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					ICE2251 Modern Physics Chapt. 6 R-B model of atom

6. The Rutherford-Bohr Model of the Atom
(Textbook. “Modern Physics”, K. Krane, 2nd ed., Chapt. 6) 6.1 Basic properties of atoms Before we construct a model of the atom, let us summarize some of the basic properties of atoms. 1. Atoms are very small, about 0.1nm (0.1x10-9m). 2. Atoms are stable. They do not spontaneously collapse: therefore the internal forces that hold the atom together must be equilibrium. 3. Atoms contain negatively charged electrons, but are electrically neutral. If we disturb an atom with sufficient force, electrons are emitted. 4. Atoms emit and absorb electromagnetic radiation. This radiation may take many forms-visible light (~500nm), X-rays (~1nm), ultraviolet rays (~10nm), infrared rays (~0.1nm), and so forth. 6.2 The Thomson model An early model of the structure of the atom proposed by J. J. Thomson. In this model, an atom contains Z electrons which are imbedded in a uniform sphere of positive charge Ze and radius r. (Fig. 6.1) The force on an electron at a distance r from the center of a uniformly charged sphere of radius R was computed using Gauss‟s law: Ze 2 F r  kr 4 0 R 3

(6.1)

This linear restoring force permits the electrons to oscillate about their equilibrium positions just like a mass on a spring subject to the linear restoring force F=kx. We expect the electrons in the Thomson atom to oscillate about their equilibrium positions with a frequency   k / m / 2 . However, this turns out not to be true; the calculated frequencies do not correspond to the frequencies observe for radiation emitted by atoms. The most serious failure of the Thomson model arises from the scattering of charge

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

particle through the atoms. We consider the passage of a single positively charged particle, (alpha particle, or example) through the atom, as Fig. 6.2. The deflected angle  by the repulsive force from the electrons in the Thomson atom can be calculated as follows. The momentum in the y direction

Py   Fy dt
Fy  F cos 

(6.2)

 b/ r
The projectile of the positive particle with a charge of q=ze experiences a force F given by the Thomson‟s Eq. 6.1.

F

zZe 2 r  zkr 4 0 R 3

(6.3) (6.4)

b Py   zkr  dt  zkb t  zkbT r
travel through the atom.

where T is the total time it takes for the projectile to
2 R2  b2

(6.5)  2 zkb R 2  b2 Then Py  (6.6) P  Py  p y p y (6.7) tan    Px px P p y 2 zkb   R2  b2 (6.8) 2 P m According to this result, when b=0 the projectile is undeflected (=0). When b=R the projectile is again undeflected. For an intermediate value b=R/2,

T

 avg

3 2 zkR 2 4 m 2

(6.9)

The calculate angle from this equation for gold target is about 0.01. This value is quite inconstant with the experimental results (Fig.4) in which the scattered particles are observed at all possible values of . The only trial to explain the large scattering angle is an increment of the scattering

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

angle by sequential scattering from many atoms as Fig. 6.5. This is an example of a „random‟ walk problem. For N times scattering, the average net scattering angle  is given by

  N avg

(6.10)

To become a large angle 90 by accumulation of one time scattering with a 0.01 step, it must happen about 104 successive scattering. If 0.01 scattering happens with a 1/2 probability, the probability of fining 104 successive scattering is 1 / 2
10000

 10 3000 . The

experiment done in the laboratory of Ernest Rutherford in 1910 showed that the probability of an alpha particle scattering at angles greater than 90 was about 10-4. It showed a remarkable discrepancy between the expected value and the observed value. The results propose that the mass and positive charge of the atom were not distributed uniformly over the volume of the atom.

6.3 The Rutherford nuclear atom In analyzing the scattering of alpha particles, Rutherford conclude that the most likely way an alpha particle (m=4u) can be deflected through large angles is by a single collision with more massive object. Rutherford proposes that the charge and mass of the atom were concentrate at its center, in a region called the nucleus. Fig. 6.6 illustrates the scattering geometry in this case. The projectile having a charge ze experiences a repulsive force due to the positively charged nucleus:

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F

ze Ze 
4 0 r 2

ICE2251 Modern Physics Chapt. 6 R-B model of atom

(6.11)

He assumed also that the nucleus is so much massive than the projectile that it does not move during the scattering process. The projectile follows a hyperbolic path; in polar coordinates

1 1 zZe 2 cos  1 (6.12)  sin   r b 8 0 b 2 K
where K is the kinetic energy of the projectile. As shown in Fig. 6.7, the initial position of the particle is =0, r, and the final position is =-, r. At the final position, Eq. 6.12 reduces to

zZ e 2 b cot / 2 2K 4 0
through larger angles.

(6.13)

This equation indicates that projectiles approaching with smaller b will be scattered

The scattering of charge particles by nuclei is called Rutherford scattering. (1) The fraction of projectiles scattered at angles greater than  Suppose a foil with a one-atom-thickness. - Each atom has a cross-section area: R2 - If the foil contains N atoms, the total cross-section area: NR2 - For scattering at angles greater than , the projectile should enter within an area: b2 - The fraction of projectiles that fall within that area: b2/R2 Suppose a foil with many atomic layers and let be t the thickness of foil, A its area, the  density and M the molar mass of the material of the foil. - The volume of the foil: At - Its mass: At - The volume of moles: At - The number of moles: At/M - The number of atoms or nuclei per unit volume: n  N A - The number of nuclei per unit area: nt = (NAt/M) -1 - For a scattering at angles greater than , the projectile must fall within an area of the

At 1
M At



NA M

(6.14)

where NA is the Avogaro‟s number (the number of atoms per mole)

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

center of atom: f b f   ntb 2

(6.15)

For a gold foil with a thickness of 2m, f>45 = 4.4x10-4 and f>90 = 7.5x10-5. These values agree well with the experimental data.

(2) Rutherford scattering formula and its experimental verification The fraction that a projectile be scattered into a small angle at  (a ring between  and

+d) is

df  nt 2b db

Differentiating Eq. 6.13,

db 

zZ e 2  csc2  / 2 d / 2 2K 4 0
2 2





(6.16)

 zZ  df  nt    2K 

 e2  2   4   csc  / 2  cot / 2d  0  





(6.17)

The area of the ring between  and +d is dA=(2rsin) rd. The probability per unit area for scattering into the ring is
2

nt N    df / dA  2 4r

 zZ     2K 

2

 e2    1   4   sin 4  / 2   0   

(6.18)

Following properties expected from above Eq. 6.18 were identified from the experiments (See Fig. 6.10 – 6.12). - N()  t - N()  Z
2 -2 -4

(t is the thickness of the foil) (Z is the number of electrons of the atoms in the foil) (K is the kinetic energy of the projectile)

- N()  K

- N()  sin (/2)

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

(3) The closest approach of a projectile to the nucleus A positively charged projectile shows down as it approaches a nucleus, exchanging part of its initial kinetic energy for the electrostatic potential energy due to the nuclear repulsion. - The potential energy that the projectile gains:

zZe 2 U   Fdr  4 0 r 1
- The total energy of the projectile, E=K+U, is maintained as a constant when it approaches the nuclear. - At initial when the projectile is far from the nucleus: U=0, E=K =m2/2 - At distance rmin , the speed is min and:

1 1 zZe 2 1 2 E  m min   m 2 2 4 0 rmin 2
Angular momentum is also conserved. mb=mminrmin  min =b/rmin Combining Eqs. 6.19 and 6.20, we find
2

(6.19)

(6.20)

1 1  b  1 zZe 2   m 2  m 2 2  rm in  4 0 rm in  
From this equation, we can solve rmin.

(6.21)

When b=0, the projectile would lose all of its kinetic energy, and thus get closest to the nucleus. At this point its distance from the nucleus is the distance d of closest approach.

d

zZe 2 4 0 K 1

(6.22)

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

6.4 Line spectra The radiation from atoms can be classified into continuous spectra and discrete spectra (or line spectra). In a continuous spectrum, all wavelengths from some minimum to some maximum are emitted. The radiation from a hot glowing object is an example of the continuous spectra. On the other hand, when lights are emitted from small amount of gas of a certain homogeneous element by electric discharge in a tube or illumination of white light into the gas, the light emits at a few discrete wavelengths. Johannes Balmer found that the group of emission lines of hydrogen in the visible region could be very accurately calculated from the formula:

  364.5

n2 n 2
2 2

(n =3, 4, 5 ….)

(6.23)

 Balmer formula (Balmer series)

where  is in units of nm and n can take integer values beginning with 3. The wavelength 364.5nm, corresponding to n, it calls the series limit. It was soon discovered that all of the groups of lines in the hydrogen spectrum could be for with a similar formula of the form

  limit

n2 n 2  n0
2

(n = n0 +1, n0+2, n0+3 ….)

(6.24)

For the Balmer series n0=2, Lyman series n0 =1, Paschen series n0 =3, Brackett series n0=4, and Pfund n0=5. Another interesting property of the hydrogen wavelength is the Ritz combination principle. If we convert the hydrogen emission wavelengths to frequencies, we find the property that certain pairs of frequencies added together give other frequencies which appears in the spectrum. Now we try to find a function that fits these data and try to find a theory that explains the derived function. This approach is the „reverse scientific method‟ which is already applied in the blackbody radiation.

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

6.5 The Bohr model Following Rutherford‟s proposal that the mass and positive charge are concentrated in a very small region at the center of the atom, Niels Bohr in 1913 suggested that the atom was in fact like a miniature planetary system, with the electrons circulating about the nucleus like planets circulating about the sun. The atom thus doesn‟t collapse under the influence of the electrostatic Coulomb attraction between nucleus and electrons for the same reason that the solar system doesn‟t collapse under the influence of the gravitational attraction between Sun and planets. In both cases, the attractive force provided the centripetal acceleration necessary to maintain the orbital motion. We consider for simplicity the hydrogen atom, with a single electron circulating about a nucleus with a single positive charge, as in Fig. 6.19. The attractive Coulomb force provides the centripetal acceleration and it balances with centrifugal force.

e 2 m 2 F  4 0 r 2 r 1
electron.

(6.25)

From this equation we can find the kinetic energy of the

1 1 e2 2 K  m  2 8 0 r
system:

(6.26)

The Coulomb potential energy of the electron-nucleus

e2 U  4 0 r 1 E 1 e2 8 0 r

(6.27)

The total energy E=K+U is obtained by adding Eqs. 6.26 and 6.27: (6.28)

We have ignore one serious difficulty with this model. Classical physics requires that accelerated electric charge like orbiting electron must continuously radiate electromagnetic energy and lose its energy eventually reaching atom collapse. To overcome this difficulty, Bohr made a hypothesis that there are certain special states of motion, called stationary states, in which electrons may exist without radiating electromagnetic energy. In these states, the angular momentum L of the electron takes values that are integer multiples of  . The angular momentum L  r  p  L  rp  mr Thus Bohr‟s postulate is mr  n (6.29)

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

From Eqs. 6.26 and 6.29, we find a series of allowed values of the radius r:

4 0  2 2 n  a0 n 2 (6.31) 2 me where the Bohr radius a0 is defined as rn  4 0  2  0.0529 nm (6.32) me 2 Combing with the energy equation 6.28, the allowed energy levels are a0 

En  
En 

1 2 32  0  n
2 2 2

me 4

(6.33)

 13.6eV n2

(6.34)

This result indicates that the electron‟s energy is quantized. Bohr also postulated that, even though the electron doesn‟t radiate when it remain in any particular stationary state, it can emit radiation when it moves to a low energy level. And the energy difference appears as a quantum of radiation whose energy h is equal to the energy difference between the levels. For example, if the electron jumps from n=n1 to n=n2 , a photon appears with energy: h  E n1  E n 2 (6.35)



 1 1   2  2 64  0   n2 n1    me4
3 2 3

(6.36)

The wavelength of the emitted radiation:



2 2 2 2 2 c 64 3 0  3c  n1 n2  1  n1 n2   2   2   n n 2  R n n 2  v me 4   1 2  2   1

(6.37)

where R is called the Rydberg constant. The wavelengths calculated in this equation is very close to the value of the longest wavelength of the Balmer series. The Bohr formulas also explain the Ritz combination principles, as exampled below.

 n n  cR  
3 2

 1 1   2 2   n3 n2 

n

2 n1

 1 1   cR  2  2  n n1   2 

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

n

  n n 3 n2 2 1

 1 1  cR  2  2 n n2  3

       cR  1  1   cR  1  1  2 2  2 2  n  n n1  n1   2   3
(6.39)

 n

3 n2

  n n 2 1

 n

3 n1

The Bohr model also helps us understand doesn‟t radiation normally at why all the and the only atom emit same in the absorb

wavelength. Isolated atoms are found ground state; the exited states live for a very short time before decaying to the ground state. The absorption contains spectrum only therefore

transitions from the ground state. From Fig. 6.22, we see that only the radiations of the Lyman series can be found in the absorption of hydrogen.

The Bohr theory for hydrogen can be used for any atom with single electron, even though the nuclear charge may be greater than 1. For a nucleus of charge Ze, the Coulomb force acting on the electrons is given by multiplication of Z to Eq. 6.25:

Ze 2 F 4 0 r 2 1
rn   4 0  2 2 n Ze 2 m

(6.40) e e (6.41) + rn ++ rn

The allowed radii and energies become

En  

1 2 32 2  0  2 n
2

m Ze 2

 

2

(6.42)

The orbits in the higher-Z atoms are closer to the nucleus and have larger energies; that is, the electron is more tightly bound to the nucleus.

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

6.6 The Franck-Hertz experiment In a vacuum tube, a small amount of Hg vapor is filled, as seen in Fig. 6.23. In this tube a filament heats the cathode C, which then emits electrons. These electrons are accelerated toward the grid G by the potential difference V, which we control. Electrons pass through the grid G and reach the plate P. When V < Vexcite=(E2-E1)/e the accelerated electron could not excite Hg atoms from the energy level 1 to 2. The electrons may make collision with Hg atoms, but lose no energy in these collision-the collisions are perfectly elastic. When V > Vexcite, the electrons can excite electrons in Hg atoms and lose energy-these collisions are inelastic. The original electron moves off with very little energy. Even if it pass through the grid, the electron might not have sufficient energy overcome small barrier to move through the vapor and thus may not reach the plate. Thus when V ~ Vexcite , a drop in the current is observed. Increasing the accelerating voltage, the voltage drops are observed at Vexcite, 2Vexcite, 3Vexcite, …, by multiple collision, as shown in Fig. 6.24. This experiment should give rather direct evidence for the existence of atomic excited state.

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

6.7 The correspondence principle Bohr‟s model is in excellent agreement with the discrete line spectra in atoms. However, Bohr‟s model has a dilemma: an accelerated charged particle radiates electromagnetic energy according to classical physics, but an electron in Bohr‟s atomic model, accelerated as it moves in a circular orbit, does not radiate. Bohr‟s solution to this serious dilemma was to propose the correspondence principle, which states that Quantum theory must agree with classical theory in the limit in which classical theory is known to agree with experiment. Or equivalently Quantum theory must agree with classical theory in the limit of large quantum numbers. Let us see how we can apply this principle to the Bohr atom. In classical theory, for a circular movement of an electron have the orbital speed   2 K / m , where K is the kinetic energy. The period to around one circle:

T

2r





2r 2K / m



16 3 0 mr 3 e

(6.43)

where we use Eq. 6.26 for the kinetic energy. The frequency is



1 e  3 T 16  0 mr 3

(6.44)

Using Eq. 6.31 for the allowed orbits rn , we find

n 

1 3 32  0  n
3 2 3

me 4

(6.45)

A classical electron moving in an orbit of radius rn would radiate at this frequency n. If we made the radius of Bohr atom so large that it went from a quantum-sized object (1010

m) to a laboratory-sized object (10 -3m), the object should behave classically.

According to Eq. 6.36, the frequency is



 1 1     n  12  n 2  64  0    me 4
3 2 3



64  0  n n  1
3 2 3 2

me 4

2n  1

2

(6.46)

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

If n is very large, then we can approximate n-1 n and 2n-12 n, which gives



2n me 4 1  2 3 4 2 3 3 3 3 64  0  n 32  0  n

me 4

(6.47)

This is identical with Eq. 6.45 for the classical frequency. The „classical‟ electron spiral slowly in toward the nucleus, radiating at the frequency given by Eq. 6.45, while the „quantum‟ electron jumps from orbit n to the orbit n-1 and then to the orbit n -2, and so forth, radiating at the frequency given by Eq. 6.45. (See Fig. 6.25)

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ICE2251 Modern Physics Chapt. 6 R-B model of atom

6.8 Deficiencies of the Bohr model In spite of its success to explain some experimental results, the Bohr model is an incomplete model. The deficiencies of the Bohr model is: (1) It neglects the motion of the nucleus. The electron does not orbit about the nucleus, but instead the electron and the nucleus both orbit about their common center of mass. This effect tends to increase the calculated wavelengths by a factor of about 1.00055. (2) it is useful only for atoms that contain one electron, but not for atoms with two electrons or more. We have to consider the force between electrons themselves. (3) It is unable to account for the double lets of spectrum. (If we analyze the spectrum in precision, we find that many lines are in fact not single lines, but very closely spaced combination of two or more lines.) (4) It can calculate the energy levels, but cannot calculate their intensities. For example, how open will an electron jump to other levels? (5) A serious deficiency of this model is that it completely violates the uncertainty relationship. (The Bohr model was presented a decade before the introduction of wave-like particle mechanics.) Applying the uncertainty relationship for radial direction of the electron moving in a circular orbit, rPr   . For the electron moving in a circular orbit, we know the value of r exactly and thus r=0. If it is moving in a circle we also know Pr exactly (in fact it is exactly zero) and so Pr=0. This violates the uncertainty principle. However, we do not discard the model completely. The Bohr model gives a mental picture of the structure of an atom that some times can be useful.



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ICE2251 Modern Physics Chapt. 6 R-B model of atom

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