a2 physics unit 9 radioactivity - animated science · 2012-06-26 · a2 physics unit 9...

A2 Physics Unit 9 Radioactivity

Mr D Powell

Mr Powell 2008 Index

Chapter Map

Common Issues Take care not to mix units when calculating. For instance, you may encounter activities given in Becquerel's and count rates stated in counts per minute; half lives are often quoted in reference works using the most relevant time unit, be it seconds or years.


9.1 Radioactivity

1. How big is the nucleus?

2. How was the nucleus discovered?

3. Why was it not discovered earlier?


Alpha Decay....

Alpha particles tracks


Probing the Nucleus – Rutherford Scattering

Rutherford alpha particle scattering experiment

scattered alphaparticles

microscope to viewzinc sulphide screen,and count alphaparticles

vary angle ofscatteringobserved

radiumsource ofalphaparticles

thin goldfoil

alpha particlebeam

zinc sulphidescreen, tiny dots oflight where struckby alpha particle

lead block toselect narrowbeam of alphaparticles

nucleus

1

2

34

5

1

2

34

5

135

90

5030

20

By firing alpha particles at a heavy gold nucleus Rutherford could easily see that atoms were mostly space with a large positive nucleus in the centre.


Scattering Formulae (extension)

Rutherford’s picture of alpha scattering

Assumptions:

alpha particle is the He nucleus, charge +2e

gold nucleus has charge + Ze, and is much more

massive than alpha particles

scattering force is inverse square electrical

repulsion

scattering angle

‘aiming error’ b

gold nucleuscharge + Ze

alpha particlescattered

equal force F butnucleus is massive,so little recoil

charge +2e

For calculations

d

force F = 2Ze2

40d2

There are complex formulae that we can use to work out the distance of closest approach. You will not have to learn these but should appreciate some of the maths involved. At the very least the idea of the forces involved


Distance of Closest Approach (extension)


Summary Question...

For an alpha particle with an initial KE of 6MeV fired at a gold nucleus find the distance of closest approach of the alpha and the nucleus...


9.2 Alpha / Beta / Gamma

1. What are alpha, beta, gamma

2. Why is it dangerous

3. What are the properties of alpha, beta, gamma


Link them up?


Guess the field?


Particle Tracks


Calibration plateau p.d...


9.3 More on Alpha / Beta / Gamma

1. What happens to the nucleus in a radioactive change

2. How does the intensity spread out

3. How do we represent the changes in the nucleus in alpha, beta, gamma.


How does the intensity spread out


Lines of stability...


Emitters....


Corrected count rates & distances....

d d0

20

0dd

kCC

We must find the source centre for inverse sq law to work correctly.


9.4 The Dangers of Radioactivity

1. Why is ionising radiation

2. What factors determine whether , , are the most dangerous

3. How can exposure to ionising radiation be reduced


Background Radiation


Sievert The sievert (symbol: Sv) is the SI derived unit of dose equivalent. It attempts to reflect the biological effects of radiation as opposed to the physical aspects, which are characterised by the absorbed dose, measured in gray. It is named after Rolf Sievert, a Swedish medical physicist famous for work on radiation dosage measurement and research into the biological effects of radiation.

1 Sv = 1 J/kg = 1 m2/s2 = 1 m2·s–2

Frequently used SI multiples are the millisievert (1 mSv = 10–3 Sv) and microsievert (1 μSv = 10–6 Sv). The millisievert is commonly used to measure the effective dose in diagnostic medical procedures (e.g., X-rays, nuclear medicine, positron emission tomography, and computed tomography). The natural background effective dose rate varies considerably from place to place, but typically is around 2.4 mSv/year. For acute (that is, received in a relatively short time, up to about one hour) full body equivalent dose, 1 Sv causes nausea, 2-5 Sv causes epilation or hair loss, hemorrhage and will cause death in many cases. More than 3 Sv will lead to LD 50/30 or death in 50% of cases within 30 days, and over 6 Sv survival is unlikely. (For more details, see radiation poisoning.) The collective dose that a population is exposed to is measured in "man-sieverts" (man·Sv).

http://en.wikipedia.org/wiki/Joule

http://en.wikipedia.org/wiki/Kilogram

http://en.wikipedia.org/wiki/Metre

http://en.wikipedia.org/wiki/Second

http://en.wikipedia.org/wiki/Metre

http://en.wikipedia.org/wiki/Second

http://en.wikipedia.org/wiki/SI_multiple

http://en.wikipedia.org/wiki/Effective_dose

http://en.wikipedia.org/wiki/X-rays



http://en.wikipedia.org/wiki/Nuclear_medicine

http://en.wikipedia.org/wiki/Positron_emission_tomography

http://en.wikipedia.org/wiki/Computed_tomography

http://en.wikipedia.org/wiki/Computed_tomography

http://en.wikipedia.org/wiki/Background_radiation

http://en.wikipedia.org/wiki/Nausea

http://en.wikipedia.org/wiki/Hair_loss

http://en.wikipedia.org/wiki/Hemorrhage

http://en.wikipedia.org/wiki/Radiation_poisoning


9.5 Radioactive Decay

What do we mean by the activity of an isotope Activity A = λN. or Energy Transfer per second = AE What is the half-life of an isotope t1/2 = 0.69/λ Does anything affect radioactive decay


Rate of Decay is the decay constant...


y = 288.79e-0.164x R² = 0.9922

0

50

100

150

200

250

300

350

0 5 10 15 20 25

Nu

mb

er

of

Ato

ms

Left

Time elaspsed arbitrary

Number of sample

Dice Results - 300


Gamma 1....

Paul Villard, a French physicist, is credited with discovering gamma rays in 1900. Villard recognised them as different from X-rays (discovered in 1896 by Roentgen) because the gamma rays had a much greater penetrating depth. It wasn't until 1914 that Rutherford showed that they were a form of light with a much shorter wavelength than X-rays. High-energy photons emitted as one of the three types of radiation resulting from natural radioactivity are the most energetic form of electromagnetic radiation, with a very short wavelength (high frequency). Wavelengths of the longest gamma radiation are less than 10 -10 m, with frequencies greater than 10 18 hertz (cycles per sec). Gamma Emission occurs primarily after the emission of a decay particle. Gamma is a form of high energy electromagnetic radiation. After a particle is ejected from a nucleus the system may have some slight excess of energy, or exist in a metastable state. This slight excess of energy is released as gamma. Gamma emission will not change the isotope or the element. The wavelength of the emitted gamma radiation will be unique to each isotope. Gamma emission is a significant health risk.

Villard

Roentgen

Rutherford


Gamma 2

After a decay reaction, the nucleus is often in an “excited” state. This means that the decay has resulted in producing a nucleus which still has excess energy to get rid of. Rather than emitting another beta or alpha particle, this energy is lost by emitting a pulse of electromagnetic radiation called a gamma ray. The gamma ray is identical in nature to light or microwaves, but of very high energy. Like all forms of electromagnetic radiation, the gamma ray has no mass and no charge. Gamma rays interact with material by colliding with the electrons in the shells of atoms. They lose their energy slowly in material, being able to travel significant distances before stopping. Depending on their initial energy, gamma rays can travel from 1 to hundreds of meters in air and can easily go right through people.

Villard

Roentgen

Rutherford


Gamma Emission


Gamma Emission

1. Gamma is a process which often happens in addition to a beta decay

2. It can often have different energies attached from differing transitions.


Alpha Decay

The nucleus is initially in an unstable energy state. An internal change takes place in the unstable nucleus and an alpha particle is ejected leaving a decay product. Alpha particles have a velocity in air of approximately one-twentieth the speed of light, depending upon the individual particle's energy. Since the number of protons in the nucleus of an atom determines the element, the loss of an alpha particle actually changes the atom to a different element. For example, polonium-210 is an alpha emitter. During radioactive decay, it loses two protons, and becomes a lead-206 atom, which is stable (i.e., nonradioactive). Examples are; Americium-241 , Plutonium-236 , Uranium-238 Thorium-232 , Radium-226 , Radon-222 , Polonium-210


Alpha Decay

The positive charge of alpha particles is useful in some industrial processes. For example, radium-226 may be used to treat cancer, by inserting tiny amounts of radium into the tumorous mass. Polonium-210 serves as a static eliminator in paper mills and other industries. The alpha particles, due to their positive charge, attract loose electrons, thus reducing static charge. Some smoke detectors take advantage of alpha emissions from americium-241 to help create an electrical current. The alpha particles strike air molecules within a chamber, knocking electrons loose. The resulting positively charged ions and negatively charged electrons, create a current as they flow between positively and negatively charged plates within the chamber. When smoke particles enter the device, the charged particles attract them, breaking the current and setting off the alarm.

Tumors on liver on the top left


Alpha Decay


Beta Decay


Protactinium Generator

This experiment uses a 'protactinium generator' to show the exponential decay of protactinium-234, a grand-daughter of uranium. It has a half-life of just over a minute, which allows the chance to measure and analyse the decay in short time. The bottle contains a complex mixture of substances but when you have shaken the bottle the and organic layer separates out which contains the protactinium-234. This decays with a half-life of 68 seconds and can be monitored.

230234234234238 ThUPaThU

U238 T0.5 = 4.47 Billion years


Protactinium Generator

The Pa234 is a beta emitter 2.3MeV (max) which escapes the tube and leaves U234 which has a very long alpha half life. We can monitor the Pa activity in the top layer and any newly spawned Pa from the bottom layer is prevented from contaminating the top layer. Hence, we have an initial sample of Pa and can see over a period of time the change in activity.

230234234 ThUPa

70s years


Graphical Decay


Real Data Time in

seconds

Activity Recorded

in Bq

0 210

20 165

40 130

60 110

80 95

100 75

120 65

140 53

160 29

180 28

200 25

Now try plotting a graph and prove some formulae

1. Plot one Loge graph or Ln graph

2. Plot one normal graph

3. Show via inspection and

mathematically the formulae below. Test each one out and see if it is true!

t1/2 = 0.69/λ = ln2/ λ N = N0e-λt

A = A0e-λt

Activity A = λN


Example Results & Proof of Formulae....

Can you now prove any of these relations by drawing a graph and taking some readings?

t1/2 = 0.69/λ N = N0e-λt

A = A0e-λt

Activity A = λN.


y = 213.38e-0.011x R² = 0.9791

0

50

100

150

200

250

0 50 100 150 200 250

Co

un

ts in

10

se

con

ds

Time in seconds (s)


Dice Exp...

76 dice were taken and thrown and then the number of dice on a 1 were counted and removed. The process continued until there were no dice left. The results were graphed in two different ways by activity and mass

Time Activity Time Mass

0 12 0 76

1 10 1 64

2 8 2 54

3 4 3 46

4 14 4 42

5 6 5 28

6 4 6 22

7 3 7 18

8 5 8 15

9 4 9 10

10 1 10 6

11 1 11 5

12 1 12 4

13 1 13 3

14 1 14 2

15 1 15 1

16 1 16 1

y = 12.379e-0.181x R² = 0.8281

0

2

4

6

8

10

12

14

16

0 5 10 15 20

Activity

y = 105.9e-0.282x R² = 0.9808

0

20

40

60

80

100

120

0 5 10 15 20

Mass

y = 43.672e-0.169x R² = 0.8918

0

5

10

15

20

25

30

35

40

45

50

0 5 10 15 20 25

Act

icit

y o

f Sa

mp

le


Activity

y = 288.79e-0.164x R² = 0.9922

0

50

100

150

200

250

300

350

0 5 10 15 20 25

Nu

mb

er

of

Ato

ms

Left


Number of sample

Time Activity Time Number

0 39 0 300

1 43 1 261

2 24 2 218

3 32 3 194

4 29 4 162

5 23 5 133

6 22 6 110

7 23 7 88

8 11 8 65

9 6 9 54

10 5 10 48

11 6 11 43

12 3 12 37

13 3 13 34

14 6 14 31

15 4 15 25

16 2 16 21

17 2 17 19

18 4 18 17

19 2 19 13

20 2 20 11

300 dice were taken and thrown and then the number of dice on a 1 were counted and removed. The process continued until there were no dice left. The results were graphed in two different ways by activity and mass


Write a simple definition for..

T0.5 =

= No = N = A = A0 = t =


9.6 Radioactive Decay Theory

Can a source decay completely What is an exponential decrease Why is the process random

t1/2 = 0.69/λ

A = λN

N = N0e-λt or A = A0e-λt


T1/2

1. The number of radioactive nuclei decaying per second (activity) is proportional to number of nuclei remaining


Decay Constant


Decay Curves...


Decay Curve This curve is an exponential decay It takes the same time for half the sample to decay i.e. Activity reduces over time.


Decay Curves...


The Decay Equation 1


Decay Equation 2


Equivalence of Formulae


Half - Life

The T0.5 time for an isotope is defined as the time taken for half the sample to decay


Using the decay equation... Show that 4 marks...

Exam Question...


Half – Life 2


Example of using Equation


T1/2 Examples...

TASK Can you work out the

decay constant for each isotope of carbon?

C9 - 5.457 s-1 C10 – 0.036 s-1 C11 – 5.7 x 10-4 s-1

C12 – 0 s-1


Example Curve


9.7 Isotopes in Use

How do we do radioactive dating What are radioactive tracers What do we use radioactivity for in hospitals.


Can you plot this data to find out half life?


Carbon Dating

pCnN 14

6

14

7

eNC decaybeta 14

7

14

6

Here is a good example of capture of a neutron into the nucleus followed by decay back again via a beta minus process.


Maths for Dating?

Imagine you start with a blob of a man and want to find out the number of original atoms in the sample which were radioactive.

t1/2 = 0.69/λ t1/2 = 5700 years so λ = 1.216 x 10-4 years-1 or work out in seconds N = N0e-λt

Now you know No from the mass spectrometer but you don’t know the time it has taken from t = 0 i.e. When it died. You now need to know the activity now (Geiger-Muller tube) and the activity of a sample of man who is living to compare. If in Bq then λ is in seconds Take the ratio of activities & work through to find age, then work back to find N0 of original sample. (see book page 168)


Tricks with 2N? T0.5 N Series

1 1/2 21

2 1/4 22

3 1/8 23

4 1/16 24

5 1/32 25

n

n

N

N

NN

5.0

5.0

0

0

If you think about the idea of the number of atoms left after a number n of half lives. You can then construct a very useful equation. This enables you to predict the number of atoms left if you know the original number of atoms and the T0.5 time. Or you can work out the number of T0.5 times it will take to reduce the number of atoms or activity to a certain amount.


9.8 More about Decay Modes

1. What can we tell about radioactive isotopes from an N-Z chart

2. Why don’t naturally occurring isotopes emit + radiation?

3. What happens to an unstable nucleus that emits radiation.


Technetium Doses...

Also, a commonly-used measure of radioactivity

is the microcurie:

1 μCi = 3.7×104 disintegrations per second =

2.22×106 disintegrations per minute

The typical human body contains roughly 0.1 μCi

of naturally occurring potassium-40.

How many Bq?

Can you estimate out the activity per second?

http://en.wikipedia.org/wiki/Potassium-40




Stable or Unstable Nuclei Stable and unstable nuclei: balance of numbers of protons and neutrons

150

100

50

0

0 20 40 60 80 100

proton number Z

stable

alpha decay

beta decay

electron capture

positron emission

fission

N=Z


Valley of Stability

Nuclear landscape: The nuclear valley of stability

Stable nuclei lie along a narrow band of values of numbers of protonsand neutrons. The more negative the binding energy, the more stablethe nucleus.

These nucleigo down hillby alphadecay

040

Bindingenergy pernucleon/MeV

neutron number NZN

Z–2N–2

‘Iron lake’Nuclei near iron in the PeriodicTable are the most strongly bound,and lie in the bottom of the nuclearvalley. Other nuclei tend to ‘rundownhill’ to become like iron.

Fusion hillLarger numbers ofneutrons and protons arebound more strongly thansmaller numbers, by thestrong nuclearinteraction.

These nuclei go down hillby nuclear fusion.

100

0

–2.5

–5.0

–7.5

0

25

50

75

80120

160

proton number Z

Alpha decay

Z–1N+1

ZN

+

ZN

Z+1N–1

–

proton number Z proton number Z

Positron decay Electron decay

Free particle plains

Binding energy taken as zero forfree protons and neutrons.

Bound nuclei have negativebinding energies, down to about–8 MeV per nucleon

Coulomb slopeMore neutrons thanprotons are neededto overcomeelectrical repulsionof many protons.

Pauli cliffsThe sides of the valley rise steeply. The Pauli exclusion principle keeps numbers ofprotons and neutrons approximately equal. Binding is less strong if either are in excess.

protonnumber Z


Decay chain from U-238

Pb-210

Po-214

Po-218

Pb-214

Rn-222

Ra-226

Th-230

U-234

Pa-234

Th-234

U-238

–

–

–

neutron number N

Main source ofnaturallyoccurringradium, half-life1600 yr. Radiumwas discoveredby Marie Curieand her husbandPierre Curie.

146

144

142

140

138

136

134

132

130

128

126

124

Bi-214

Bi-210

Po-210

–

–

–

Pb-206 stable

half-life 4.5 109 yr

88 89 90 91878685848382

Pb Bi Po At Rn Fr Ra Ac Pa U

proton number Z

92

Th

Decay Chains ..... Decay chain from Th-232

142

140

138

136

134

132

130

128

126 Pb-208

Po-212Tl-208

Pb-212

Bi-212

Po-216

Rn-220

Ra-224

Th-228

Ac-228

Ra-228

Th-232

–

–

–

–

neutron number N

87 88 89 90868584838281

Tl Pb Bi Po At Rn Fr Ra Ac Th

proton number Z

Radon gas, half-life 56 s,from building materials isan important part of thenatural backgroundradiation

Marie Curie, discoverer withPierre Curie of polonium,named the element after hernative Poland


Review...

Why don’t naturally occurring isotopes emit + radiation? 1. they are all manmade by a process of bombarding stable nuclei with

protons.

2. Hence, protons need to be fast enough to overcome coulomb repulsion.

3. If elements gain a proton they actually become more unstable and not more stable and move away from line. So Copper for example will never give on + as it would become more unstable.


9.9 Nuclear Radius

1. Are more massive nuclei wider?

2. How does the radius of a nucleus depend on its mass number A?

3. How dense is the nucleus?


Small Nucleus....

Nucleons....


Estimate Nuclear Diameter with “Electron Diffraction”


Quick Question

d

E

hc

22.1sin


Atomic Scale

Can you estimate the distance of the outer shell electron for a gold atom? If you are not sure just pick a figure......

3.3 miles


Atomic Radius....

Various types of scattering experiments suggest that nuclei are roughly spherical and appear to have essentially the same density. The data are summarised in the expression called the Fermi model; r0 = 1.2 x 10-15m = 1.2fm r = atomic radius A = mass number (nucleons)

3

1

Arr o


Element A A1/3 R in fm

Carbon 12 2.29 2.75

Oxygen 16 2.52 3.02

Silicon 28 3.04 3.64

Calcium 40 3.42 4.10

Vanadium 50 3.68 4.42

Strontium 88 4.45 5.34

Indium 115 4.86 5.84

Atomic Radius....

Various types of scattering experiments suggest that nuclei are roughly spherical and appear to have essentially the same density. The data are summarized in the expression called the Fermi model; r0 = 1.2 x 10-15m = 1.2fm r = atomic radius A = mass number (nucleons) Can you plot a graph and show that this formulae is correct and the constant is 1.2fm

3

1

Arr o FACT The diameter of the nucleus is in the range of 1.6fm (1.6 × 10−15 m) (for a proton in light hydrogen) to about 15 fm (for the heaviest atoms, such as uranium).


Finding R0

y = 1.2x + 7E-14 R² = 1

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

8.00

0.00 2.00 4.00 6.00 8.00

R in

fm

cuberoot of A ( Nucleon Number)

Nuclear Radius


Summary & Question

1. Calculate the radius of an oxygen nucleus which has 16 nucleons..

r=r0A 1/3= 1.4 x 10-15 x (16)1/3 r=3.5 x 10-15m (3.5fm)


Nuclear Density – is massive!


Reflection high-energy electron diffraction (RHEED)

Reflection high-energy electron diffraction (RHEED) is a technique used to characterise the surface of crystalline materials. RHEED systems gather information only from the surface layer of the sample, which distinguishes RHEED from other materials characterization methods that rely also on diffraction of high-energy electrons. Transmission electron microscopy, another common electron diffraction method samples the bulk of the sample due to the geometry of the system. Low energy electron diffraction (LEED) is also surface sensitive, but LEED achieves surface sensitivity through the use of low energy electrons.

http://en.wikipedia.org/wiki/Reflection_high_energy_electron_diffraction

http://en.wikipedia.org/wiki/Crystalline

http://en.wikipedia.org/wiki/Electrons

http://en.wikipedia.org/wiki/Transmission_electron_microscopy

http://en.wikipedia.org/wiki/Transmission_electron_microscopy

http://en.wikipedia.org/wiki/Electron_diffraction

http://en.wikipedia.org/wiki/Low_energy_electron_diffraction

http://en.wikipedia.org/wiki/Low_energy_electron_diffraction






Reflection high-energy electron diffraction (RHEED)

The image shows the setup of the electron gun, sample and detector /CCD components of a RHEED system. Electrons follow the path indicated by the arrow and approach the sample at angle θ. The sample surface diffracts electrons, and some of these diffracted electrons reach the detector and form the RHEED pattern. The reflected (specular) beam follows the path from the sample to the detector.

Specular reflection is the mirror-like reflection of light (or of other kinds of wave) from a surface, in which light from a single incoming direction (a ray) is reflected into a single outgoing direction

http://en.wikipedia.org/wiki/Mirror

http://en.wikipedia.org/wiki/Reflection_(physics)

http://en.wikipedia.org/wiki/Light

http://en.wikipedia.org/wiki/Wave

http://en.wikipedia.org/wiki/Ray_(optics)


Wave nature of the electron

• Louis de Broglie (1892 - 1987)

• If light can be modeled as a particle or as a wave, can an electron be modeled as a wave?

• The wavelength of a matter wave (1923) is given by:

• Everyday objects are too massive to give observable wavelengths; however, electrons are light enough to give observable wavelengths. Diffraction of electrons was observed by two groups in 1927, Davisson & Germer and George Thomson.

• The Bohr model could also be explained using standing waves.

• Whole numbers (1,2,3,etc.) of de Broglie wavelength give the allowed radii found in the Bohr model.


Experiments of GP Thompson


Electron Waves Two specific examples supporting the wave nature of electrons as suggested in the DeBroglie hypothesis are the discrete atomic energy levels and the diffraction of electrons from crystal planes in solid materials. In the Bohr model of atomic energy levels, the electron waves can be visualized as "wrapping around" the circumference of an electron orbit in such a way as to experience constructive interference.

Bragg diffraction. Two beams with identical wavelength and phase approach a crystalline solid and are scattered off two different atoms within it. The lower beam traverses an extra length of 2dsinθ. Constructive interference occurs when this length is equal to an integer multiple of the wavelength of the radiation.

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


Microstructure of Nickel Superalloys

• These pictures are taken using electron diffraction. They are trying to establish the structure and if there are any problems in the structure of Nickel alloys used in aeroplane manufacture. It is important that any micro fissures are detected early on.


Experiments of Davisson and Germer 1927


De Broglie Summary / Questions

Exam Questions


Q2 June 07.... The table shows data for some nuclei. 1 eV = 1.6 ×10-19 J & speed of electromagnetic radiation = 3.0 × 108 ms-1

(a) (i) Show that these data support the rule that where R0 is a constant;

R = R0A(1/3) (ii) The mass of a nucleon is about 1.7 × 10-27 kg. Calculate the density of nuclear matter. (6 marks) (b) (i) Explain what is meant by the binding energy of a nucleus. (ii) Show that the total binding energy of a sodium-23 nucleus is about 3 × 10-11 J. (iii) Calculate the mass-equivalent of this binding energy. (5 marks)


Q4 June 05....

Americium-241 ( Am) is a common laboratory source of alpha radiation. It decays spontaneously to neptunium (Np) with a decay constant of 4.8 × 10-11s-1. A school laboratory source has an activity due to the presence of americium of 3.7 × 104Bq when purchased. Avogadro constant = 6.0 × 1023mol-1 one year = 3.2 × 107s (a) (i) Calculate the half-life, in years, of americium-241. (2 marks) (ii) Calculate the number of radioactive americium atoms in the laboratory source when it was purchased. (2 marks) (iii) Calculate the activity of the americium in the laboratory source 50 years after being purchased. (3 marks) (iv) Suggest why the actual activity of the sources is likely to be greater than your answer to part (iii). (1 mark)


Q4 June 05.... (answers)

(a) (I) half life = 1.44 x 1010 s or half life = 0.69/λ or ln2/λ Or 450 (449) y (or 460(456)y if they use 3.15 x 107 s = 1y)

II) A = λN = 7.7 x 1014

III)

IV) the decay products/neptunium may also be radioactive


Q4 June 05....

(b) (i) Use the following data to deduce the energy released in the decay of one Americium-241 nucleus.

mass of americium-241 nucleus = 4.00171 × 10-25kg mass of an alpha particle = 0.06644 × 10-25kg mass of neptunium nucleus = 3.93517 × 10-25kg

speed of electromagnetic radiation in free space = 3.00 × 108ms-1

3 marks

(ii) Explain what is meant by decays spontaneously and how consideration of the masses of particles involved in a proposed decay helps in deciding whether the decay is possible.

2 marks


Q1 June 06....

1 (a) Radioactive lead-214 changes to lead-206 by a series of decays involving alpha (α) and negative beta (β.) emissions. Explain clearly how many alpha and beta particles are emitted during this change. (4 marks) (b) The half-life of lead.214 is 26.8 minutes. (i) Explain what is meant by half-life. (ii) Show that the decay constant of lead 214 is approximately 0.026 minute-1. (iii) Calculate the percentage of the original number of nuclei of lead-214 left in a sample after a period of 90 minutes. (7 marks)


Q2 June 07.... (c) Nuclear structure can be explored by bombarding the nuclei with alpha particles. The de Broglie wavelength of the alpha particle must be similar to the nuclear diameter. Calculate the energy of an alpha particle that could be used to explore the structure of manganese-56. (4 marks)

Planck constant = 6.6 × 10-34 J s mass of an alpha particle = 6.8 × 10-27 kg

wavelength = nuclear diameter of manganese = 9.2 × 10-15 m momentum = h/λ = 7.17 × 10-20 Js m-1 or (N s)

=h/p or h/mv as p = mv so as Ek = ½mv2 we can sub in; Ek = p2/2m or can use = h/mv so v = h/m

Using velocity = 1.05-1.06 × 107 ms-1

3.7(4) -3.8(1) × 10-13 J


Q2 June 07....

(d) (i) State the proton number and nucleon number of the nucleus formed by the decay of manganese-56.

Proton number......... Nucleon number......... (ii) The activity of a sample of manganese 56 varies with time according to the equation

A = A0e-λt

What value should be used for λ in calculations involving manganese-56 when t is in seconds? (4 marks)

Z = 26 A = 56 λ = 0.69/half life or 0.69/2.6 7.37 - 7.41 × 10-5 (s-1)

a2 physics unit 9 radioactivity - animated science · 2012-06-26 · a2 physics unit 9...

Documents