wileyplus assignment 4 review of course next week

WileyPLUS Assignment 4

Chapters 28 - 31

Due Thursday, April 9 at 11 pm

Review of Course Next Week

Monday, after completing chapter 31,

and Wednesday

Send questions!

1Friday, April 3, 2009

PHYS 1030 Final Exam

Wednesday, April 15

9:00 - 12:00

Frank Kennedy Gold Gym, seats 1 - 290

30 multiple choice questions

Formula sheet provided


Chapter 31: Nuclear Physics & Radioactivity

• Nuclear structure, nuclear size

• The strong nuclear force, nuclear stability, binding energy

• Radioactive decay, activity

• 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:A

ZX

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 =mass

volume! AmN

43!r3

(mN = average mass of a nucleon)

Density =AmN

43!r30A

=3mN

4!r30

The radius of a nucleus of mass number A is:

r = r0A1/3, r0 = 1.2×10−15 m

pn

=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.


Prob. 31.-/6: One isotope (X) contains an equal number of neutrons

and 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.


The Strong Nuclear Force

Nuclei 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

MassZm

p + Nm

n

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

Alternatively, a neutral atom with Z electrons is broken up into N

neutrons and Z hydrogen atoms. Then, Δm = [(ZmH + Nmn) – matom].

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

ParticleElectric

ChargeKilograms

Atomic Mass

Units (u)

Electron -e 9.109382!10-31 5.485799!10-4

Proton +e 1.672622!10-27 1.007276

Neutron 0 1.674927!10-27 1.008665

Hydrogen 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.


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=B

c2=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. 31.47/10: 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

ParticleElectric

ChargeKilograms

Atomic Mass

Units (u)

Electron -e 9.109382!10-31 5.485799!10-4

Proton +e 1.672622!10-27 1.007276

Neutron 0 1.674927!10-27 1.008665

Hydrogen atom 0 1.673534!10-27 1.007825


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 Decay

A

ZX → A−4

Z−2Y + 4

2He

Parent Daughter (!)

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 number

Charge

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. 31.20/50: 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.


Beta (β–) Decay

Nucleon number: A = A + 0

Charge: # Z = (Z + 1) + (–1)

Nucleon number

Charge

A

ZX → A

Z+1Y + e−

Nucleon number: A = A + 0

Charge: # Z = (Z – 1) + 1

(or "–)

Positron (β+) Decay

A

ZX → A

Z−1Y + e+Nucleon number

Charge

(or "+)

234

90Th → 234

91Pa + e

−

2211Na → 22

10Ne + e+


Beta (β–) Decay

Z = 90

90 electrons

Nucleus

Neutral atom Neutral atom

has 91 electronsTo 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

234

90Th → 234

91Pa + e

−

234

90Th

Z = 91

90 electrons

234

91Pa

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


Prob. 31.27: Find the energy released when "+ decay converts 22Na

(Z = 11, atomic mass = 21.994434 u). Notice that the atomic mass

for 22Na includes the mass of 11 electrons, whereas the atomic mass

for 22Ne (Z = 10, atomic mass = 21.991383 u) includes the mass of

only 10 electrons.

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

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


http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta.html

Beta-decay – a problem

X (parent, at rest)

Y (daughter)

e– or e+

Kinematics: if energy and momentum are conserved, the electron (e– or e+)

should have a well-determined kinetic energy following the beta-decay.

But, the electron does not have a well-determined energy, as seen above.

Is energy not conserved?! No, energy is conserved...

Expected energy of the e+

Beta-decay

X → Y + e


The Neutrino

6429Cu → 64

28Ni + e+ + ν

A third particle, a neutrino, is also emitted in the decay, so that

the released energy is shared between three particles instead

of two:

Expected energy of the e+

The neutrino is very difficult to detect.


Neutrino detector in Japan


Neutrino detector in Japan - X-Files version


Gamma (γ) Decay

A

ZX∗ → A

ZX+ !

Excited state

of the nucleus

Nuclear

Energy

# ray photon

X*

X

Gamma rays are produced in the decay

(de-excitation) of a nuclear state.

This is similar to the production of a

photon by an atom, except that the energy

levels are associated with the nucleus

itself, not with electrons in the atom.

Gamma rays are generally of higher energy and are even more

penetrating than x-rays.


Gamma Knife – to zap a tumour

Gamma rays from 60Co sources are

channeled through collimators in a

metal helmet.

Gamma rays are concentrated at the

site of the tumour, to selectively

destroy the malignant tissue.

60Co gamma ray source:

60

28Ni∗ → 60

28Ni+ ! (# 1.2 MeV)

60

27Co → 60

28Ni∗+ e

−+ !̄ Tumour

Half of the 60Co decays away in 5.3 years, so has to be replaced...


Winnipeg Free Press,

April 4, 2004


Winnipeg Free Press, March 19, 2008


wileyplus assignment 4 review of course next week

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