WileyPLUS Assignment 5 is available

WileyPLUS Assignment 5 is availableChapters 28, 29, 30

Due Friday, April 9 at 11 pm

Friday, April 9Review - send questions!

PHYS 1030 Final ExamFriday, April 23, 1:30-4:30 pm

Frank Kennedy Brown Gym, 30 questions, the whole course, formula sheet provided

1Monday, April 5, 2010

The Laser

Light amplification occurs when:

• there is a “population inversion”,• the excited state is long-lived – a “metastable state”,• a photon of energy Ei - Ef triggers an electron in state Ei to fall to state Ef, emitting a second photon of energy Ei - Ef : (stimulated emission).

The result is a build up of photons of the same wavelength that are in phase – coherent light.

If there is no population inversion, the initial photon is more likely to be absorbed by an electron that moves from Ef to Ei.

Metastable state (~10-3 s)Photon:

E = Ei – Ef

Population inversion


Lasers

Many uses:

• playback/recording of CDs, DVDs• checkout scanners• welding metal parts• accurate distance measurement from time of travel of light pulses• telecommunications• study of molecular structure• medicine – selective removal of tissue, example, shaping the cornea of the eye – removal of “port wine stain” birth marks – destroying tumours with light-activated drugs – reattaching detached retinas


Laser Altimetry of Mars

Red: highPurple: low

Volcano

Impact crater~7 km deep!


Offers an alternative treatment for nearsightedness and farsightedness that does not rely on lenses

Photorefractive keratectomy (PRK)

Correcting nearsighted Correcting farsighted

UV laser - light

absorbed by cornea,

causing vaporization of material


HolographyA laser beam is split into two parts. The first, a reference beam, travels directly to the film. The second reflects from the object and recombines with the reference beam at the film, generating interference fringes there.

Hologram (interference fringes, not an image)


A Hologram

Image encoded in the hologram

http://hyperphysics.phy-astr.gsu.edu/Hbase/optmod/holfilm.html#c1


Viewing the image by shining laser light on

the hologram

for first-order maximum (the

brightest)sin!=

"d

d = spacing of the fringes

The fringes act as slits of varying spacingd

O

P


http://hyperphysics.phy-astr.gsu.edu/Hbase/optmod/holog.html

Holograms and ParallaxThe images were taken through a transmission hologram. The hologram was illuminated from behind by a helium-neon laser which has been passed through a diverging lens to spread the beam over the hologram.

This recorded interference pattern actually contains much more information than a focused image, and enables the viewer to view a true three-dimensional image which exhibits parallax. That is, the image will change its appearance if you look at it from a different angle, just as if you were looking at a real 3D object.


Summary of Chapter 30

• The positive charge of the the atom is concentrated in a nucleus that is ~10-15 - 10-14 m in radius.

• The angular momentum of the electron in an atom is quantized, leading to quantized energy levels in hydrogen-like atoms: En = –13.6 Z2/n2 eV

• Photons are emitted and absorbed by atoms at the same wavelength ! identification of elements in the “atmosphere” of stars, discovery of helium.

• The energies of K" x-rays can be calculated by replacing Z by Z–1 ! identification of some chemical elements for the first time.

• Stimulated emission, the laser.


Chapter 31: Nuclear Physics & Radioactivity

• Nuclear structure, nuclear size

• The strong nuclear force, nuclear stability, binding energy

• Radioactive decay, activity, half life

• The neutrino

• Radioactive age measurement

• Decay series


The Nucleus

Protons and neutrons (“nucleons”) are closely packed together in nuclei that are roughly spherical in shape.

Proton: q = +eNeutron: q = 0

Number of protons, Z = atomic numberNumber of neutrons, N = neutron number

Total number of nucleons, A = mass number, or nucleon number

A = Z + N

Nuclei are specified by: AZX

chemical symbol of the element

Example, 146CZ is sometimes omitted, as the chemical symbol gives the same information

neutrons and protons have almost the same mass


That means that nuclei have the same density:

Density =massvolume

� AmN43!r3

(mN = average mass of a nucleon)

Density =AmN43!r

30A

=3mN

4!r30

The radius of a nucleus of mass number A is:

r = r0A1/3, r0 = 1.2×10−15 m

p n

=3×1.67×10−27 kg4!(1.2×10−15 m)3

= 2.3×1017 kg/m3 !!!

Comparable with the supposed density of a black hole or a neutron star.

Isotopes: Nuclei of the same chemical element (same atomic number, Z), but different A and N.

Example: 12C, 13C, 14C. Only 12C, 13C are stable.


One isotope (X) contains an equal number of neutrons and protons. Another isotope (Y) of the same element has twice the number of neutrons as the first.

Determine the ratio rY/rX of the nuclear radii of the isotopes.

1.145


The Strong Nuclear ForceNuclei are held together by the strong nuclear force.– gravity is much too weak– the Coulomb force between proton charges is repulsive and decreases nuclear stability.

Stable nuclei

The strong nuclear force is:– attractive– extends over only ~10-15 m (a short-range, nearest-neighbour force)

The repulsive Coulomb force between protons favours nuclei with slightly more neutrons than protons.

Effect of Coulomb repulsion between protons

The “valley of stability”


Binding energy, mass defect

MassZmp + Nmn

mnucleus

Δm = mass defect

Z protons, N neutrons

!m= (Zmp+Nmn)−mnucleus

Mass defect:

Binding energy = energy to break up the nucleus into its constituent nucleons.

Binding energy: B= !m c2

Mass tables give the mass of neutral atoms with electrons in orbit. Then, Δm = [(ZmH + Nmn) – matom], mH = mass of H-atom

Z protons + N neutrons

mnucleus = (Zmp +Nmn)−B/c2


Atomic mass unit (u): 12 u = mass of 12C atom (including the 6 electrons)

Atomic and Nuclear Mass

MassMass

Particle Electric Charge Kilograms Atomic Mass

Units (u)Electron -e 9.109382!10-31 5.485799!10-4

Proton +e 1.672622!10-27 1.007276Neutron 0 1.674927!10-27 1.008665Hydrogen atom 0 1.673534!10-27 1.007825

1 u is equivalent to a mass energy, mc2, of 931.5 MeV.

Atomic mass (including Z electrons) = nuclear mass + mass of Z electrons.

1 u = 1.66054x10-27 kg


Example

The mass defect is: !m = 2 ! (mass of H atom + mass of neutron) – (mass of 4He atom) = 4.0330 – 4.0026 u!m = 0.0304 u

Mass defect, !m=Bc2

=28.4×106×1.6×10−19 J

(3×108)2 = 5×10−29 kg

Using the energy equivalent of the atomic mass unit, the binding energy is: B = (0.0304 u) ! 931.5 MeV/u = 28.4 MeV.


Binding energy per nucleon, B/A

Peak value, 8.7 MeV per nucleon

Sharp fall due to small number of nearest neighbour nucleons – short range nuclear force

Decrease due to Coulomb repulsion between protons

Unstable beyond 209Bi

Why are certain nuclei unstable? Because neighbouring nuclei have lower mass energy. Decay is possible to the lower mass nuclei while releasing kinetic energy.

Fission ⇒ energy release


Prob. 31.13/47: Mercury 202Hg (Z = 80) has an atomic mass of 201.970617 u. Obtain the binding energy per nucleon.

• Work out the mass defect knowing the mass of the atom.

• Convert the mass defect to a binding energy.

MassMass

Particle Electric Charge Kilograms Atomic Mass

Units (u)Electron -e 9.109382!10-31 5.485799!10-4

Proton +e 1.672622!10-27 1.007276Neutron 0 1.674927!10-27 1.008665Hydrogen atom 0 1.673534!10-27 1.007825

7.90 MeV/nucleon20Monday, April 5, 2010

Radioactivity

Three forms:• Alpha (") – the nucleus of a 4He atom is emitted from the “parent” nucleus• Beta (#) – an electron (+ or – charge) is emitted• Gamma ($) – a nucleus falls from one energy level to another and emits a gamma ray photon


Conserved quantities in radioactive decay

Conserved quantities:

• number of nucleons (nucleon number)

• charge

• energy

• linear momentum

• angular momentum


Alpha DecayAZX → A−4

Z−2Y + 42He

Parent Daughter "-particle

Nucleon number: A = (A – 4) + 4 !

Charge: Z = (Z – 2) + 2 !

Daughter and "-particle: greater binding energy, lower combined mass than parent ⇒ energy is released in the decay.

Energy released = [mX – (mY + m")] ! 931.5 MeV, if masses in atomic mass units (u).

The "-particles have a kinetic energy of typically a few MeV.

Nucleon numberCharge

23892U → 234

90Th +42He


Alpha Decay in a Smoke Detector

Alpha particles from a weak source collide with air molecules and ionize them, which allows a current to flow between the plates.

In the presence of smoke, ions colliding with smoke are generally neutralized (i.e. neutral atoms are formed), so that the current decreases and the alarm is tripped.

When the battery is low, the current is low, which also trips the alarm!


Prob. 31.23/20: Find the energy (in MeV) released when alpha-decay converts radium 226Ra (Z = 88, atomic mass = 226.02540 u) into radon 222Rn (Z = 86, atomic mass = 222.01757 u).

The atomic mass of an alpha particle is 4.002603 u.

4.87 MeV


Beta (β–) Decay

Nucleon number: A = A + 0

Charge: # Z = (Z + 1) + (–1)

Nucleon numberCharge

AZX → A

Z+1Y + e−

Nucleon number: A = A + 0

Charge: # Z = (Z – 1) + 1

(or #–)

Positron (β+) DecayAZX → A

Z−1Y + e+Nucleon numberCharge

(or #+)

23490Th → 234

91Pa + e−

2211Na → 22

10Ne + e+


Beta (β–) Decay

Z = 90

90 electrons

Nucleus

Neutral atom Neutral atom has 91 electrons

To calculate the energy released using tabulated masses of neutral atoms, the e– that is generated in the beta decay is lumped in with the 90 existing atomic electrons to form a neutral Pa atom and then,

!E = [mTh−mPa]×931.5 MeV

Beta decay

23490Th → 234

91Pa + e−

23490Th

Z = 91

90 electrons

23491Pa

e–

(protactinium)

atomic masses, in atomic mass units

Neutron into proton


Beta (β+) Decay

Z = 11

11 electrons

Nucleus

Neutral atom Neutral atom has 10 electrons

Using tabulated atomic masses, the energy released in the decay is,

% &E = [mNa - (mNe + 2me)] x 931.5 MeV

Beta decayZ = 10

10 electrons

e+

atomic masses, in atomic mass units

2211Na → 22

10Ne + e+

1 electron, e–

2211Na 22

10Ne

Proton into neutron


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WileyPLUS Assignment 5 is available