1

Atoms

Atomic Constituents The work of J.J. Thomson, E. Rutherford and others toward the end of the 19th century and into the beginning of the 20th century established that atoms consist of a positively charged nucleus and negatively charged electrons. In a neutral atom, the positive charge on the nucleus exactly balances the total negative charge of the electrons. All experiments up to the present time indicate that the electron is an elementary particle, i.e., it has no internal structure. However, the nucleus was found to have structure. Rutherford discovered that the positive charge on the nucleus is carried by protons. Each proton has a charge of +1e, where e is the magnitude of the electron’s charge. The nucleus also contains neutrons, which carry no charge. Modern elementary particle theory and recent experiments indicate that protons and neutrons are not elementary, but are made up of quarks. These quarks carry charges that are a fraction of the electron's charge ( 1

3 e± or 23 e± ). However, free fractional charges have never been observed,

because quarks are always confined within other particles (such as protons and neutrons). Free single quarks do not exist. Thus, all charge in nature is found to be integer multiples of e. Protons and neutrons are collectively called nucleons. The mass of a neutron is only slightly greater than that of a proton. Both are approximately 2000 times as massive as the electron. The number of protons in the nucleus is called the atomic number (or proton number) Z.1 The number of neutrons (neutron number) is given the symbol N. The total number of nucleons is the mass number (or nucleon number) A. Evidently, A = Z + N. The label mass number is appropriate because the electrons contribute very little mass to the atom. To good approximation, the mass of the electrons can be neglected. Symbols for the various atoms are usually written in the form XAZ , where X is the one- or two-letter symbol for the atom. For example, 126 C denotes a carbon atom with a mass number of 12. (The text omits the atomic number from the symbol). Each atom has a different atomic number Z. This is because the characteristics of a neutral atom are determined by the number of electrons in the atom. Atoms of the same element (same Z) may have different neutron numbers and therefore different mass numbers. Such atoms are called isotopes. Examples: 126 C , 136 C , 146 C ; 32He , 42He . Atomic Mass Units and the Mole An atomic mass unit is defined in the following way: one atom of the isotope 126 C is defined to have a mass of exactly 12 atomic mass units (12 u). Using this definition we find that the atomic masses of the other elements are not whole numbers. This is mainly because of the binding energy associated with nuclei. Since nuclei represent bound states of nucleons, the mass of a nucleus is less than the total mass of the individual nucleons by an amount

1 In a neutral atom Z = number of electrons.

2

mΔ = binding energy/ c2. Note that 1 u = 1.66×10-27 kg. One mole (of any substance) contains the same number of objects as there are atoms in exactly 12 g of the isotope 126 C . This number is Avogadro’s number NA = 6.022×1023. Note: NA × (mass of 126 C atom) = 12 g. Thus, NA × (12 u) = 12 g, i.e., 1 u = (1 g)/ NA = (10-3 kg)/ NA = = 1.661×10-27 kg, as given above. Atomic and Nuclear Sizes The size of an atom is dictated by the size of the “orbit” of the outermost electrons. Roughly speaking, this size is of the order of 1 angstrom unit, i.e., 10-10 m. The symbol for the angstrom

unit is A

. The typical size of a nucleus on the other hand is of the order of 10-15 m or 1 fermi. One fermi is equal to 1 femtometer (fm). Probing Atomic Structure Discovery of the Electron With his cathode ray tube2 apparatus, J. J. Thomson was able to measure the charge to mass ratio of the electron. You will do a similar experiment in the Modern Physics Laboratory course.

Important elements of the apparatus include: (1) cathode – emits of electrons; (2) anode – accelerates and collimates electrons; (3) deflector plates – a voltage is applied between these plates to deflect electrons; (4) low-pressure gas – for viewing electrons' path (due to ionization and light emission)3 or fluorescent screen for seeing position at which the electron beam impacts the screen. Consider plates of length L and separation d. If a potential difference V is applied to the plates, then the electric field E is given by E = V/d. If the electrons have a horizontal velocity u, then

2 Cathode rays are beams of what we now know as electrons. 3 The apparatus has to be evacuated anyway, to permit the electrons to move freely.

3

the time taken to travel the length of the plates is given by t = L/u. After emerging from the region between the plates, the y-component of the velocity is given by

.y yeE L eELu a tm u mu

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ (4.1)

Note that the electrons are non-relativistic. Also, the acceleration of an electron due to the gravity is negligible compared to that due to the electric field. Now,

2tan ,y y

x

u u eELu u mu

θ = = = (4.2)

where θ is the deflection angle. Note that in this experiment, the deflection is usually very small; therefore, the approximation tanθ θ≈ (where θ is in radians) is a very good one. From Eq. (4.2), we can solve for e/m:

2

tan .e um EL

θ= (4.3)

Thus, e/m could be measured if θ , E, L, and u are known. To find u, Thompson added a magnetic field B

at right angles to the electric field and the path of the electrons. The magnetic

field was adjusted so that the electron beam was no longer deflected. In this case, ( ) 0.F q E u B= + × =

(4.4) Because the electric field, the magnetic field, and the velocity are all perpendicular to one another, Eq. (4.4) reduces to

.EuB

= (4.5)

Substituting in Eq. (4.3) gives

2 2

tantan .e E Vm LB dLB

θθ= = (4.6)

(Show picture of Thomson's data.) Thomson's result was low by a factor of approximately 2 compared to the currently accepted value because he did not explicitly account for other sources of magnetic field in his apparatus. (Note that the derivation of Eq. (4.6) given above is essentially the solution to Prob. 3.40 in the textbook.) Millikan Oil Drop Experiment Knowing e/m, one could find m if e were known. Robert Millikan devised his famous oil drop experiment in order to obtain the value of e. [For a brief video clip, see: http://www.youtube.com/watch?v=XMfYHag7Liw] Tiny droplets of oil (about 5 microns in diameter) were sprayed in the region between two conducting plates, and were viewed by a telescope.

dragF

mg 0E =

dragF

mg

elecF qE=

4

Without an electric field, the droplets fall under the influence of gravity alone. A drag force opposes their motion and eventually the drop reaches terminal velocity when 0 ,drag Tmg F bu= = (4.7) where uT0 is the terminal velocity with no electric field and b is the drag constant given by 6b Rπη= (4.8) for a sphere. In the expression for the drag constant, η is the viscosity of air and R is the radius of the droplet. Now, the mass m of the drop is given by 34

3 ,m V Rρ π ρ= = (4.9) where ρ is the density of the oil. Substituting Eqs. (4.8) and (4.9) into Eq. (4.7), one finds that

09 .2TuRgηρ

= (4.10)

Hence, the radius of the drop can be calculated and from that, the mass can be obtained. (Typical size of a drop: ~3×10-6 m or 3 mµ .) In the process of spraying, a droplet may acquire a net charge. Also, x-rays can be used to bombard the region containing the droplets, thereby producing charged droplets (positive or negative). If a downward electric field is applied, a negatively charged droplet will rise if elecF mg> . The drag force now points downward. When terminal velocity is reached, we have

,TEqE mg bu= + (4.11) where TEu is the terminal velocity with the electric field in place. But we can eliminate b using Eq. (4.7):

0

,TET

mgqE mg uu

⎛ ⎞= + ⎜ ⎟

⎝ ⎠

i.e.,

01.

T

TE

umgu

qE

⎛ ⎞+⎜ ⎟

⎝ ⎠= (4.12)

Thus, by measuring 0Tu and ,TEu q can be calculated. Millikan found that the charge on a droplet was always an integer multiple of 1.6×1019 C. From this he deduced the quantization of electric charge. Show picture of Millikan’s data (p. 14 Rohlf). Mention Begeman's role and history. The Rutherford Nuclear Atom Play: Mech. Univ. video “Atoms” Ch. 15. After the discovery and characterization of the electron, Thompson proposed a model of the atom in which electrons were dispersed like raisins in a uniform distribution of positive charge

5

(“plum pudding" model). In order to test this model the following experiment was carried out by Geiger and Marsden. Alpha particles (He-4 nuclei) resulting from radioactive decay were used to bombard thin (10-6 m) gold foil, and the angles at which the alpha particles were scattered were measured. It would be expected that the scattering (or deflection) angles would be very small (less than a few degrees) because the electrons were too small to affect the alpha particles significantly. Also, the positive charge is distributed over a relatively large region, so the repulsive electrical force between the atom's positive charge and that of the alpha particle is relatively weak. When Geiger and Marsden performed the experiment, most of the alpha particles were deflected through small angles as expected. However, very surprisingly, there were a few large-angle scattering events with the scattering angle > 90º. On the basis of these results, Rutherford proposed his nuclear model of the atom, in which the positive charge of the atom is concentrated in an extremely small region at the center of the atom. The electrons orbit the nucleus at relatively large distances away. Thus, the atom’s volume is mostly empty space. This would explain why most alpha particles are not scattered. However, on the rare occasions when an alpha particle approaches a nucleus closely, very large electrical forces are experienced by the alpha particle, which is then scattered through a large angle.

[Show: Figs 6.8 & 6.9 & 6.10, p. 155, Krane's Modern Physics] The trajectory is in general a hyperbola. The scattering rate for each angle θ can be found using classical mechanics. In polar coordinates, the equation of the hyperbola is

( )2

20

1 1sin cos 1 ,

8zZe

r b b Kφ φ

πε= + − (4.13)

where z is atomic number of the scattered particle, Z is the atomic number of nucleus, and K is the initial kinetic energy of the incoming particle. Now, for r→∞ , φ → π −θ. Substituting these values yields

2

0

1cot .

2 4 2zZ ebK

θπε

= (4.14)

Incident alpha particles θ

Foil

b r φ

θ

b

Nucleus

6

For foil with n atoms per unit volume and thickness t, the number of atoms per unit area = nt. Particles that are incident within an area 2bπ centered on the scattering nucleus will be scattered at angles greater than θ . The fraction f of projectiles that are scattered at angles θ≥ is given by (number of scattering centers/unit area) 2bπ× . Thus, 2.f nt bπ= (4.15)

We wish to find the number of particles per unit area scattered in an annular (ring-shaped) region between θ and dθ θ+ . The fraction of projectiles entering between b and b+db is given by 2 .df nt bdbπ= (4.16) By differentiating b [given in Eq. (4.14)] with respect to θ , we find

( )( )2

2 1 12 2

0

csc2 4zZ edb dK

θ θπε

= − (4.17)

so that

22

2 1 12 2

0

csc cot .2 4zZ edf nt dK

π θ θ θπε

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ (4.18)

The area of the ring shaped region into which the particles are scattered is given by (2 sin ) ,dA r rdπ θ θ= (4.19) where r is the distance of the detector from the foil. The total number of particles scattered per unit area at θ is given by

( ) ,sc

N dfN

dAθ = (4.20)

where N is the total number of projectiles. Substituting for |df| and dA, we get

22 2

2 4 10 2

1 .4 2 4 sinscNnt zZ eNr K πε θ

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ (4.21)

This is the Rutherford scattering formula. This formula was found to be consistent with many experiments. [Show data and fit (Fig 6.16 pg.159 Krane)] Problem 3.48, Textbook

Assume particles re non-relativistic. Energy conservation gives

b rmin

b

Nucleus

umin v

7

2

2 21 1min2 2

0 min

1 ,4

zZemu murπε

= + ------- (i)

where u is the speed of the projectile when it is far from the nucleus. Now, angular momentum is also conserved in this interaction because the force is a central force; therefore, it exerts no torque. Thus, min min ,mub mu r= i.e.,

minmin

.ubur

= ------- (ii)

Thus,

2 2 2

21 12 2 2

min 0 min

1 .4

b u zZemu mr rπε

⎛ ⎞= +⎜ ⎟

⎝ ⎠ -------- (iii)

If b = 0,

2

212

0

1 ,4

zZemu Kdπε

= =

where d = rmin for b = 0. Thus,

2

0

,4e zZd

Kπε=

which is the answer obtained in the text. [Also, do Prob. 3.47. From Gauss' Law, 30

14

QrR

E πε= inside a uniformly charged sphere.]