Particle Nature of The Nucleus -All nuclei have mass that is multiple of specific number. -t.f....

Particle Nature of The Nucleus

• -All nuclei have mass that is multiple of specific number.

• -t.f. nucleus made of smaller particles: protons, neutrons.

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Nuclides – elements same p+Isotopes – specific types of nuclides

• Mass number A

• X• atomic proton# Z

• Proton Z number identifies Z type of element.

• Ex: all H has one proton. Oxygen has 8 protons in nucleus. Proton number is equal to number of e- in neutral element. 2

Mass or Nucleon Number

• Nucleons live in nucleus -protons and neutrons.

• Mass number (A) is equal to the total all neutrons and protons in the nucleus.

• If p+ and no considered as atomic weight 1, the mass number gives the atomic weight in amu - u.

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Isotopes• Elements come in different isotopes.

• Two isotopes of the same substance will have equal proton numbers, but different numbers of neutrons.

• For the same element, the Z number must always be the same but the A number may vary.

• FYI for the most part dif isotopes of the same element behave the same way chemically. 4

Nuclides

• Nuclide is a specifically defined isotope.

• Nuclides are defined by many different aspects, such as half life, mode of decay, percent abundance, and so on.

• More specific than isotope. Isotopes are sets of nuclides having the same number of p+, but different number of no - . Each individual isotope is a separate nuclide.

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Calculations of nucleons:

• For a given isotope: protons (Z) + neutrons = mass number (A)

• Its easy to find the number of no in isotope.

A – Z is total neutrons.

• Find the number of no in 37 Cl17

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Mass Spectrometerseparates isotopes by mass

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• Ionization (thermal)• Accelerate ions E field• Velocity detector• Magnetic Deflector• Detector.

Velocity SelectorCrossed E and B fields

• Velocity selector

• Felc = Fmag• Eq = qvB v = E/B

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Entering Magnetic Field

Heavier ions are deflected less than lighter ones.

Fmag = Fc. qvB = mv2/r. qB = mv/r.

qBr = mvr = mv

qB

Detectors Count How many land at each point.

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• Read Hamper 7.2 and the Mass Spectrometer pg 255-256

• Do Hwk worksheet w reading.

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Nuclear Force

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The strong nuclear force holds nucleons together. Heating metals or shining EM radiation can strip e- from atom. Not so to pull apart nucleus.

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Binding E (BE) = work (J or eV) needed to pull apart nucleus.

• When work is added to nucleus to split it up, where does the E go?

• If they are not in motion no KE, or PE chm or PE elc what happens to work put in?

• Einstein -separate nucleons have greater mass

than when they are bound in nucleus.

E from W to split atom was

converted to mass! • E = mc2.

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The higher the BE between nucleons, the more work needed to split it up, the more stable the nucleus – the less likely to decay (fall apart)

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Binding E per Nucleon. Total E of Nucleus Num. Nucleons.

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Higher binding E per nucleon = stable elements.

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Mass Defect/Deficit

• The difference between the mass of the atom and the mass of the individual constituents.

• Since the nucleus has less mass than the sum of its parts, the difference is called mass defect and it equals the BE.

• Look at the table, which are the most stable?

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What happens when nucleus in not stable?

• Strive for stability.

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Mass Defect

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Because E and mass are equivalent it can be shown that 1 u = 931.5 MeV.

• We can use the above relationship to calculate the BE per nucleon for each element.

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Calculation of BE per nucleon

1. Calculate the BE of 54Fe which has an average mass of 53.9396 u.

•Mass p+ = 1.00782 u

•Mass no = 1.00866 u

Calculate mass defect & BE/nucleon• 26 p+ x 1.00782 u = 26.20332• 28 no x 1.00866 u = 28.24248• 26 e- x 0.000549 0.014274• Mass constituents 54.4458 u

mass Fe nucl 53.9396 u

subtract

Mass defect 0.5062 u

energy x 931.5 MeV/u

• BE = 471.5 MeV ÷ 54 nucl = 8.7 MeV/nucl 23

2. Find the binding energy of He-4 which has a known mass of 4.002602 u. (I’m including the e- this time).

Neutral He-4 consists of 2e-, 2p+, and 2no.

Find the sum of the parts in u.

Look up in table:e- rest massp+ rest massno rest mass

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2 x (0.000549u) 0.001098 +2 x (1.007277u) 2.014554 +2 x (1.008665u) 2.017300

constituents 4.032982 u

nucleus 4.002602 u

defect 0.3038 u x 931.5 MeV/nucl

28.29897 MeV ÷ 4 nucl = 7.0747 MeV/nucl

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Mass Defect

The nucleus of a deuterium atom consists of a proton and a neutron. If the mass of deuterium is 2.014102 u, calculate the BE in MeV.

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Binding Energy

• The amount of work needed to pull apart nucleus.

• The nucleus converts the missing mass to BE, it’s like a glue to hold nucleus together.

• When nucleus transitions to lower energy state, mass defect E is released!

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BE and atomic size 3 min

• http://www.youtube.com/watch?v=UkLkiXiOCWU

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To calculate the BE/nucl:

1. Find mass defect - the difference between the mass of the separate nucleons (unbound) and the mass of the bound nucleus.

2.Calculate the mass in atomic mass units.

3. Each unit has an energy equal to 931.5 MeV.

4. Multiply the mass defect by 931.5 MeV to convert the mass to energy.

5. Divide by # nucleons.

Do Hamper pg 154 #9. and addl prob’s Holt

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Radioactive DecayRadioactivity – emission of energy.

• 1896 Antoine Becquerel discovered that certain U salts emitted rays that could penetrate dark paper to expose a photographic plate. Called penetrating rays radiation.

• Marie Curie followed up & discovered other elements with the same properties (Th).

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Two types of Transmutation

• Spontaneous

• Unstable Nucleus

• Artificial/Induced

• Bombard Nucleus with a particle.

Emitted From Nucleus

• Alpha

• Beta

• Positron

• Gamma

• Neutrino

• Antineutrino

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Spontaneous Radioactive Decay is random process

• The type of decay & rate of decays do not depend on pressure, temperature, chemical bonds.

1899 Rutherford found U emits two dif types of “radiation” and .

rays were more penetrating.

Later, Villard found another type of radiation with even more penetrating power - gamma .

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Alpha Rays

rays are helium nuclei, (2p+ and 2no), that are emitted from nucleus. They are positively charged since e- missing.

When nucleus istoo large, it may emitalpha particle.

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Parent Daughter

Alpha Decay of Americium-241 to Neptunium-237

He nucleus

particles are easily stopped by skin or thin sheet of paper.

Likely to knock e- from orbits if they hit them. loses all its KE at once when it’s stopped.

Charge = +2e.

Mass = 4 units.

Energy is KE = ½ mv2. 38

Alpha Decay

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Alpha have KE ~5 MeV

Beta Particles

• Fast moving electrons emitted from nucleus.

• More penetrating than alpha because they are smaller.

• Think of B as a no that emits e-, becomes a p+.40

Less capable of ionizing (knocking out e-) -

Charge = -1 or -e.

mass = e, but considered massless in mass # calcs.

KE = ½ mv2. v can be sig portion of c.

Need a few mm of Al to stop them.

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Positron Decay like beta decay but p+ loses + (positive electron )becomes a neutron:

A highly penetrating type of nuclear radiation, similar to x-rays and light, except that it comes from within the nucleus of an atom, and, has a shorter . Gamma ray emission is a decay mode by which excited state of a nucleus de-excite to lower (more stable) state in the same nucleus. In diagrams, a gamma ray is represented by this:

In equations = .

Can pass thru human body, concrete, and lead.

Gamma Radiation

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Gamma Radiation

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Gamma rays are EM waves. They have:

Lowest ionizing power.

No charge. No mass.

Energy described by E = hf.

Travel with vel of light in vacuum.

No maximum stopping range.46

Neutrinos are particles that are emitted with beta and positron decay.

No charge.

Almost no mass.

They do not interact with matter.

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Which emitter would be safer to swallow?

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How could we distinguish the different types of radiation? What could we observe?

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7.2.4 Biological Effects of Ionizing Radiation

Outline them.

Becquerel Rays 9 min.

• https://www.youtube.com/watch?v=INF9y154EZA

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Products of Decay

When a parent nucleus decays, a daughter product, is produced, a new element could be produced.

We can identify products by balancing mass and atomic numbers.

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1. U – 232 decays by alpha emission. Write the nuclear equation and determine the daughter of the decay.

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alpha232 4 A

U He + X92 2 Z

X will have 4 less nucleons than U-232.X will have 2 less protons than U-232.

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Daughter will be: 228X 90

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Look up on periodic table element with 90 p+ (Thorium). Answer is:

228 Th 90

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Positron/Beta Decay

• During beta decay a beta, - (e-) or a positron + (+e) is emitted from the nucleus.

• When an e- (-) emitted a no changes to a p+, the atomic number increases by 1.

• When an e+ (+) emitted a p+ changes to a no, the atomic number decreases by 1.

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A neutrino v or antineutrino v is also emitted in positron or beta decay.

+ decay emits neutrino v.

decay emits an antineutrino v.

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Positron Decaylike beta decay but:

• Proton no + + +

• 90 Ru + + 90 Tc + .• 44 +1 43

A positron particle + is the antimatter e-.

neutrino

• 90 Ru + + ___ + .• 44 +1

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2. C-14 undergoes beta decay. What will be the daughter. Remember a beta particle is an e- ejected from the nucleus so that a neutron becomes a proton.

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14 0 A C e + X + v.6 -1 Z

A beta particle/e- has atomic number –1 and is considered massless.

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Balance – make right side add up to left side.

14 0 A C e + X + v.6 -1 Z

A will be 14. Z will be ??

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Z will be 7 since 7 and -1 = 6. So:14 0 14

C e + X + v6 -1 7

Z=7 is nitrogen. So daughter is :

14N

7

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3. Radium-226 decays by alpha emission. What is the resulting daughter element?

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226 Ra X + 4He88 2

Mass X = 226 – 4 = 222.Atomic number X

88 – 2 = 86.

222 Rn 86

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4. Sulfur 35 emits - particles when it decays. Write the equation. What will be the daughter product?

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35 S 0 e + A X + v.16 -1 Z

Atomic number X is 17 since neutron went to proton, so daughter is 17Cl – 35.

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5. Complete the equation:

• 23 Mg + + __ __ + .• 12 +1 __

• 23 Mg + + 23Na+ .• 12 +1 11

Nuclear StabilityThe more stable the nucleus, the less likely that it will decay.

Electrostatic repulsion in the positively charged nucleus makes it want to decay.

The strong nuclear force holds the nucleus together.

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The extra neutrons increase the strong force & help shield against electric repulsion.

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As more p+ added to nucleus more no needed for stability (for nucleus not to decay). There seems to be a ratio between p+ and no for each element to maintain enough strong force to keep nucleus from flying apart.

In general heavier elements require more no.p+– no ratio determines stability.

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Nuclides above the band are too large -decay by . To the left decay occurs.

Nuclides below the band have too few no, positron decay occurs. A p+ becomes a no.

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Summery: Natural Decay occurs spontaneously.

decay reduces the mass # by 4 and the atomic # by 2.

decay does not affect the mass #, but increases daughter's the atomic # by 1.

decay does not affect the mass #, but decreases the daughter's atomic # by 1.

decay affects neither the atomic nor the mass # but returns excited nucleus to ground state.

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http://www.youtube.com/watch?v=INF9y154EZAhttp://www.youtube.com/watch?v=I7WTQD2xYtQ

Properties of Becquerel Rays 9 min

Hwk Read Hamper 7.3 do #11-13 But Write out the balanced nuclear reaction for each.

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Nuclear Reactions: Natural & Artificial Transmutation

A process changing the nucleus =nuclear reaction.

Typical reactions are nuclear fission & fusion.

Reactions can occur spontaneously or be artificially induced.

Transmutation – can be artificially induced nuclear reaction.

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Fission We looked at fission by spontaneous radioactive decay.

Fusion

nucleus splits to 2 or more parts.Parent mass # decreases.

material is added and “fused” to the nucleus.

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1919 Rutherford bombarded gas with particles. New particles were produced in the gas which were not particles. From the deflection he concluded they were p+.

He later discovered that the nuclei of the gas absorbed the particle & emitted a p+ .

The fused to form a larger nucleus.

Fusion occurs naturally in stars.

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Nuclear Fusion

Equation for Fusion Rx.Mass # of parent increases

4 He + 14 N 17 O + 1H2 7 8 1

Proton

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Other types of particles can be used to bombard the nucleus.

Neutrons, protons, and H-2 are common.

Which rx is this?83

16O + 1n AX + 2H 8 0 z 1

• Identify X if a 2H is emitted in the reaction.

• Balance mass numbers & proton numbers on the left must and the right first.

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16O + 1n AX + 2H 8 0 z 1

• 16O + 1n 15X + 2H 8 0 7 1

Element must have 7 p+. It will be 15N. 7

Solve for the mass & proton number for X by balancing.

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Energy Considerations in Fission & Fusion

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Nuclear decay seeks to stabilize nucleus. When nucleus goes from lower to higher binding energy ratio/nucleon, energy is released in process.

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Release of E occurs when element goes to more stable state.

• Use BE table to predict the total energy release.

• 235 U 2 117 Pd fragments• 92 46

• Find the energy released. Use the BE/nucleon table.

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• 235 U ~ 7.6 MeV per nucleon

• 117 Pd ~ 8.4 MeV per nucleon.

• Take difference. This E is released per nucleon.

• 8.4 – 7.6 = 0.8 MeV/nucleon.

• But 235 U has 235 nucleons so:

• 235 x 0.8 MeV = 188 MeV released! 89

BE Released in Rx can be converted to KE and/or heat.

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3: Lithium can be bombarded with no to induce the following rx:

• Mass Li-6 6.015126

• Mass no = 1.008665

• Mass H-3 3.016030

• Mass He-4 4.002604

6 Li + 1n 3 H + 4He 3 0 1 2

How much E is released?

m = (7.023791 – 7.018634) u

• m = 0.005175 u x 931.5 MeV/u = • 4.804 MeV• Available as KE • some may be released as heat.

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Calculate mass for each side.

(7.023791u) (7.018634)

6 Li + 1n 3 H + 4He 3 0 1 2

9. Energy From Reactions 10 min

• http://www.youtube.com/watch?v=-YMgacsJyD0&playnext=1&list=PL0191606751B22A12&feature=results_main

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8.1 Natural transmutations and half lives 10 min

http://www.youtube.com/watch?v=I7WTQD2xYtQ

Read Hamper 7.4 and 7.5

• ForHwk. Hamper pg 162 15-17, pg 163 #18 – 19, 20 and pg 166 #21,22, pg

Decay Series

Heavier elements – even a decay often does not decrease the mass of heavy elements enough for stability. These elements go through a series of decays before reaching a stable state.

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Radioactive Decay Series

All elements above Z=83 are radioactive. Most of these decay through a series of transitions. The daughters are radioactive and decay a number of times before they become a stable element.

Many elements decay through many and steps.

U-238 decay series

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Spontaneous Random Decay & Half life

Radioactive decay is random & not affected by outside factors. There is no way of knowing whether a particular nucleus will decay.

Changing temperature or pressure does not change chances of a decay, a chemical rx does not alter its activity.

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We can only know the chances of a decay happening.

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Rate of Decay is proportional to number (N) of atoms in the sample.

If large number of atoms, the number decaying would be larger.

Radioactive decay is an exponential process.

The number, N, of parent atoms decreases exponentially over time so do the number of decays/ unit time.

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As number of parent atoms decreases, the rate in units per second decreases exponentially.

The daughter product grows as a mirror image. 102

Half Life t1/2 defined as:time it takes for half of original parent in a

sample to decay, or time for original mass to be halved, or time for activity to be halved.

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Activity in counts/sec (Bq) easier to measure. It’s the decay rate of the sample.

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What is the half life of this sample?

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Ex 1: The half life of Quinnium is 4 days. If we start with 160 g of pure Quinnium , how much will be left after 12 days??

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160 g

80 g

40 g

20 g

4 days

8 days

12 days

3 half lives have passed107

Exact multiple of half lives:

Rd Hamper 7.4

• Do 15-17 pg 162 and finish #2 pg 167

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Radioactive Decay EquationsThe activity defined av # of disint/ sec

for a particular sample unit = becquerel (Bq).

Activity of sample:

A = activity (Bq) counts or decays/s

N =# disintegrations

T = time (s)

Negative sign b/c disintegrations are decreasing as N decreases.

t

NA

Ex 1: If a sample starts with 1000 parent atoms and decreases to 100 parent atoms in 3600 seconds, what is the activity of the sample?

• 900 atoms/ 3600 s = 0.25 Bq.

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t

NA

Activity is proportional to N (parent atoms) at any time so:

NA is radioactive decay constant s-1.

We can’t predict which nucleus will decay but we can predict the number of nuclei decaying in a specific period of time.

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Ex 2: A sample has an activity of 2 x 103 Bq and a of 5 x 10-5 s-1. What is the number of remaining parent atoms at this time?

• A/ = N

• 2 x 103 Bq/ 5 x 10-5 s-1 =

• 4 x 107 parent atoms remain.

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Derive the half life equation.

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The graph below shows the fraction of original atoms remaining with the passage of ½ lives.

The number of parent atoms decays exponentially over time. The Y axis is the activity of the sample.

N

The equation of the graph is:

N = Noe-t.

No original # atoms (parent) = decay constant.

N # (parent)atoms remaining e is natural log

Radioactive Decay Law

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Relating ½ life to decay constant

N = Noe-t. eNN t

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When t is t1/2, then

N = No. 2

Rearrange:

2 = No. N

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But when t = t1/2

2 = No. N

Take ln both sides:

ln 2 = T1/2.

T1/2 = ln2

eNN t

o

eNN to

eT 2/12

so

flip

N = Noe-t. can be written in terms of original sample activity rather than number of atoms.

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A = oe-t

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Since:

• N = Noe-t.

• Multiply both sides by .

N = Noe-t.

NA &

A = Noe-t

A = oe-t

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 Ex 3: A sample of radioactive isotope originally contains 1.0 x 1024 atoms and has a half-life of 6.0 hours. Find the:

a. decay constant

b. initial activity

c. number of atoms remaining after 12 h.

d. number of atoms remaining after 30 min.

How to determine the half life.1. Measure the activity using ionizing

properties (Geiger Counter)

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Short Lived Nuclei – read it off

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Longer Lived Isotopes (long half life)

• Plot N vs t on semi log graph• Find gradient

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Detection Devices

Electroscope - charged electroscope can be discharged by oppositely charged ions. Rate of discharge related to greater radioactivity.

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Other Detection Methods

• Ionization ChamberGeiger-Muller Detector

• Gas filled tubes. Use the fact that radiation can ionize other atoms thereby giving them a charge. Current is generated and detected. 125