectrum
When a cool gas is placed in the path of a continuous spectrum of light, dark absorption lines will appear in the resulting spectrum. Conversely, if we observe the gas in an oblique angle, we can see emission lines produced by the gas too. The most interesting thing about the spectrum of an element is that, the wavelengths of absorption lines and emission lines produced by an element match exactly each other. This is the one of the evidences for electrons in an atom are situated in some kind of resonant columns.
When a photon falls on an electron and the resonant frequency of the shell (electron shell or transitory shell) in which the electron exists matches with the frequency of the photon, the electron will oscillate with the frequency of the photon and the photon will be efficiently absorbed by the atom. In the resultant spectrum, the absorbed photon will not be present. This is the way an atom produces its absorption lines.
When an electron oscillates, it will emit a radiation with the frequency of its oscillation. Therefore an atom produces emission lines with exactly matching wavelengths of the absorption lines that produced by the atom.
Questions
1. Calculate the energy in joules of ultraviolet light of wavelength 3 x 107 m. Take the velocity of light as 3 x 108m/s and Planck’s constant as 6.6 x 10-34 Js
A. 6.6 x 10-19 B. 9.8 x 10-2 C. 3 x 10-8 D. 4.5 x 108.
2. An electron jumps from one energy level to another in an atom radiating 4.5 x 10-19 Joules. If Planck’s constant is 6.6 x 1034 Js, what is the wavelength of the radiation? Take velocity of light = (3 x 108 m/s).
A. 3.4 x 10-7m B. 4.4 x 10-7 C. 5.9 x 10-8 D. 2.5 x 10-6
3. If an electron in the ground state were to absorb a photon with energy greater than 10 eV, what would happen?
A. the electron will stick to the atom B. the electron would be ejected from the atom C. the electron wil be neutralised D. there will be increase in the electron in the atom.
4. Which of the spectra Line spectra occur when there is transmission of electron from one region to another usually from higher to lower energy level or vice-versa?
A. Line spectra B. Continuous spectra C. Absorption spectra D. Emission spectra
5. Which of these is not correct?
A. Thus Bohr suggests that electrons in the atom exist in discrete (or quantized) energy states.
B. The discreteness of atomic energy levels was first shown directly in the Franck-Hertz experiment.
C. Franck-Hertz experiment is one of the classic demonstrations of the quantization of atomic energy levels
D. To increase the energy of the electron, to move the electron into a higher energy level—energy of the electron must be stable.
Answer
1. A 2. B 3. B 4. A 5. D
