Physics 1201 Final Exam information – Spring 2013

1)  Format: 35 multiple-choice questions, 5 points each à 175 points total

2)  Closed book and closed notes

3)  Equations and constants are provided on the final exam

4)  Cumulative – it covers all of the material covered in the course

5)  Approximate weightings of material: ~10 questions from material covered for midterm 1 ~10 questions from material covered for midterm 2 ~15 questions on material covered since midterm 2 6) See the Assignment Sheet for the date and time of your final exam

PRELIMINARY

PRELIMINARY

Chapter 30

Nuclear Physics and Radioactivity

Nuclear Structure

The atomic nucleus consists of positively charged protons and neutral neutrons.

1 atomic mass unit ≡1u ≡ 931.5 MeVc2

Nuclear Structure

neutrons

ofNumber protons

ofNumber neutrons and

protons ofNumber

NZA +=

atomic number

atomic mass number

i.e. number of nucleons

Nuclear Structure

Nuclei that contain the same number of protons but a different number of neutrons are known as isotopes.

For example, two isotopes of carbon are 612C 6

14C

Nuclear Structure

r ≈ r0A1 3, r0 ≈1.2×10

−15m =1.2 fm1 fermi ≡1 fm ≡10−15m

r ≈ 1.2×10−15m( )A1 3

Approximate empirical radius of a nucleus of mass number, A

Nuclear Structure

r ≈ r0A1 3 ≈ 1.2×10−15m( )A1 3

r 16O( ) = 1.2×10−15( ) 16( )1 3 = 3.0×10−15m = 3.0 fm

r 208Pb( ) = 1.2×10−15( ) 208( )1 3 = 7.1×10−15m = 7.1 fm

Example: Nuclear radius and density Find the radii of 16O and 208Pb nuclei. Estimate and compare their densities.

ρnucleus =Mnucleus

Vnucleus≈Amnucleon

43πr3

"

#$

%

&'≈Amnucleon

43πr0

3A"

#$

%

&'

≈mnucleon

43πr0

3"

#$

%

&'

à Nuclear densities are approximately the same for all A

The Strong Nuclear Force and the Stability of the Nucleus

The mutual repulsion of the protons tends to push the nucleus apart. What then, holds the nucleus together? The strong nuclear force.

The Strong Nuclear Force and the Stability of the Nucleus

As nuclei get larger, more neutrons are required for stability. The neutrons act like glue without adding more repulsive force.

The Mass Deficit of the Nucleus and Nuclear Binding Energy

Binding energy of ZAX = Mass deficit( )c2 = Δm( )c2

Δm = Z ×m 11H( )+ N ×m 0

1n( )−m ZAX( )

The Mass Deficit of the Nucleus and Nuclear Binding Energy

Example: The Binding Energy of the Helium Nucleus The atomic mass of helium is 4.0026u and the atomic mass of hydrogen is 1.0078u. Using atomic mass units, obtain the binding energy of the helium nucleus.

The Mass Deficit of the Nucleus and Nuclear Binding Energy

Δm = Z ×m 11H( )+ N ×m 0

1n( )−m ZAX( )

Δm = 2 1.0078 u( )+ 2 1.0087 u( )− 4.0026 u= 4.0330 u− 4.0026 u = 0.0304 u

1 u = 931.5 MeVc2

Binding energy = Δm( )c2 = 0.0304( ) 931.5 MeVc2

"

#$

%

&'c2 = 28.3 MeV

The Mass Deficit of the Nucleus and Nuclear Binding Energy

Radioactivity

A magnetic field separates three types of particles emitted by radioactive nuclei. The decay takes the general form:

parent→ daughter + x

Q ≡mparentc2 − mdaughter +mx( )c2 ⇒Q− value→ KE released

Q > 0⇒ exothermic Q < 0⇒ endothermic

Radioactivity α DECAY

ZAP → Z−2

A−4 D + 24 He

Q =mPc2 − mD +mα( )c2

= 238.0508 u− 234.0436 u+ 4.0026 u( )"# $% 931.5MeVu

&

'(

)

*+

= 4.3MeV

Radioactivity

A smoke detector

Radioactivity

β DECAY

e D P 011 −+ +→ A

ZAZ

Radioactivity

γ DECAY

γ P P +→∗ AZ

AZ

excited energy state lower energy

state

88226Ra*→ 88

226Ra+γ Eγ = 0.186 MeV

a photon that originates from the energy-level transition of the nucleus

The Neutrino

During beta decay, energy is released. However, it is found that most beta particles do not have enough kinetic energy to account for all of the energy released. The additional energy is carried away by a neutrino à

90234 Th → 91

234 Pa + −10 e + ν

mass ≈ 0, q = 0

A free neutron beta decays in about 12 minutes,

01n → 1

1p + −10 e + ν