Link to return to Modern Physics front page
Links to specific sections in the text:
3.a. Atoms and electrons
3.b. The Rutherford Atom
3.c. Spectral Series
3.d. The hydrogen atom according to Bohr
3.e. Heavier atoms
3.f. Bohr's Correspondence principle
3.g. Franck-Hertz experiment
The ancient Greeks: Some great thinkers among the ancient Greeks (like Democritos) postulated on phylosophical grounds that atoms must exists. Atom means indivisable. They perforfmed a thought experiment of dividing matter into smaller and smaller pieces, so that the pieces would have the same properties and decided that could not be done forever but there are building blocks of matter that are its smallest representations.
Scientific evidence came only in the age of enlightment when, following Galileo, experiments were accepted as a basis of all physical sciences. Dalton was the first who pointed out that the law of multiple proportions in chemical experiments can be interpreted if matter consisted of molecules and molecules from atoms. Then came Avogadro who said that different gases have the same number of molecules in the same volume, provided their temperature and pressure coincided. Maxwell derived the thermodynamic properties of matter from its molecular nature. Finally, Einstein explained the Brownian motion by the motion of molecules. His brilliant paper finally mede it possible to see the action of molecules through a microscope.
The law of electrolysis by Faraday: There were a few experiments that had great influence on the development of atomic thought. First to be mentioned will be Faraday's discovery of the law of electrolysis. Faraday experimented with electrolylites like molten salt. He found that the same amount of charge, 96,500 C, passed throgh the electrolyte when one mole of salt was separated into one atomic weight of sodium and one atomic weight of chlorine. For n-valent elements he needed n times as much charge to produce tge same amount of free element. In other words the mass of the neutral matter liberated was equal to
m = q ¥(molar mass) / [96,500C¥(valence)].
He had a simple explanation: atoms are neutral, but in the molecule they take a charged form, called ion. When we separate them using the electric field created by the electrodes the process would stop very soon because the charged electrode would repel the deposit of further like charges. If allow the charges to be equalized through an outside circuit then the atoms become neutralized and new ions can be deposited on the electrodes. Since every ion carries a definite charge (96,500C/ NA) = e (or ne for n-valent ions) the total charge needed to separate NA (Avogadro's nummber) atoms is 96,500C. Faraday's result prooved the existence of atoms and molecules and determined the size of the elementary charge, e.
Thomson's Experiment to measure e/m: Cathode rays were seen for a long time before Thomson succeeded in proving the existence of electrons an measured the ratio e/m. The discovery of the electron put to rest the Greek idea of invisible atoms. The electorn, 2000 times lighter then the lightest atom was discovered. Previously it was not clear cathode rays consisted of pure radiation or of particles and if that is true what the mass of the particles was. Thomson used a vacuum tube in which the cathode rays were collimated into narrow beams. The cathod ray passed through a region in which there was a vertical electric field. After passing through this region a deflected beam was observed on anappropriate screen. Thomson immediately saw that the beam must consist of negatively charged particles. Measuring the deflection of the beam he could also calculate the ratio e/m. If we assume that the negative particles consituting the beam are the ones responsible for the charge of the ions then the electrolysis experiments determines their charge. Knowing e/m allows one to determine the mass as well. The geometry of Thomson's experiment was the following:
An image will be inserted later.
The electrons are in a homogeneus electric field Ey, that is perpendicular to their original momentum. The electric field is created by the plates of a capacitor. Suppose that after traveling distance d the electrons leave the capacitor and enter a region free of electric field. If we know the velocity of electrons, vx, then the angle of deflection can be calculated. The vertical acceleration is Fy/ m = a = Eye / m, constant. Thus the vertical velocity is
vy = Eyet / m = Eyed / mvx
where t = d / vx is the time spent by the electrons in the capacitor. Then the angle of deflection is calculated from
tan q = vy / vx = Eyed / mvx2 = (Eyd / vx2) (e / m )
Measuring the angle of deflection one can calculate the ratio e / m. Of course, the electric field is nothing else but Ey = V / y, where V is the voltage on the plates of the capacitor and y is the distance of the plates from each other. But how does one know the velocity of incoming electrons, vx? Thomson devised an ingeneous way to measure this velocity. He created a magnetic field perpendicularly to the direction of the electric field andof the beam, along the z axis. This magnetic field deflects charges moving along the x axis in the y direction. The Lorentz force acting on such a charge is the combination of electric and magnetic forces,
Fy = Ey e - Bz evx
He changed the magnetic field until the electrons were not deflected at all. Then the force along the y axis was zero and vx could be calculated as
vx = Ey / Bz
Thomson found a value of e / m = 1.1¥ 1011C/kg. about 50% less then the currently accepted value. Still, knowing the elementary charge from electrolysis, it was immediately clear that the mass of the negatively charged particle, discovered by Thomson, was less then a thousandth of the mass of the lightest atom, the Hydrogen atom. The atom must indeed be dividible. The electron was the first elementary particle discovered.
Millikan experiment of measuring e: Though the electron charge could be indirectly calculated by elctrolysis experiments, physicists strived to find a direct way to measure this all-important physical quantity. The first successful experiment is tied to the name of Millikan. He measured the motion of positively charged minsicule oil-droplets between horizontal plates. Observing the droplets through microscope was easy when they were iluminated form the sides by a powerful light source. The oil droplets would fall under the influence of gravity. Observing single droplets he was able to determine their masses. He used Stokes' law for objects moving through medium. Stokes law states that the friction force (drag foce) acting on a sphere of radius r is
F = 6 p h r v
where v is the velocity of the object and h is the coefficient of viscosity in air. Since in gravitational field the velocity of the sphere would increase rapidly, the drag force also increases and soon equals the gravitational field,
m g = 6 p h r v
At this point there no force acts upon the oil droplet and moves with a constant velocity. In practice, this "terminal" velocity is attained extremely rapidly and all that one can observe is oil droplets moving with a constant velocity. Knowing the density of oil one knows the relation between the radius and the mass of droplets m = 4rp r3/ 3. Measuring the velocity one is able to determine the radius of the droplets from the equation:
4rp g r3/ 3 = 6 p h r v,
or equivalently
r = [ 9 h v / 2 g r ]1/2
and
m g = 4rp g [ 9 h v / 2 g r ]3/2/ 3 = 9 p [ h v ]3/2 [ 2 / g r ]1/2
Millikan, dropping the oil spheres through a small hole which was in the charged upper plate, automatically charded the the droplets by a certain charge. When a potential diffrence is created between the plates charged droplets will experiemce athird force, beyond the gravitational and drag forces. After an extremely short transition they will attain a different terminal velocity, dictated by the balance of forces (here we assume that the droplets start to move up after turning the electric field on)
q E - m g - 6 p h r v' = 0,
or equivalently
q = (1/E) { 9 p [ h v ]3/2 [ 2 / g r ]1/2 + 6 p h [ 9 h v / 2 g r ]1/2v' }
= (1/E) 9 p h 3/2 [ 2 v/ g r ]1/2 [ v + v' ]
Measuring the velocities v and v' this formula allows for the determination of the charge on the droplets. Following many droplets with electric field on and off he was able to generate a large number of measured charge values of his droplets. In every case the charge was an integer multiple of a fixed elementary charge, e µ1.7¥10-19C, proving the quantization of charge.
Millikan devised an even more precise and spectacular proof of charge quantization in his experiments, for which he did not even need the viscosity of tair of the density of his droplets. He used X-rays to irradiate the air around his droplets during his experiments. The X-ray ionized certain air molecules (Compton scattering) and the ionized molecules sometimes collided with his charged oil droplets. Upon such a collision, the droplets having been negatively charged loose some of their negative charge. Every time when the charge of a droplet changes its velocity also changes abruptly, as shown by the above relation between charge and velocity. Now, using three velocities for each droplet (two velocities with electric field and one without) he was able to eliminate all parameters but the three velocities and get an equation for the ratio of two charges, before and after the irradiation by X-rays. This is easily seen if one takes the ratio of two equations for the charge at changed v' velocities. All other parameters, E, h, r unchanged one obtains
q1 / q2 = (v + v1')/ ( v + v2 ),
where v1 and v2 are the two velocities in the presence of the electric field, while v is the velocity in the absence of the electric field. The ratios q1 / q2 for various oil droplets were always the ratios of small integer numbers showing the existence of an elementary charge. Of course, from this equation alone he would not have been able to determine the value of the elementary charge.
Example: Suppose n a Millikan experiment the distance between the plates is 1cm. The potential is 31.5V. A plate with a marking in every 1/500 cm behind the droplets is used to measure the drop and rise of oil droplets. A droplet rises 10 notches in 10s in the presence of the electric field. It drops 10 notches in 100s under the influence of the gravitational field alone. The density of the oil used is 0.9g/cm3=900kgm-3. The coefficient of viscosity is h = 1.8¥10-4 gs-1cm-1 = 1.8¥10-5 kgs-1m-1. How many free electrons are on the droplet?
Solution: The two velocities are v' = 0.002 cm/s= 2¥10-5 m/s and v= 0.0002cm/s = 2¥10-6 m/s. The electric field is E = V / d = 3150 Vm-1 Then we should use the formula
q = (1/E) 9 p h 3/2 [ 2 v/ g r ]1/2 [ v + v' ]
= 9 p / 3150 [1.8¥10-5]3/2[ 2 ¥ 2¥10-6 / 9.1¥900]1/2 2.2¥10-5C
= 3.33¥10-19C µ 2e.
Thus there are two electrons on the droplet.
Based on his experiment, Thomson decided that atoms were not indivisible as it was thought before, but rather contained electrons in some way. Beyond the individisible atom the simplest picture he could imagine was the following: Atoms are neutral, because in a normal state matter is neural. Even if one charges matter the charge is always much smaller than eNA= 96,500C for each molar mass. Then if electrons were part of the atom then one needs a postive charge to counterbalance the negative charge of electrons. Thomson assumed that electrons are distributed in a uniform postively charged sphere inside the atom. This was the atom model of Thomson that Rutherford set out to investigate by experiments.
Even before Rutherfords experiment, in 1903, Philip Lenard questioned the validity of Thomson's atom model. He noticed that cathode rays (electrons) could penetrate thin foils of matter, which were still many thousands of atoms thick, practically without a change of direction. It is easy to see that an electron going through a maze of Thomson atoms would hopelessly lose its way and completely change its direction. So Lenard postulated that atoms should be mostly empty.
Rutherford was fascinated by Lenard's model and decide to investigate it. He decided to use a-rays for his investigations. The reason for using a-rays was that they appeared to be very penetrating, traversing matter to larger distances then electrons. Using experiments like Thomson's it could be shown that a-rays consisted of positively charged particles (+2e) of mass similar to that of a He atom. So it was clear that they were positively charged He ions. The experiments Rutherford deviced (1906-1911) seem deceptively simple. a-rays emanating from a radioactive material were collimated into a narrow beam, which was used to bombard a thin gold foil. A photographic plate behind the foil detected the scattering if any of the a-ray particles. If one expects the Thomson model of a positively charged medium with electrons, as raisins in a cake, swimming in it then most a-ray particles should go through the atom undisturbed. Some would collide with electrons, but considering that a-ray particles were over 7000 times as heavy as electrons, such a collision would deflect an a-ray particle as much as a whale would be deflected by a collision with a goldfish. Indeed, the photographic plate showed a large blob where the beam traversing the gold foil hit. The blob was slightly smeared by the gold foil, in agreement with calculations assiming collisions with atomic electrons. What came as a surprise, was that ever so rarely, once in every few thousand collisions a-ray particles were substantially deflected by as much or more than 90o. Some of them were even scattered backwards. This cannot be explained by any other way but by the presence of small massive charged objects in matter. There mass should be at least as large as that of the He atoms, possibly much larger. Since electrons are negative, to insure neutrality of matter, these objects must be positively charged, but also from the formula derived by Rutherford for the pattern of scattering it also followed that the target particles must be positve.
Example: Let us find the maximal angle of deflection for the elastic scattering of particles of different mass. Suppose that the mass and velocity of incoming particles is ma and va, while the same for the target particles is mt and vt, in the CMS. Suppose the scattering angle is q in the CMS. Momentum consrvation tells us that mava = mtvt. Then the longitudinal and transverse momenta in the lab system are
vaL = va cosq + vt = va ( cosq + ma / mt ),
vaT = va sin q.
The ratio of these determines the tangent of the scattering angle in the lab system
tan q L = sin q / ( cosq + ma / mt )
If the mass of the target particle is smaller than that of the alpha particle then the denominator is always positve and the tangent is always positive as well. In other words the lab scattering angel cannot reach 90o. If the mass of the alpha particle is much larger than the mass of the target than the cosine in the denominator is negligible and the maximum value for the tangent of the lab angle is attained when sin q = 1 corresponding to a lab angle of q L µ tan q Lµ mt / ma which is vary small if the target is an electron. In other words alpha particles scattered by electrons can only scatter at very small angles.
After the initial observations Rutherford and his collaborators performed a long series of painstaking experiments in which they measured and calculated the expected angular distribution of a-ray particles. Later on, after 1911, it was also possible to use the newly discovered Wilson chamber to see that path of a-ray particles directly. Rutherford, measured the angular distribution of the alpha particles scattered at large angles. Assuming that a small positively charged object of large mass scattered the alpha particles, using simply Coulomb repulsion as the force between the two he calculated the trajectory of the alpha rays at various impact parameters. The impact parameter is distance the incoming particle is off from central collision. At a given value of the impact parameter, p, the equation of motion can be solved and the trajectory calculated. It turns out that the relation between the impact parameter and the scattering angle, q , is the following (one can find the trajectory by using energy and angular momentum consevation for the a-ray particle):
cot(q / 2 )= M v2p / (2Ze2k ),
or
p = 2Ze2k cot(q / 2) / (M v2).
Note that very far from the target particle, either before or after the collision, the particle in the beam (a-ray particle) does not feel the Coulomb force of the target, consequently it will travel along a straight line. The relative angle of these two straght lines is defined as the scattering angle. We will not derive this relation because it is fairly lengthy. The relation between the impact parameter and the angle shows that at large impact parameter the angle is very small. It should be like that because if the alpha particle travels very far from the target then it should not be deflected at all. The closer it passes to the target the larger the deflection becomes. At p=0 cot q = 0 as well, implying q = p/2, perpendicular deflection.
The chance of having the impact parameter between p and p+dp is proportional to the area of a ring of thickness dp, which is ds = 2p p dp, as shown on the diagram below
ds is called a cross section. Suppose that there is a single target atom in a target area A. Then the probability that a target particle has an impact parameter between p and p+dp is ds/A. Since the impact parameter has a one to one relation with the scattering angle, one can express ds in terms of the scattering angle as well. One gets
ds = 2p p dp = 2p (2Ze2k )2cot(q / 2 ) / (M v 2 sin (q / 2)) 2 dq
= 2p (2Ze2k )2cos(q / 2 ) sin (q / 2)/ [(M v 2)2 sin4 (q / 2)] dq
= 2 (Ze2k )22p sin q dq / [ (M v 2 ) 2sin4 (q / 2 )]
= 2 (Ze2k )2dW / [ (M v 2 ) 2sin4 (q / 2) ].
The multiplier d W = 2p sin q dq is just the infinitesimal solid angle. In other words, it is an infinesimal fraction of the surface of the unit sphere, corresponding to a ring of thickness dq at an angle q from the x axis. This is exactly the area where the projectile will scatter if has an impact parameter between p and p+dp. Then ds can be interpreted as the area that is to be hit so that the alpha particle would scatter into solid angle d W . Such an area is called a differential cross section. It is cross section, because it plays the role of a cross section of a ball that blocks the path of a projectile. It is called differential, because we specify a differential angle range for the scattered particle. If one integrates a differential cross section over all angles then one obtains the total cross section. The differnetial cross section is the quantity an experimentalist would measure. It can be easily translated to numbers an experimentalist would measure. Suppose a detector has an area of A and is a distance R away from the target. As we mentioned earlier dW is segment of the unit sphere at angle q from the direction of the incoming beam. To get the solid angle corresponding to area A one has to consider the ratio A / dW = 4p R2 / 4 p 1 = R2. In other words, dW = A / R2. Also ds = Dn / n, where Dn is the number of particles detected in area A per unit time and n is the number of particles falling on unit area in unit time. Using these relations we can write
Dn = (Ze2k )2nAN / [2R2 (M v 2/ 2) 2sin4 (q / 2)]
which is a direct relation between among measured quantities. N was introduced in the above equation because without N the formula would tell us the scattering probability for the case if there is only one target atom per unit area. N is the number of atoms in the targer per unit area.
Rutherford's results showed a remarkable agreement with this formula. At the largest angles, however there was a deviation. Rutherford realized the reason of the deviation. The formula was derived assuming the scettering of pointlike objects. If the objects are extended then there are deviations from the above derived Rutherford formula. This would happen if the alpha particle goes as close as to penetrate the extended positive charge inside the atom. He devised a clever condition for this, which helped him to estimate the size of the positive charge. An alpha particle can penetrate the positive charge of radius r only if its kinetic energy is larger then the Culomb repulsion energy at the surface
M v 2/ 2 > 2Ze2k / r.
By changing the kinetic energy or the type of atom (Z) one can reach a boundary where the kinetic energy is no longer sufficient to reach the central positive charge. This critical value of the kinetic energy or Z allows one to calculate the radius r. Rutherford concluded that the size of the positive charge in aluminum atoms was about r=5¥10-15m, 4 or 5 order of magnitude smaller then the size of the aluminum atom. Rutherford called the central area where the charge was concentrated at atomic nucleus. His atom model consists of a central, extermely small atomic nucleus and electrons orbiting the nucleus, but otherwise the atom is void of matter.
Most of these questions were not answered for a long time after Rutherford. Question number four was soon answered by Bohr. We will first discuss atomic spectra then the atom model of Bohr, which is a great improvement over the model of Rutherford.
It has been known for a long time that light can be decomposed into its monochromatic components using a prism. The spectrum of a hot liquid or solid has a continuous spectrum, because the atoms interact and emit light cooperatively. The system has many degrees of freedom and these degrees have all different modes of oscillation. The number of modes is so large (as Avogadro's number) that the spectrum is continuous for all practcal purposes. The situation is different with hot gases. Interaction among gas molecules is negligible and so gas molecules radiate independently. They have a small number of degrees of freedom, consequently a small number of oscillation modes. Each molecule of a given gas is an exact copy of the others and radiates at the same frequency. The result is a line spectrum: Large dark intrvals are interpersed wit
The spectrum we discussed so far is a so-called emission spectrum. There is an alternative way to measurre the spectum of gases, the absorbtion spectrum. Suppose light rays of a continuous spectrum traverse colder gas. Then the same frequencies of light that are emitted by hot gases are absorbed from the continuous spectrum and we observe dark lines in a light continuous spectrum. This is the way most stellar spectra are obseved. In the dense interior of the star a continuous spectrum of light is produced. In the outer layers colder gases absorb the lines characteristic to these gases from the continuous spectrum..
In the latter half of the nineteenth century a large number of data were collected concerning spectral lines, but until Niels Bohr there was no theoratical explanation for them. A phenomenological formula was derived for the lines of the hydrogen atom. This formula has the form
1 / l = f / c = R [ 1/ n2t - 1/ n22 ]
where nt and n2 are integers and R is a constant, R = 1.10¥107 m-1, is Rydberg's constant. At a fixed value of nt one obtains a "series if one takes all the integer values n2 > nt. In particular one obtains the Lyman series (ultraviolet light) for nt = 1, the Balmer series (visisble light) for nt = 2, and various infrared light series for nt > 2.
Bohr gave a brilliant explanation for the emisison spectrum of the hydrogen. He had some ad-hoc assumptions though which were explained only after the discovery of quatum mechanics. He was fully aware of Planck's and Einstein's discovery concerning the quantized nature of radiation. He assumed, correctly, that spectral lines are well defined because atoms can emit light at certain well defined frequencies only. Using the Balmer formula the photon energy corresponding to spectral lines can be written as
E = hf = h c R [ 1/ nt2 - 1/ n22 ] = h c R / nt 2- h c R / n2 2= E2 - E1
a difference of two energies. He assumed that hydrogen atoms can be in a series of stationary states labeled by the "quantum number" n. The binding energy of these states is
E = - h c R / n 2
He further assumed that when an atom goes from one state into another state, then energy difference is radiated in the form of a photon. This immediately explains absorbtion spectra as well: transitions between two states are allowed in both directions, the emission of a photon allows the atom to go into a lower excited state, while the absorbtion of a photon of the same frequency allows the atom to go back to the higher excited state.
Bohr was aware of the Rutherford atom model as well, so he ommediately saw the possibility of improving on that model. In the Bohr atom model the lectrons orbit the nucleus on circular orbits. There are certain stable orbits in which the electron does not radiate. Bohr understood that our everyday experience, based on the Maxwell equation, requiring that accelerated charges radiate, cannot be applied to atomic physics. He also realized that the formula for energy levels can be derived from a simple postulate: The angular momentum of elctrons inside the atom is quantized as follows
L = m r v = nh / 2p.
The quantity h / 2p is usually denoted by an h with its top crossed, called h-bar. I will use the form h / 2p because I cannot write the appropriate character on the web. The quantization condition for the momentum allows us to derive the expression for the energy level of the hydrogen atom. Consider that on a stable orbit the mass times the centrifugal acceleration is exactry equal to the force keeping the electron in orbit, the Coulomb force.
m v2/ r = e2 k / r 2
or equivalently
m v2 r = e2 k.
Divide now both sides by L=mvr and use L = nh/2 p to get
v = e2 k / L = 2p e2 k / nh
and
r = L / m v = nh / (2p m v ) = (nh ) 2/ (4p2m e2 k ) = n 2 r 0,
where r 0 is the Bohr radius,
r 0 = h 2/ (4p2m e2 k )
In other words, the radius of the atom increases quadratically with the quantum number n. Its lowest value is attained at n=1, where it equals to the Bohr radius.The Bohr radius is 0.0529 nm. It is the size of the hydrogen atom in its ground state. Let us now calculate the total energy. The total energy is the sum of the kinetic energy and the potential energy. The kinetic energy is
K = m v2 / 2 = m ( 2p e2 k / nh ) 2 / 2 = e2 k / ( 2 r 0 n 2 ),
while the potential energy is
V = - e2 k / r = - e2 k / ( r 0 n 2 ).
Combining K and V we obtain
E = K + V = - e2 k / ( 2 r 0 n 2 ).
This is indeed of the form
E = -h c R / n 2
where we can identify the Rydberg constant R as
R = e2 k / ( 2 r 0 h c ) = e 4 k 2 2p2m / ( h 3 c ) .
Notice that R is expressed by constants that where very well known by 1913, when Bohr discovered his atom model. Since he obtained a precise value for R, in terms of universal constant also well known, in complete agreement with experiments, his model was accepted instantaneosly.
The energy levels of the hydrogen atom can also be written as
En = - |E0 | / n2 = - 13.6 eV/ n2
In other words the energy levels become denser and denser for large n, they approach zero. All transitions among energy levels are possible, they corespond to the energies of emitted photons.
The generalization of the Bohr energy formula for nuclei with charge Ze is very simple. One only needs to substitute e2 k in the Coulomb force by Ze2 k . Then of course this subtitution must be performed in all the formulas. In particular, one obtains
r =n 2 r 0 / Z
and
En = - |E0 | Z 2/ n2 = - Z 2 13.6 eV/ n2
The formula applies to Z-1 times ionized atoms only, because with less ionization the other electrons would generate a screening effect that is very hard to take into account. This formula applies to the spectrum of the singly ionized He+ ion, which has a single electron except the nuclear charge is 2e.
Problem: Consider the following transitions of the Li+ ion:
1. ni=1 ---> nf=2
2. ni=3 ---> nf=4
3. ni=2 ---> nf=6
4. ni=3 ---> nf=8
5. ni=4 ---> nf=2(a) Is radiation absorbed or emitted in these transitions?
(b) What is the ratio of wavelengths in these processes?
Solution:
(a) The first four transitions are either from the ground state to an excited state (1), or from an excited state to a higher excited state. Thus, the atom always absorbs radiation to use the energy of radiation to raise it to a higher excited state with lower binding energy. In the last process the atom goes into a lower state, so it radiates a photon.
(b) Wave lengths are given by the formula
1 / l = RZ2 [ 1/ n2i- 1/ n2f ]
The prefactors cancel in ratios of wavelengths. Thus,
l2 / l1 = [ 1/ n21t - 1/ n212 ] / [ 1/ n22t - 1/ n222 ] =[ 1 - 1/4] / [ 1/9 - 1/16] = 108/7
l3 / l1 = [ 1/ n21t - 1/ n212 ] / [ 1/ n23t - 1/ n232 ] =[ 1 - 1/4] / [ 1/4 - 1/36] = 27/8
l4 / l1 = [ 1/ n21t - 1/ n212 ] / [ 1/ n24t - 1/ n242 ] =[ 1 - 1/4] / [ 1/9 - 1/64] = 432/55
l5 / l1 = [ 1/ n21t - 1/ n212 ] / [ 1/ n24t - 1/ n242 ] =[ 1 - 1/4] / [ 1/4 - 1/16] = 4
Bohr also realized that the chemical elements and their properties can be explained if one considers that more thenone electrons can occupy the same energy level, En . In particular two n=1 electrons complete the electron shell of the neutral He atom. He realized that no more then two electrons can occupy this shell, so the n=1 shel of He is complete and so it cannot easily receive or give electrons to other atoms. Therefore we say that the He has a closed shell. Chemical reactions are related to the exchange and sharing of electrons. It is hardto remove electrons from a closed shell. Therefore He, an inert gas is very resistant to chemical readtion.
The next heavier atom, Li must contain three electrons. In the ground state of the Li atom one of the electrons must be in the n=2 state, which forms the next shell. This electron is much weaker bound (the real binding of this electron is even weaker than the value provided by a H-like formula, with Z=3, because of the screening effect of the two n=1 electrons). The outer electron in the n=2 shell is much easier to separate from the Li atom, therefore the Li atom redily enters into a chemical reaction. It shares or gives away its n=2 electron.
Bohr realized that the valence of atoms is related to the number of electrons in the outer electron shell. Atoms have similar chemical properties if the occupation of their outer shell is similar. For example, the alkali elements, starting with Li, all have a single electron in their outer shell and are all very reactive.
As we saw earlier, the binding energyof electrons in heavy atoms is proportional to Z2. Obviously, for large values of Z this energy can be huge. Therefore, if an elctron is knocked out from a low shell of a heavy atom then a very large amount of energy is freed when another electron "falls" into this "hole" left by the removal of the first electron. Then a very high energy photon, an X-ray is produced. In fact, this is the mechanism for producing X-rays as it was pointed out by Bohr.
Bohr found an explanation for the quantization rule of angular
momentum. He theorized that for very large orbit (which is equivalent
to large n) the rules of classical physics must be valid, because the
angular momentum becomes macroscopic. He said that the atom should
correspond to a classical system. Classically, the atom
should radiate energy: there are no stable orbits. In other words the
radius should decrease continuously. On the quantum language we would
expect that the classical picture is an approximation of the atom
going from level n to n-1 to n-2, etc. These jumps are so small that
in the large n limit they seem to be continuous.
It is easy to express all physical quantities by the angular
momentum. Note that Newton's law implies that
mv 2 / r = k e 2 / r 2 .
This can be written as
L v = k e 2
allowing to express v and r by L. Let us do that. We obtain
v = k e 2/ L
and r = L2 / (m k 2e 4 )
Substituting these into the expression of the total energy we obtain
E = - m k 2e 4 / 2 L 2
Classically the energy is radiated. The freqency of radiation of a rotating charge is be equal to f = v /2rp, the velocity divided by the circumference of the orbit. Again, expressed by the angular momentum we obtain
f = m k 2e 4 / 2 p L 3
Use now the fact that radiation of frequency f is radiated in quanta of
hf = h m k 2e 4 / 2 p L 3
When energy is radiated the atom drops to a lower energy state. The diffence of the energies of these states, provided the change is small, is
E' - E = - (m k 2e 4 / 2 ) ( 1/ L '2- 1 / L 2)
µ (m k 2e 4 / 2 ) (L '2- L 2) / L 4 µ m k 2e 4 (L ' - L ) / L 3
Subtituting the expression of hf, we can cancel m k 2e 4 / L 3 and we obtain
L ' - L = h / 2 p
This equation tells that the angular momentum jumps by a unit of h bar. Of course, this was derived at high energy, or angular momantum only. Bohr ingeneously generalized this relation and postulated that the angular momentum can always change by this unit, i.e. in general it is an integer multiple of h / 2 p . The way the quantization of the anglar momentum was "derived" it automatically implies that the correspondence principle is satisfied. In the classical, high angular momentum, limit the atom radiates according to classical rules. If we were able to calculate the lifetime of high angular momentum states (this can be done using quantum electrodynamics) then we would obtain that the decaying of high excited states would approximate the spiraling in of a classical electron rotating around the proton. The jerkiness of this spiral would not be observed due to the smallness of the energy (or angular momentum) quantum. The lower the energy of the state gets the more the system deviates from the classical rules. When one reaches the ground state, which is stable, classical physics gives a completely false answer.
The Franck-Hertz experiment was the first direct proof of quantization of atomic levels. A vacuum tube, with a drop of mercury inside was used to investigate the passage of electrons from a heated cathode filament to a collector anode (which was at a negative voltage). On the way to the anode the electrons were accelerated by a grid at a potential that was positive copared to that of the cathode. The current observed at the anode should be an increasing function of the accelerating voltage on the grid. The observation was different. The experimenters saw, superimposed on the rising tendency of the current, alternating minima and maxima.
The explanation if this phenomenon is simple, if one considers the Bohr atom. As the voltage on the grid increases, at some point the kinetic energy of the electrons is larger than the excitation energy of a Hg atom (from ground state to excited state). Colliding with a Hg atom sometimes the electron transfer the excitation energy to the Hgand loses substantial kinetic energy. After the loss of kinetic energy the electron will not have sufficient kinetic energy to reach the anode. Therefore the current will drop. If we further increase the potential then the current will start to rise again because even electrons that transfered energy to a Hg atom once will gain enough energy to reach the anode. If we increase the voltage even further then the electrons that collided inelastically once will have enough energy to collide inelastically again. Then the current will drop again. The result is equidistant maxima and minima in the current. The distance of these maxima, 4.9 eV, is just the excitation energy of the Hg atom. To complete the proof that they really saw Hg atoms excited by electrons Franck and Hertz also detected the 4.9 eV photons emitted by the excited Hg atoms when they went back to the ground state.