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JJC 2011 JC2/H2 Physics (9646) Nuclear Physics CKM Page 1 of 29 Nuclear Physics 1. The nucleus 2. Isotopes 3. Mass defect and nuclear binding energy 4. Nuclear processes 5. Radioactive decay 6. Biological effect of radiation Learning Outcomes (a) infer from the results of the α-particle scattering experiment the existence and small size of the nucleus. (b) distinguish between nucleon number (mass number) and proton number (atomic number). (c) show an understanding that an element can exist in various isotopic forms each with a different number of neutrons. (d) use the usual notation for the representation of nuclides and represent simple nuclear reactions by nuclear equations of the form 14 4 17 1 7 2 8 1 N+ He O+ H (e) show an understanding of the concept of mass defect. (f) recall and apply the equivalence relationship between energy and mass as represented by E = mc 2 in problem solving. (g) show an understanding of the concept of binding energy and its relation to mass defect. (h) sketch the variation of binding energy per nucleon with nucleon number. (i) explain the relevance of binding energy per nucleon to nuclear fusion and to nuclear fission. (j) state and apply to problem solving the concept that nucleon number, proton number, energy and mass are all conserved in nuclear processes. (k) show an appreciation of the spontaneous and random nature of nuclear decay. (l) infer the random nature of radioactive decay from the fluctuations in count rate. (m) show an understanding of the origin and significance of background radiation. (n) show an understanding of the nature of α, β and γ radiations. (o) define the terms activity and decay constant and recall and solve problems by using A = λN. (p) infer and sketch the exponential nature of radioactive decay and solve problems by using the relationship x = x o exp(-λt), where x could represent activity, number of undecayed particles and received count rate. (q) define half-life. (r) solve problems by using the relation 1 2 λ t 0.693 = . (s) discuss qualitatively the effects, both direct and indirect, of ionizing radiation on living tissues and cells.

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• JJC 2011 JC2/H2 Physics (9646) Nuclear Physics

CKM Page 1 of 29

Nuclear Physics 1. The nucleus 2. Isotopes 3. Mass defect and nuclear binding energy 4. Nuclear processes 5. Radioactive decay 6. Biological effect of radiation Learning Outcomes (a) infer from the results of the α-particle scattering experiment the existence and small size of

the nucleus. (b) distinguish between nucleon number (mass number) and proton number (atomic number). (c) show an understanding that an element can exist in various isotopic forms each with a

different number of neutrons. (d) use the usual notation for the representation of nuclides and represent simple nuclear

reactions by nuclear equations of the form →14 4 17 17 2 8 1N + He O + H (e) show an understanding of the concept of mass defect. (f) recall and apply the equivalence relationship between energy and mass as represented by

E = mc2 in problem solving.

(g) show an understanding of the concept of binding energy and its relation to mass defect. (h) sketch the variation of binding energy per nucleon with nucleon number. (i) explain the relevance of binding energy per nucleon to nuclear fusion and to nuclear

fission. (j) state and apply to problem solving the concept that nucleon number, proton number,

energy and mass are all conserved in nuclear processes. (k) show an appreciation of the spontaneous and random nature of nuclear decay. (l) infer the random nature of radioactive decay from the fluctuations in count rate. (m) show an understanding of the origin and significance of background radiation. (n) show an understanding of the nature of α, β and γ radiations. (o) define the terms activity and decay constant and recall and solve problems by using

A = λN.

(p) infer and sketch the exponential nature of radioactive decay and solve problems by using the relationship x = xoexp(-λt), where x could represent activity, number of undecayed particles and received count rate.

(q) define half-life.

(r) solve problems by using the relation 1

2

λt

0.693= .

(s) discuss qualitatively the effects, both direct and indirect, of ionizing radiation on living

tissues and cells.

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(a) infer from the results of the α-particle scattering experiment the existence and small size of the nucleus.

Some background information

• An atom is neutral – of equal positive(from nucleus) and negative(from electron) charges • An alpha-particle (symbolised by α) is a helium nucleus without orbiting electrons • It comprises of 2 protons (positively charged) and 2 neutrons (neutral); thus the alpha

particle is positively charged Rutherford scattering: Energy and closest approach Scattering of alpha particles Rutherford used alpha particles, typically of energy 5 MeV, from radioactive decay and directed them at a very thin gold foil. He then attempted to detect the subsequent path of the alpha particles. How close an alpha particle can get to different nuclei? An alpha particle has charge 2e, where e = 1.60 x 10–19 C. A nucleus has charge Ze, where Z is the number of protons in the nucleus (and the number of electrons in the atom). The electrical potential energy of the two charges at a distance r is:

rZee

042 =energy potential electrical

πε×

where 0ε = 8.85 x 10–12 C2 J–1 m–1.

When Electric Potential Energy = KE of alpha particles, r = distance of closest approach which gives us a rough estimate of the size of the nucleus. Alpha scattering by gold experimental results. This graph shows the energy in MeV of an alpha particle at distances r from a gold nucleus, Z = 79.

2 5

2 0

1 5

1 0

5

00 2 4 6 8 1 0

d is ta n ce fro m n u c le u s / 1 0 –1 4 m

A p p ro a c h o f a lp h a pa rtic le to n u c le u s

Z = 7 9 (g o ld )

Make an arithmetical check to show that at distance r = 1.0 x 10–14 m the electrical

potential energy is between 20 MeV and 25 MeV, as shown by the graph.

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Hence we may conclude that the size of the nucleus is approx.1.0 x 10–14 m! Besides, obtaining an estimate of the size of the nucleus, we can also find out about many properties of the nucleus from other observations. The α-particle scattering experiment

Fig. 1 An arrangement (top view) used in Ernest Rutherford’s lab in 1911-1913 to study the scattering of α-particles by thin gold foils

The Experimental Set-up

The whole experimental set-up is in a vacuum. A beam of α-particle from a radon gas source is incident onto a thin gold foil (~10-6 m thick).

A detector is used to view the beam. This detector may comprise of a fluorescent screen in front of a telescope, and it is mounted on an axis so that it can be moved in either direction from θ = 0° up to almost 180°. Every α-particle that hits on the fluorescent screen in front of the telescope would be seen, through the telescope, as a small speck of light.

The experiment involves counting the number of specks seen per unit time n at values of θ from 0° to almost 180°.

http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/ruther14.swf

http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/ruther14.swf

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Results The results of the experiment are shown below:

Observations: 1. A small number of α-particle suffers large deflection (i.e. with θ > 90°)., and some are even

reflected back (i.e. with θ = 180°).

Inference: There exist massive particles in the gold foil that can deflect the α-particles appreciably, or even reflect them backwards. Rutherford named these massive particles nuclei. (Singular: nucleus)

It cannot be the case of electrons deflecting/reflecting the α-particles since its mass is about 7000 times that of an electron1. The nucleus should be positively charged to repel the positively charged α-particle when it comes near (Coulomb’s electrostatic repulsion).

2. Almost all the α-particles pass straight through the thin gold foil without being deflected (θ = 0°).

Fig. 2.1 Paths of typical α-particles

passing by a gold nucleus http://www.waowen.screaming.net/revision/nuclear/rsanim.htm

Incident α-particles gold nucleus

Fig. 2.2 Paths of typical α-particles as they pass through the atoms of the gold foil

http://physics-animations.com/Physics/English/top10.htm

From Fig. 2.1, it is seen that α-particles that pass closer to the nucleus suffer a larger deflection in their paths and vice versa.

1 Imagine a stone (α-particle) being thrown into a swarm of mosquitoes (electrons), the mosquitoes will not deflect the stone appreciably.

http://www.waowen.screaming.net/revision/nuclear/rsanim.htmhttp://www.waowen.screaming.net/revision/nuclear/rsanim.htmhttp://physics-animations.com/Physics/English/top10.htmhttp://physics-animations.com/Physics/English/top10.htm

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From Fig. 2.2, only a small number of α-particles are deflected by these gold nuclei. This shows that most of the α-particles have missed hitting them. Inference: The gold nuclei are very small in size (i.e. the foil has a lot of empty space)

such that the α-particles pass straight through without hitting anything, It cannot be the case of nuclei in the gold foil being very light. Otherwise, they cannot deflect the α-particles appreciably as inferred from the 1st observation.

General information (some numerical meaning): • Spacing of atoms in the gold foil ≈ 10-10 m (≈ diameter of an atom). • Spacing of nucleus ≈ 10-15 m (approximately 5000 times smaller than overall atomic size)

• If the thin gold foil has a thickness of 0.001 mm (=10-6 m, 100 times thinner than this piece of paper), it would have 10,000 layers of atoms.

• Unified atomic mass constant, u = 1.66×10-27 kg. (a unit of mass, in atomic scale)

• Mass of an electron, me = 9.11×10-31 kg

• Mass of an α-particle = 4u = 4 × 1.66×10-27 kg = 6.64×10-27 kg ≈ 7000 me.

• Mass of a gold nucleus = 197u = 197 × 1.66×10-27 kg = 3.27×10-25 kg

• In the α-particle scattering experiment, most of the α-particles passed through this 10,000 layers of atoms in the gold foil without hitting any nucleus! If they hit any of the electrons, they would have bulldozed through without any appreciable deflection, being about 7000 times more massive.

Worked Example 1: The deviation of α-particles by thin metal foils through angles that range from 0° to 180° can be explained by A scattering from free electrons. B scattering from bound electrons. C diffuse reflection from the metal surface. D scattering from small but heavy regions of positive charge. Explanation: (D) A & B: Electrons would be too light to deflect the α-particles.

C: Most of the α-particles pass straight through, and a small number reflected.

D: The nuclei are positively charged, so that α-particles are deflected even when they pass near, but not physically hitting, a nucleus. The deflection is more when they pass nearer a nucleus, due to bigger electrostatic repulsive force.

Thus, the α-particle scattering experiment shows the existence of positive nuclei (inferred from deflection of small number of α-particles) in the foil, and that these nuclei are relatively small in size (inferred from large number of α-particles passing straight through).

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Worked Example 2: Two α-particles with equal energies are fired towards the nucleus of a gold atom. Which diagram could represent their paths?

Explanation: (A) B: The path near the gold nucleus is considered near enough, and should show deflection. It

should not be straight. C: Different distance from nucleus should have different electrostatic repulsive force and

should be deflected differently;

D: Both α-particle and the gold nucleus are positive, and should be repulsive, should not be attractive as shown.

(b) distinguish between nucleon number (mass number) and proton number (atomic number).

A simple atomic model consists of a small, dense and positive nucleus, surrounded by negative orbital electrons. The nucleus consists of 2 types of particles, known as protons and neutrons. Each has about the same mass (~1u) while the magnitude of charge of a proton is equivalent to that of an electron (1.60 x 10-19 C). Protons and neutrons in a nucleus are collectively known as nucleons. The total number of protons and neutrons in a nucleus is known as its nucleon number (which is an integer, without unit). Nucleon number is also known as mass number2.

Proton number of a nucleus is the number of protons in it. Each element in the Periodic Table has a unique proton number in the nucleus of its atom, starting with hydrogen having proton number = 1, helium having proton number = 2, etc. It is the proton number in the nucleus of its atom that determines the element. Hence, proton number is also known as atomic number of the nucleus of the atom of that element. All atoms are neutral. In an atom, the proton number of its nucleus and the number of electrons orbiting this nucleus must be the same. Worked Example 3: (Ans: A) Which statement correctly describes a nucleon? A a neutron or a proton B a neutron, proton or an electron C any atomic nucleus D a radioactive atomic nucleus

2 Since a proton or a neutron has a mass of about 1u, so in this unit of mass, the total mass of a nucleus is about the same as its nucleon number, thus nucleon number is also known as mass number. E.g., an α-particle (which is actually a helium nucleus, consisting of 2 protons and 2 neutrons) has a nucleon number = 4. It also has a mass of about 4u.

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(c) show an understanding that an element can exist in various isotopic forms each with a different number of neutrons.

An element is determined by the proton number of the nucleus in its atom. Atoms whose nuclei have the same proton number belong to the same element. This is despite the fact that they may carry different number of neutrons.

Isotopes are atoms having the same number of protons but different number of neutrons.

Quite a number of elements have isotopes. Take for instance the element oxygen (of atomic number 8); nuclei of oxygen atoms must each have 8 protons. Some of the oxygen nuclei have 8 neutrons each, some have 9 neutrons and some even have 10 neutrons; but they are all nuclei of oxygen atoms. Atoms with these nuclei are known as isotopes of oxygen. Another element that has isotopes is hydrogen. An ordinary hydrogen atom has 1 proton in its nucleus, with no neutron. Another isotope which has 1 neutron in its nucleus is known as deuterium, and yet another which has 2 neutrons in its nucleus is known as tritium. Chemically, these isotopes are identical (since chemical reactions only involve orbital electrons, and all isotopes of the same element have the same number of orbital electrons), but their nuclei have different masses. Therefore, all isotopes have identical chemical properties but different nuclear and physical properties. http://lectureonline.cl.msu.edu/~mmp/kap30/Nuclear/nuc.htm (d) Part 1. Use the usual notation for the representation of nuclides…… A nuclide refers to a nuclear species with a given number of protons and neutrons; for example 23Na, 24Na, 24Mg are all different nuclides. The usual notation used for the representation of a nuclide is as follows:

XAZ where: X is the chemical symbol of that element (e.g. H for hydrogen, He for helium) A is the nucleon number (mass number); and Z is the proton number (atomic number).

The isotopes of oxygen (mentioned in learning outcome (c) previously) can be represented by 16 , and 18 . 8 O

178 O 8 O

Likewise, the ordinary hydrogen atom, with 1 proton and no neutron in its nucleus, can be represented by . Isotopes of hydrogen like deuterium (1 proton, 1 neutron) and tritium (1 proton, 2 neutrons) can be represented by and respectively.

11H

21H

31H

E.g., each lithium nucleus has 3 protons and 4 neutrons (i.e. Z = 3, A = 7), the notation for this nuclide is 73 L i

http://www.mpdocker.demon.co.uk/as_a2/topics/atomic_structure/quiz_1.html

The number of eutrons n in a nuclide of element X can be determined from by subtracting from A i.e. number of neutrons = A – Z

AZ X

Z

http://lectureonline.cl.msu.edu/%7Emmp/kap30/Nuclear/nuc.htmhttp://www.mpdocker.demon.co.uk/as_a2/topics/atomic_structure/quiz_1.html

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Worked Example 4: The number of nucleons in the nucleus of the atom 3517 is Cl

A 17 B 18 C 35 D 52 Explanation: (C) A: Number of protons B: Number of neutrons (35 subtract 17) C: Number of nucleons (protons and neutrons) D: Students who know nothing may just try their luck by choosing this option. (d) Part 2. … and represent simple nuclear reactions (e.g. fusion or fission) by nuclear equations of the form 14 →4 17 17 2 8 1N + He O + H

PS 123! For any nuclear equation, ensure that the nucleon number and proton number are both conserved i.e. total number of nucleons on LHS = total number of nucleons on RHS,

Using the nuclear equation , 14 4 17 17 2 8 1N + He O + H→ For the conservation of nucleon number: 14 + 4 = 17 + 1 For the conservation of proton number: 7 + 2 = 8 + 1 The conservation of mass-energy (more on this in learning outcome (f)) cannot be seen from such a nuclear equation. (e) show an understanding of the concept of mass defect. The mass of a single nucleus is less than the total mass of its individual nucleons. For example, when a single helium nucleus is split up into 2 protons and 2 neutrons, the mass of the single helium nucleus is less than the total mass of its 4 individual nucleons (as seen below), thus there is a shortage of mass (mass defect). The mass defect is 0.03035u (= 4.03188u – 4.00153u).

http://www.eas.asu.edu/~holbert/eee460/massdefect.html

Total number of protons on LHS = total number of protons on RHS

PS 123!

ass defect of a nucleus = ( ) − (mass of the nucleus) Total mass of itsindividual nucleonsM

In actual fact, energy has to be supplied to the single helium nucleus to split it up into 4 individual nucleons. This energy supplied is converted to mass to make up the mass defect of 0.03035u and is known as binding energy (to be discussed in detail in learning outcome (g)).

http://www.eas.asu.edu/%7Eholbert/eee460/massdefect.html

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(f) recall and apply the equivalence relationship between energy and mass as represented by E = mc2 in problem solving

According to Albert Einstein, mass is a form of energy. Just like other forms of energy: light, heat, electrical or elastic potential energy, which can be transformed from one form to another; mass can also be regarded as energy and can also be transformed to other forms of energy. Conversely, other forms of energy can also be transformed to mass. This may seem confusing, since energy is measured in joules whereas mass is measured in kg. Einstein's equation associates mass with energy. E.g., if a mass of 1 kg can be completely transformed to energy, the amount of energy would be E = (1 kg)(3.00 × 108 m s-1)2 = 9 × 1016 J. 3 Worked Example 5: In a nuclear reaction, energy equivalent to 10-11 kg of matter is released. The energy released is approximately A 4.5 x10-6 J B 9.0 x 10-6 J C 3.0 x 10-3 J D 4.5 x 105 9.0 x 105 J J E

Solution

: (E) Energy released, E = mc2 =(10-11)( 3.00×108 )2 = 9.0×105 J Worked Example 6: The masses of the particles in the equation are as follows.

9 4 1 124 2 0 6Be + He n + C

9.01212 4.00260 1.00867 12.00000u u u→

u

(i) Calculate the loss of mass which appears to take place in the reaction. (ii) Hence find the energy equivalence of this mass.

Solution:

(i)

Loss of mass = 4.00260u + 9.01212u - (1.00867u + 12.00000u) = 6.05 x 10-3u

At least 6 sig.fig. for this topic because of the closeness of values

To recall and apply E = mc2

where: E represents energy in J;

m represents mass in kg; and c represents the speed of light in vacuum (c = 3.00 × 108 m s-1)

3 In Singapore, electrical energy costs about \$0.30 per unit. A unit of electrical energy here is 1 kWh = (103 W)(3600 s) = 3.6 × 106 J. If an average household spends about \$150 a month on electricity bill, that would be about 500 units = 1.8 × 109 J. Since (9 × 1016 J) ÷ (1.8 × 109 J) = 50 000 000, the amount of energy transformed from 1 kg of mass can last a household 50 million months, or about 4.2 million years. This also means that if a mass of 0.1 g (like 2 drops of water) can be transformed to energy that would last an average household about 4200 years!

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(ii) Energy equivalence of this mass = mc2 = ( 6.05 x 10-3u ) c2 =( 6.05 x 10-3 x 1.66 x 10-27 )( 3.00 x 108 )2 = 9.04 x 10-13 J Atomic scale: Instead of joules, unit of energy in atomic scale is the ‘electron-volt’ (unit symbol: eV). From the equation U = qV, when one electron is passed through a p.d. of 1 V, its energy changes by 1 eV, so

1 eV = (1.6 × 10-19 C)(1 V) = 1.6 × 10-19 J 1 MeV = 106 eV = 1.6×10-13 J

For mass, 1 u = 1.66 × 10-27 kg If 1 u of mass is transformed to energy, E = (1.66 × 10-27 kg)(3.00 ×108 m s-1)2 = 1.494 × 10-10 J = 9.34 × 108 eV = 934 MeV

PS 123! 1u of mass is equivalent to 934 MeV of energy. This relationship is useful in converting a certain mass (in u) to its energy equivalence (in MeV) and vice versa.

In the Data Booklet, the mass of an electron (or a proton) is given as ‘rest mass’. This implies that the value of the mass of an electron given there is its mass when the electron is at rest (stationary, with no kinetic energy). If the electron were to be in motion, it would have some kinetic energy (which is equivalent to some mass!) and the mass of the moving electron would then be more than when it was at rest. In fact, the faster it moves, the more mass it possesses (because the kinetic energy it possesses is regarded as its additional mass)! Note the order of magnitude of both energies is close. Rest mass of electron = 9.11×10-31 kg which is equivalent to 0.5 MeV (check it out!) (g) show an understanding of the concept of binding energy and its relation to mass

defect.

Binding energy of a nucleus is the minimum energy required to break the nucleus into its individual nucleons.

It can also be understood as the amount of energy that would be released if the nucleus is formed from its individual nucleons. It is commonly mistaken as the energy stored in the nucleus or the potential energy of the system, i.e. energy needed to hold the nucleus together4. http://www.eas.asu.edu/~holbert/eee460/massdefect.html 4 Because the nucleus is viewed as a closely packed collection of protons and neutrons, it is surprising that it can exist. Like charges (protons) in close proximity should exert large repulsive electrostatic forces on each other; these forces should cause the nucleus to fly apart. However nuclei can remain stable because of the presence of nuclear force. This force, within a short range (~10-15 m), is an attractive force that acts between all nucleons and it is able to overcome the electrostatic force.

http://www.eas.asu.edu/%7Eholbert/eee460/massdefect.html

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For the case of the helium nucleus mentioned in learning outcome (e), it has 2 protons and 2 neutrons (hence 4 nucleons), its binding energy is 28.3 MeV and mass defect is 0.03035u. If a single helium nucleus is to be broken up into 4 individual nucleons, there is a mass defect (shortage) of 0.03035u to form those 4 individual nucleons. Hence, 28.3 MeV of energy (binding energy) is required to be converted to 0.03035u of mass to make up the mass defect before those 4 nucleons can be formed. In other words, 28.3 MeV of energy is required to break a helium nucleus into its 4 individual nucleons. Conversely, if these 4 individual nucleons were to be combined to form a helium nucleus, there is a mass “excess” of 0.03035u. This mass excess would be converted to 28.3 MeV of energy and released in the process. http://www.phys.unsw.edu.au/einsteinlight/jw/module5_binding.htm Worked Example 7: A nucleus has a nucleon number A, a proton number Z, and a binding energy B. The masses of the neutron and proton are mn and mp, respectively, and c is the speed of light. The mass of the nucleus is given by

A (A - Z)mn + Zmp - 2Bc

B (A + Z)mn + Zmp + 2Bc

C Amn + Zmp - 2Bc

D Amn + Zmp + 2Bc

Solution: (A) Number of neutrons = nucleon number – proton number = A - Z Mass defect of a nucleus = Total mass of its individual nucleons – mass of the nucleus Since number of protons = Z, the number of neutrons = A - Z Mass defect of a nucleus = mass of neutrons + mass of protons – mass of the nucleus = (A – Z)mn + Zmp – mass of nucleus Binding energy of a nucleus = (mass defect of nucleus)c2 ⇒ B = (mass defect) c2 B = [(A – Z)mn + Zmp – mass of nucleus]c2 B/c2 = (A – Z)mn + Zmp – mass of nucleus Mass of the nucleus = (A – Z)mn + Zmp – B/c2

From E = mc2, binding energy of a nucleus = (mass defect of the nucleus) c2 Binding energy is the energy equivalent of its mass defect.

Mass defect of a nucleus = ( ) − (mass of the nucleus) Total mass of itsindividual nucleons

http://www.phys.unsw.edu.au/einsteinlight/jw/module5_binding.htm

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(h) sketch the variation of binding energy per nucleon with nucleon number (mass number).

Binding energy as an absolute amount is not an accurate indicator of the stability of a nucleus due to the dependence on the number of nucleons in each element. Having more nucleons increase total binding energy but it also means stronger repulsive electrostatic forces between each other. A more accurate indicator would therefore be the binding energy per nucleon.

Binding energy per nucleon of a nucleus is the (average) energy per nucleon needed to separate a nucleus into its respective individual nucleons.

The nuclide with a higher value is more stable than one with a lower value because they need more energy per nucleon to split into individual nucleons. For e.g. the binding energy per nucleon for a helium nucleus = (28.3 MeV) ÷ (4 nucleons) = 7.1 MeV.

Fig. 3 Typical variation5 of Binding energy per nucleon against mass (nucleon) number, A. Some important features of the graph: • The maximum binding energy per nucleon for all nuclei is nearly 9 MeV

• The nuclei with the maximum binding energy per nucleon is in the range 50 < A < 100 (i.e. the region of greatest stability is atoms with these values of A)

• The smallest nucleus with binding energy is the deuteron ( 21 H , the nucleus of deuterium), with

2.224 MeV, so its binding energy per nucleon is 1.112 MeV. The smallest of all nuclei is 11 H, which has no binding energy (only 1 proton).

• The largest nucleus in this particular graph is uranium ( 23892 U ), which has a binding energy per nucleon of about 7.5 MeV.

• The slope of the graph on the left is very steep and positive belonging to the series of “light” elements from hydrogen to sodium (increasing binding energy per nucleon as mass number increases); the slope of the graph on the right is gradual and negative (decreasing binding energy per nucleon as mass number increases). • Nickel, 62Ni is the most stable and tightly bound nucleus, followed by iron, 58Fe & 56Fe because they have the highest binding energy per nucleon.

5 A detailed explanation on the general shape of the graph is available at http://www.tpub.com/content/doe/h1019v1/css/h1019v1_77.htm

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(i) explain the relevance of binding energy per nucleon to nuclear fusion and to nuclear fission

Before discussing fusion and fission, we need to clarify the meaning and implication of some terms used in this section of the notes. Anything that has a high energy content (meaning it possesses large amount of energy) is unstable and hence likely to change to another state. On the other hand, anything that has low energy content is stable and hence is difficult to change.

high binding energy per nucleon → nucleus is stable → nucleus has low energy content

low binding energy per nucleon → nucleus is unstable → nucleus has high energy content

Relevance of binding energy per nucleon:

Nuclei with mass numbers on either side of the peak are not as stable as those at the peak. The higher values of binding energy per nucleon near A ~ 60 imply that energy is released when a heavy nucleus (A ≥ 120) i.e. right of the peak, splits into two lighter nuclei. This is because the two lighter nuclei are more stable (lower in energy content) than the original nucleus. Likewise, it is more likely for light nuclei (A ≤ 30) i.e. left of the peak, to combine into heavier nuclei. For both cases, the final binding energy per nucleon will be greater than the original value. The final nuclei will be more stable and energy will be released in the process. For both processes, total mass of products < total mass of initial nuclei Hence this ‘loss’ in mass is converted to energy using E = mc2 in the form of kinetic energies of the products and/or electromagnetic radiation.

Nuclear fission is the splitting of a large nucleus into two or more smaller nuclei, with the emission of a few neutrons and/or other radiations. It is usually instigated by a neutron. http://www.classzone.com/books/earth_science/terc/content/visualizations/es0702/es0702page01.cfm?chapter_no=visualization E.g., when a large nucleus like uranium-235 ( ) reacts with a neutron, nuclear fission takes

place in which the large uranium nucleus is broken into two smaller nuclei: lanthanum ( ) and

bromine ( ), with the emission of three neutrons. The reaction may be represented by

23592 U

14857 La

8535 Br

235

92 U + → 14 + + 3 10 n

857 La

8535 Br

10 n

http://www.eas.asu.edu/~holbert/eee460/fission.html

http://www.classzone.com/books/earth_science/terc/content/visualizations/es0702/es0702page01.cfm?chapter_no=visualizationhttp://www.classzone.com/books/earth_science/terc/content/visualizations/es0702/es0702page01.cfm?chapter_no=visualizationhttp://www.eas.asu.edu/%7Eholbert/eee460/fission.html

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Another possible nuclear fission with 235 which gives rise to a rhodium nucleus ( ) and a

silver nucleus ( 11192 U

12345 Rh

47 Ag ) may be represented as follows:

23592 U + → +

11110 n

12345 Rh 47 Ag + 2

10 n

With reference to the graph above, note that the large uranium nucleus is represented on the right side of the graph. The smaller nuclei resulted from the nuclear fission of uranium are represented nearer the central region of the graph which has a higher binding energy per nucleon. This means that they are more stable, and hence they have lower energy content than the original uranium nucleus (per nucleon). This implies that nuclear fission results in the release of energy (from uranium nucleus with a higher energy content to the smaller nuclei which have lower energy content). It also implies that the total mass of all fission products must be less than the total mass of the particles undergoing fission (in this case, 235 and ) since the released energy has to be converted from the loss of mass during the fission process.

92 U10 n

chain reaction Nuclear fusion is the formation of a larger nucleus from two small nuclei, with the possible emission of other radiation. E.g., when two deuterons fuse to form a helium nucleus, a neutron is released in the process, represented by

21H + +

21H

32 He

10 n

Another possible nuclear fusion is the formation of a helium nucleus from a deuterium nucleus (deuteron) and a tritium nucleus, represented by

21H + +

31H

42 He

10 n

With reference to the same graph above, note that the small hydrogen nuclei are represented on the left side of the graph. The larger nucleus resulted from this nuclear fusion (helium) is represented slightly nearer the central region of the graph which has higher binding energy per nucleon. This means that it is more stable, and also that it has lower energy content than the original hydrogen nuclei (per nucleon). http://www.atomicarchive.com/Movies/Movie5.shtml This implies that nuclear fusion results in the release of energy (from hydrogen with a higher energy content to the larger helium nucleus which has a lower energy content). It also implies that the total mass of all fusion products must be less than the total mass of the particles undergoing fusion (in this case, 2 and ) since the released energy has to be converted from some loss of mass during the fusion process.

1H31H

http://www.eas.asu.edu/~holbert/eee460/fusion.html

http://library.thinkquest.org/17940/texts/fission/fission.htmlhttp://www.atomicarchive.com/Movies/Movie5.shtmlhttp://www.eas.asu.edu/%7Eholbert/eee460/fusion.html

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Supplementary Info: Since all nuclei are positive, the fusion of two nuclei requires some energy to overcome their initial electrostatic repulsion when they are brought together to be fused. This requires the two nuclei to have considerable kinetic energy, so they must be at a considerably high temperature (like 108 K), and so it is easier for smaller nuclei to undergo fusion (also known as thermonuclear reaction). Comparing fission and fusion: In fission, the binding energy per nucleon increases from about 7.5 MeV to about 8.5 MeV (about 1 MeV) whereas in fusion, the binding energy per nucleon increases from about 1 MeV to about 7 MeV (about 6 MeV). So fusion would result in a release of energy about six times the amount released from fission, for the same mass involved in these processes (because the values are all based on ‘per nucleon’).

At present, atomic bombs and nuclear reactors (which are like atomic bombs, but with the release of energy under control, instead of a blast) use fission to produce energy. They require nuclear fuels with large nuclei, like uranium and plutonium, which are highly radioactive and limited in supply from ores, mined in places like Russia. The Sun and all hydrogen bombs use fusion to produce energy (which involves hydrogen nuclei, hence the name ‘hydrogen bomb’ which are reputed to be much more powerful than atomic bombs; so now you should know why. The high temperature required to initiate this thermonuclear bomb may be provided by the explosion of an atomic bomb.) To date, fusion can only be achieved in this uncontrolled way of a bomb. It cannot be done within a nuclear reactor which releases its energy in a controlled manner. If attempts to control the fusion process were to succeed and be used in reactors, the hydrogen fuel required would be virtually inexhaustible, in the form of hydrogen nuclei found in the water of our oceans! The on-going release of energy from the Sun (and the other stars) due to this fusion process is converted from mass, resulting in the continuing loss of mass from the Sun (many tonnes per second).

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Worked Example 8: The rest mass of the deuteron, 2 , is equivalent to an energy of 1876 MeV; the rest mass of a proton is equivalent to 939 MeV and that of a neutron to 940 MeV. A deuteron may disintegrate to a proton and a neutron if it

1H

A emits a γ-ray photon of energy 2 MeV. B capture a γ-ray photon of energy of energy 2 MeV. C emits a γ-ray photon of energy 3 MeV. D captures a γ-ray photon of energy 3 MeV. E emits a γ-ray photon of energy 4 MeV. Solution: (D)

2H→ 1n + 1H The total mass of the products of this process is equivalent to 939 MeV + 940 MeV = 1879 MeV. The mass of the original deuteron is equivalent to 1876 MeV

For the deuteron to disintegrate to a proton and a neutron, there is a mass defect equivalent to

1879 MeV – 1876 MeV = 3 MeV

Therefore, the deuteron needs to capture a γ-ray photon of energy 3 MeV.

This 3 MeV is the binding energy of the deuteron, which is the minimum energy required for the deuteron to break into a proton and a neutron. Worked Example 9: Which of the following statements is a correct deduction from the graph?

A Of the nuclides plotted, is the most stable. 21HB Energy will be released if a nucleus with a mass number greater than about 80 undergoes

fusion with any other nucleus. C Energy will be released if a nucleus with a mass number less than about 80 undergoes

fission as a result of particle bombardment. D 2713 Al will not spontaneously emit an alpha particle to become

23 . 11Na

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Solution: (D) A: The most stable nuclei should have the highest binding energy. B: If a nucleus with a mass number > 80 were to undergo fusion, the product would have a larger mass number, represented by a point further to the right of the graph. This would mean that the product would have a lower binding energy per nucleon, which implies that it has higher energy content, or more energy per nucleon than the initial nucleus. This should result in energy being absorbed, instead of being released. C: If a nucleus with a mass number < 80 were to undergo fission, the products would have smaller mass number, represented by points further to the left of the graph. This would mean that the products would have lower binding energy per nucleon, which implies that they have higher energy content, or more energy per nucleon than the initial nucleus. This should result in energy being absorbed, instead of being released. D: This is true. From Fig. 3, 23 has a lower binding energy per nucleon (and therefore higher energy content) than

11Na2713 Al . This means that energy needs to be supplied to enable

2713 Al to

become 23 . Hence it cannot occur spontaneously 11Na Worked Example 10: The relative atomic masses, Ar, of , and 42 He

2311Na

2713 Al are 4.0026, 22.9898 and 26.9815

respectively. Discuss whether it is possible for 2713 Al spontaneously to emit an alpha particle. Solution: This question is closely related to Worked Example 9 above. The process may be represented by

2713 Al

4 + 2 He2311Na

The total mass of the products = 4.0026u + 22.9898u = 26.9924u This is more than the mass of 2713 Al (which is 26.9815u) by 26.9924u – 26.9815u = 0.0109u Recall E = mc2 : E = (0.0109 u )x c2 = 0.0109 x 1.66 x 10-27 x (3 x 108)2 = 1.63 x 10-12 J Thus 0.0109u is equivalent to 1.63 x 10-12 J of energy. So 1.63 x 10-12 J of energy is required for this process to occur, and it would not occur spontaneously.

• JJC 2011 JC2/H2 Physics (9646) Nuclear Physics

Worked Example 11: (a) Write down the number of electrons, protons and neutrons in a neutral atom of . 126 C (b) The masses of the electron, proton and neutron are 0.00055 mu, 1.00728 mu and 1.00867

mu respectively. Find the sum of the masses of the constituents of the 12 atom. (Give your answer in mu.)

6 C

(c) By definition, the mass of the 126 C atom is 12 mu exactly. Why is your answer to (b) not 12

mu exactly? Solution: (a) Number of electrons = 6 ; number of protons = 6 ; number of neutrons = 6. 6. (b) Sum of the masses = 6 × (0.00055 + 1.00728 + 1.00867) mu (b) Sum of the masses = 6 × (0.00055 + 1.00728 + 1.00867) mu = 12.099 mu = 12.099 mu (c) The reason is because in forming the neutral atom, some mass is lost. This loss in mass,

known as the mass defect, is manifested as the energy released(c) The reason is because in forming the neutral atom, some mass is lost. This loss in mass,

known as the mass defect, is manifested as the energy released in order to bind the electrons, protons and neutrons together to form the neutral atom.

(j) state and apply to problem solving the concept that nucleon number, proton number, energy and mass are all conserved in nuclear processes.

Nuclei involved in a nuclear process must have a total number of protons and a total number of neutrons. For a nucleus, these numbers may be changed after the nuclear process, but the total number of protons and neutrons for all nuclei involved in the same process must remain constant. A nuclear process may also result in a change of mass and a change of energy, separately. But according to Einstein's equation of E = mc2, mass and energy should be regarded as equivalent, as 'mass-energy'. This is to say that a change of mass should be considered as equivalent to a change of energy (with the value of energy calculated using the above equation). Likewise, a change of energy should also be considered as equivalent to a change of mass (with the value of mass calculated using the above equation). Hence, the total of 'mass-energy' should be constant in a nuclear process. Worked Example 12: The isotope 22 decays into with the emission of an α-particle and a γ-ray photon of frequency f. The principle of conservation of energy is expressed in this decay by the equation

688 Ra

22286 Rn

A 88 = 86 + 2 B 226 = 222 + 4 C [mRa – mRn] c2 = hf D [mRa – (mRn + mα)] c2 = hf + 21 α α2 m u +

21Rn Rn2 m u

[Where appropriate, m represents the mass of a particle, u the speed of a particle and c the speed of electromagnetic radiation.]

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Solution: (D) This process may be represented by 22 → + 688 Ra

42 He

22286 Rn

The energy released in this process must be converted from the missing mass. Missing mass Δm = mRa – (mRn + mα) The energy converted from this mass, E = (Δm) c2 = [mRa – (mRn + mα) ] c2 This energy is released in the form of the kinetic energies of the product particles and the γ-ray

= hf + + [mRa – (mRn + mα) ] c2 = hf + +

(k) show an appreciation of the spontaneous and random nature of nuclear decay. A radioactive nucleus is one that has high energy content and is unstable. It will try to become more stable by reducing its energy content by emitting radiation. This process of spontaneous emission of radiation is known as nuclear decay (or radioactive decay, or disintegration). A nucleus decays spontaneously (i.e. occurs without any instigation or stimulation) and randomly (unpredictable as to when and how), independent of external conditions such as pressure, temperature or its chemical combination with other nuclei.

(l) infer the random nature of radioactive decay from the fluctuations in count rate.

Since radiations from radioactive disintegrations cannot be seen with the naked eye, we need detectors. A detector, for e.g. a Geiger-Muller tube (G.M. tube), is always connected to a counting device, known as a counter. Whenever a radiation is detected, the counter would obtain a signal which would be counted. Of course, not all radiations can be detected and counted. In fact, most of the emitted radiations would be radiated in other directions and would not be counted. The number of counts per unit time recorded by a counter is known as ‘count rate’. The fact that count rate fluctuates indicates that radioactive decay must be emitting radiations in a fluctuating manner that is unpredictable, i.e. it is random (Compared outcome (o).

http://www.csupomona.edu/~pbsiegel/www/Geiger_Counter/Geiger.html http://www.csupomona.edu/~pbsiegel/www/vertuallab/vertlab.html

21α α2 m u

21R2 m n Rnu

21α α2 m u

21Rn Rn2 m u

Fig.4a. Example of a Geiger-Muller tube, connected to a counting device

Fig.4b. An example of 137Ba decay data, counting numbers of decays observed in 30s intervals. The best-fit exponential curve is shown. The points do not fall exactly because of fluctuations, i.e. random. For a large sample size, statistically, its activity is directly proportional to the number of radioactive nuclei present in the sample.

http://www.csupomona.edu/%7Epbsiegel/www/Geiger_Counter/Geiger.htmlhttp://www.csupomona.edu/%7Epbsiegel/www/Geiger_Counter/Geiger.htmlhttp://www.csupomona.edu/%7Epbsiegel/www/vertuallab/vertlab.htmlhttp://www.csupomona.edu/%7Epbsiegel/www/vertuallab/vertlab.html

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(m) show an understanding of the origin and significance of background radiation.

When there is no radioactive source near a detector, it still registers counts of radiation, known as background counts. This is due to the presence of radioactive isotopes in all the matter near the detector. Radioactive material is found throughout nature. It occurs naturally in the soil, rocks, water, air, and vegetation. Primary contributions come from:

• sources in the Earth. These include sources in our food and water, which are incorporated in our body6, and in building materials and other products that incorporate those radioactive sources;

• sources from space, in the form of cosmic rays, This radiation primarily consists of

positively charged ions from protons to iron nuclei derived from the sun and from other sources outside our solar system. The dose of radiation varies in different parts of the world based largely on the Earth’s magnetic field and altitude7;

• sources in the atmosphere. One significant contribution comes from the radon gas that is

released from the earth's crust (e.g. the decay of radium in Worked Example 12).

Fig.5. Different sources of background radiation (by approximate proportion)

As seen in Fig. 4, artificial sources account for about 15 per cent of the average background radiation dose. Nearly all artificial background radiation comes from medical procedures such as receiving X-rays for X-ray photographs. In calculations, the so-called background count rate should be deducted from the received count rate to give the correct readings due to the source only.

6 Some of the essential elements that make up the human body, mainly potassium and carbon, have radioactive isotopes that add significantly to our background radiation dose. 7 This radiation is much more intense in the upper troposphere (10 km altitude) and is thus of particular concern for airline crews and frequent passengers, who spend many hours per year in this environment.

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(n) show an understanding of the nature of α, β and γ radiations.

The following table8 shows their natures:

Alpha (α) Beta (β) Gamma (γ) Charge +2 e − e No charge Mass 4 u 1/1840 u No mass

Typical speed 0.1 c 0.3 c to 0.9 c c Nature Particle Particle EM waves

Penetrating ability Stopped by 5 cm of air or a piece of paper

Stopped by a few mm of aluminium

Reduced significantly by

α-particles are helium nuclei ( ). β-particles are electrons. γ-rays are electromagnetic waves with very short wavelengths ∼ 10-12 m and very high frequencies; so γ-photons have very high energies, and hence are very penetrating.

42 He

Some radioactive nuclides emit one kind of radiation, α, or β, or γ radiation only. Some nuclides emit two kinds of radiation, either α and β, or α and γ, or β and γ radiations. Some nuclides emit all 3 kinds.

(o) define the terms activity and decay constant and recall and solve problems by using A = λN.

The activity, A, of a radioactive sample is the number of radiations emitted per unit time. It is also known as the rate of decay. The number of radioactive nuclei in a sample, N, is usually very large. Since radioactive decay is a random process, for such a large sample size, statistically, its activity is directly proportional to the number of radioactive nuclei present in the sample. Mathematically, this relationship is expressed as

To recall and apply

A ∝ N ⇒ A = λN

where λ is the constant of proportionality, known as the decay constant.

We assume that a radioactive decay is a one-shot process, i.e. whenever a radioactive nucleus has decayed, it would become non-radioactive, and no longer able to emit another radiation. Hence, from a sample of N radioactive nuclei, when one of them emits a radiation, the number N would decrease by 1 and become (N − 1). If the number of radioactive nuclei decreases by dN per unit time, it would have emitted dN radiations per unit time. So, its activity, A, can be expressed as

A = − dNdt

.

where the negative sign indicates a decrease in N. The units of A are s-1, which has an equivalent unit known as Becquerel (symbol: Bq, named after Henri Becquerel, who shared the Nobel prize with Marie Curie). http://www.walter-fendt.de/ph14e/lawdecay.htm 8 c = speed of light; u = atomic mass constant; e = elementary charge

http://www.walter-fendt.de/ph14e/lawdecay.htm

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Substituting A = λN, the above equation can be rewritten as

λN = − dNdt

λ = ( )dNN-

dt

where ⎛⎜⎝ ⎠

dN-N

⎞⎟ may be interpreted as the fraction of the radioactive nuclei in a sample that has

decayed. So, the decay constant λ is defined as below:

Worked Example 13: The rate of decay, dN/dt, of the number, N, of nuclei present in a sample of a radioactive element at time t is A is proportional to t. B is proportional to N. C is proportional to 1/t. D is proportional to 1/N. Solution: (B)

The rate of decay, dN/dt, is the activity A which is proportional to N.

Worked Example 14: A radioactive isotope has a decay constant λ and a molar mass M. Taking Avogadro’s constant to be L, what is the activity of a sample of mass m of this isotope? A λmML B λmL

M C λML

m D

λmLM

Solution: (B) A sample of mass M would contain L nuclei (from meaning of Avogadro constant). So a sample of mass m would contain mL

M nuclei, i.e. N = mL

M.

Hence the activity, A = λN = λ. mLM

.

The decay constant λ of a radioactive nuclide may be defined as the fraction of the radioactive nuclei in a sample of the nuclide that has decayed per unit time.

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(p) infer and sketch the exponential nature of radioactive decay and solve problems by using the relationship x = xoexp(-λt), where x could represent activity, number of undecayed particles and received count rate.

NB: exp = e From A = λN, λN = −

dNdt

. . . . . (1)

The equations9 ln N = ln No − λt . . . . . (2) and N = No e-λt . . . . . (3) are possible solutions to equation (1) above. No is the original number of radioactive nuclei in the sample at t = 0 s. From equation (2), a graph of ln N against t would be a straight line graph, with a gradient of −λ and a y-intercept of ln No.

t

ln N

ln No

From equation (3), a graph of N against t would be an exponential decay graph, with an initial value of No at t = 0.

N

No

t http://lectureonline.cl.msu.edu/%7Emmp/applist/decay/decay.htm http://www.vjc.moe.edu.sg/fasttrack/physics/radioactivity.htm 9 Transposing, we get

dN

N = −λ dt

o

N

N

dN

N∫ = −λ t

0dt∫

Integrating, ln N − ln No = −λt ln N = ln No − λt

Take the anti-log of the equation, we will obtain a solution of the differential equation (1), expressed as N = No e-λt, where No is the original number of radioactive nuclei in the sample, at t = 0. A graph of N against t would be an exponential decay graph.

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Worked Example 15: The relationship between Nt and t is A Nt = 1000 e-(20 t/s). B Nt = 20 e(20 t/s). C Nt = 3 e-(0.05 t/s). D Nt = 20 e-(0.05 t/s). (ln 20 = 3.00) Solution: (D) This is a graph of ln N against t. The gradient of the graph is −λ = − 360 = −0.05 s

-1. The y-intercept is ln No = 3. Given ln 20 = 3.00, so No = 20. Substitute into N = No e-λt, and compare with answers. Since A = λN, so Ao = λNo. Hence equation (3) can be changed to

A = Ao e-λt . . . . . (4) Any detector placed at a certain fixed position to measure the activity of a radioactive sample, will receive a certain fraction of the radiations emitted from that sample. So the received count rate, C, must be directly proportional to its activity, i.e. C ∝ A. Hence equation (4) can be changed to

C = Co e-λt . . . . . (5) Similarly, N, the number of radioactive nuclei present in a sample at time t is proportional to its mass of the radioactive nuclei, m: m = mo e-λt (6) The same can also be said about equation (2), so ln A = ln Ao − λt ln C = ln Co − λt ln m = ln mo − λt

where Ao is the initial activity, Co is the initial received count rate and mo is the initial mass at time t = 0

t

ln N

3

The graph shows the number of particles Nt emitted per second by a radioactive source as a function of time t.

60

To solve problems by using x = xo e-λt

where x can be count rate C, activity A, mass m or number N of radioactive nuclei; xo is the initial value of the corresponding physical quantity.

• JJC 2011 JC2/H2 Physics (9646) Nuclear Physics

(q) define half-life.

There are several ways to define half-life, t1/2. It can be defined as the time taken for (1) half of the original number of radioactive nuclei in a sample to remain undecayed; or (2) the activity of a radioactive sample to decrease to half its original value; or (3) the received count rate to reduce to half its original value. http://www.7stones.com/Homepage/Publisher/halfLife.html Worked Example 16: What is the time T? A τ /2 B ln (τ /2) C τ D ln (2τ ) Solution: (C) Since the time T in the graph is the time for the number of atoms to become half its original value, T must be the half-life.

(r) Solve problems by using the relation12

0.693λ=t

.

When t = t1/2, N = 12 No. From ln N = ln No − λt, (refer to outcome (p) above), ln ( 12 No) = ln No − λt1/2

→ ln ( 12 ) = − λt1/2

→ t1/2 =ln2λ

→ λ = ln2 0.693=t t1/2 1/2

Worked Example 17: The radioactive decay of a certain nuclide is represented by the equation below. Find its half-life.

dNdt

= − λn where λ = 2.4 x 10-8 s-1

A 2.9 x107 s B 1.3 x 107 s C 1.2 x 10-8 s D 3.4 x 10-8 s Solution: (A)

t1/2 =λ693.0 =

8104.2693.0

−×= 2.89 × 107 s.

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The graph represents the decay of a newly-prepared sample of radioactive nuclide X to a stable nuclide Y. The half-life of X is τ. The growth curve for Y intersects the decay curve for X after time T.

time

number of atoms

T

X Y

Half-life of a radioactive nuclide is defined as the time taken for half of the original number of radioactive nuclei in a sample to decay.

http://www.7stones.com/Homepage/Publisher/halfLife.html

• JJC 2011 JC2/H2 Physics (9646) Nuclear Physics

Worked Example 18:

The 14C : 12C ratio of living material has a constant value during life but the ratio decreases after death because 14 C is not replaced. The half-life of 14C is 5600 years.

The 14C content of a 5 g sample of living wood has a radioactive count rate of about 100 per minute. If the count rate of a 10 g sample of ancient wood is 50 per minute, the age of the sample is about A 1400 years B 2800 years C 5600 years D 11200 years Solution: (D) For a fair comparison, the same mass of wood should be used. So a 5 g sample of ancient wood should have a count rate of 25 per minute. Collecting data:

Co = 100 per minute, C = 25 per minute, λ =0.693

5600 years= 1.2375 × 10-4 year-1.

Using equation ln C = ln Co − λt

t =λ

−ln lno

C C=

×4

ln(100) ln(25)

1.2375 10

= 11200 years.

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simpler form of the decay law equation, as follows: There is a ⎛ ⎞⎜ ⎟⎝ ⎠

1=2

n

o

xx

where x can be C, A, m or N;

n is the number of half-lives that occur, i.e. n =1/2

tt

.

To re-do Worked Example 18: = =

n = 2

Worked Example 19: The activity from a radioactive source is found to fall by 0.875 of its initial activity in 210 s. What is the half life of the source? A 30 s B 60 s C 70 s D 105 s

⎛ ⎞⎜ ⎟⎝ ⎠

12

n

o

C =C

25 12

n( ) 1100 4

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Solution: (C)

Activity falls by 0.875 of its initial value → activity is (1 – 0.875) of its initial value. → A = (1 – 0.875) Ao = 0.125 Ao

Using ln A = ln Ao − λt, ln (0.125 Ao) = ln Ao − λ(210 s)

→ λ =ln - ln(0.125 )

210o oA A = 9.902 × 10-3 s-1

→ t1/2 =× -3

0.6939.902 10

= 70 s.

Using the simpler equation, ⎛ ⎞= ⎜ ⎟⎝ ⎠

12

n

o

AA

= 0.125 = 18

n = 3 = 210

1/2t → t1/2 =

2103

= 70 s.

Worked Example 20:

The table shows the count-rate recorded at a point in a laboratory at various times, with and without a source in position.

time/days count-rate/ s-1

with source without source 10 30 90

60 30 20

20 20 20

From these readings, what is the half-life of the source?

Solution: (A) At day 10, Co = 60 – 20 = 40 s-1. At day 30, C = 30 – 20 = 10 s-1.

= = n = 2

The source has undergone 2 half- lives. ⇒ 2 t1/2 = 20 days ⇒ t1/2 = 10 days (s) discuss qualitatively the effects, both direct and indirect, of ionizing radiation on

living tissues and cells.

Neutral atoms or molecules may become charged ions when radiation (α, β and γ radiations – refer to outcome (n)) collides with them. We say that they have been ionized. Radiation that can ionize atoms or molecules is called ionizing radiation. Effects of ionizing radiation on living matter Ionizing radiation can cause damage to different parts and systems of the human body.

(a) At the molecular level, irradiation of biological tissues may result in direct damage to proteins, nucleic acids, and other vital molecules found in the tissue cells. The molecular structures of enzymes, which are organic catalysts responsible for controlling many physiological mechanisms in the body, and of DNA (deoxyribonucleic acid) which bears the genetic blueprint of an individual, may be irreversibly altered. This indirectly causes deformity and / or genetic characteristics changes which may be hereditary (passed over to subsequent generations).

⎛ ⎞⎜ ⎟⎝ ⎠

12

n

o

CC

=10 140 4

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(b) At the sub-cellular level, ionizing radiation may cause direct damage or changes to chromosomes which are structures that bear the DNA. Damages may also be caused to other structures within the cell, e.g. mitochondria which are the power houses of the cells. This indirectly causes the breakdown of cellular functions, thus body functions.

(c) At the cellular level, cell division may be inhibited. E.g. a pregnant woman exposing her

developing foetus to ionizing radiation runs the risk of producing a deformed baby. This is because parts of the body may suffer from the effects of ionizing radiation and therefore fail to develop fully. Ionizing radiation may also cause the normal pattern of cell division to go irregular, resulting in the initiation of cancers such as leukaemia. The greater the dose of radiation a cell gets, the greater the chance that the cell will become cancerous. The cancerous cell then goes out of control, divides rapidly and causes serious health problems by spreading to other cells.

(d) At the tissue or organ level, certain parts of our body are vulnerable to ionizing radiation. It

can cause cataract, a condition characterized by opacity or clouding of the eye lens. Reproductive organs are also very susceptible to ionizing radiation, resulting in sub-fertility or even sterility. Blood-forming or related organs such as the liver also suffer the effects of radiation, sometimes with delayed effects over time.

(e) Radiation can cause immediate damage to living flesh and skin which, according to the

dose, is accompanied by radiation burns (i.e. redness of the skin followed by blistering and sores which are slow to heal), radiation sickness, loss of hair and, in extremely severe cases, immediate death.

At community level, exposure to ionizing radiation in the long term may lead to damages or changes to micro-organisms and plant life of the surroundings. As a result, this will introduce changes to the other organisms in the food chain. With the long term retention of the affected substances in our bodies, there is uncertainty to what changes can happen, for instance, premature ageing and a shortening of life. However, at a positive level, ionizing radiation can also have its uses on living tissue. For instance, very high doses of radiation can kill the cell completely. This idea is applied positively to kill cancer cells, and also harmful bacteria, germs and other “bad” microorganisms. Another instance is to serve as radioactive tracers to aid diagnosis or rehabilitation. For example, to trace the location of a blood clot (thrombosis) or to track the activity of internal body functions like thyroid glands. http://www.atomicarchive.com/Movies/index_movies.shtml

----------------------------- The End -------------------------------

http://www.atomicarchive.com/Movies/index_movies.shtml

• JJC 2011 JC2/H2 Physics (9646) Nuclear Physics

CKM Page 29 of 29

References: Oh T.H., Tan C.H.,Chin, F.S., Damien Ler et al. (2005, 2007). Nuclear Physics, Notes for “A” Level Physics (9246), Jurong Junior College.

Halliday, D., Resnick, R., & Walker, J. (2005). Fundamentals of Physics (7th Edition). New Jersey: John Wiley & Sons, Inc.

Hutchings, R. (1991). University of Bath. Macmillan Science 16-19 Project Physics. London: Macmillan.

Serway, R.A., & Jewett, J.W., Jr. (2004). Physics for Scientists and Engineers (6th Edition). Belmont, CA: Brooks/Cole – Thomson Learning Websites (to be used in conjunction with this set of notes):

Page 2 http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/ruther14.swf

Paths of typical α-particles passing by a gold nucleus

Page 3, Fig. 2.1

http://www.waowen.screaming.net/revision/nuclear/rsanim.htm http://physics-animations.com/Physics/English/top10.htm

Paths of α-particles as they pass through the atoms of the gold foil

Page 3, Fig. 2.2

Isotopes Page 6 http://lectureonline.cl.msu.edu/~mmp/kap30/Nuclear/nuc.htm

Notation for the representation of nuclides

Page 6 http://www.mpdocker.demon.co.uk/as_a2/topics/atomic_structure/quiz_1.html

Concept of mass defect Page 7 http://www.eas.asu.edu/~holbert/eee460/massdefect.html http://www.phys.unsw.edu.au/einsteinlight/jw/module5_binding.htm

Binding energy and its relation to mass defect

Page 10 http://www.eas.asu.edu/~holbert/eee460/massdefect.html

Fission Page 12 http://www.classzone.com/books/earth_science/terc/content/visualizations/es0702/es0702page01.cfm?chapter_no=visualization

http://www.eas.asu.edu/~holbert/eee460/fission.html Chain reaction Page 13 http://library.thinkquest.org/17940/texts/fission/fission.

html Fusion Page 14 http://www.eas.asu.edu/~holbert/eee460/fusion.html

http://www.atomicarchive.com/Movies/Movie5.shtml Random nature of radioactive decay from the fluctuations in count rate.

Page 18 http://www.csupomona.edu/~pbsiegel/www/vertuallab/vertlab.html

http://www.csupomona.edu/~pbsiegel/www/Geiger_Counter/Geiger.html

Exponential nature of radioactive decay and solve problems by using the relationship x = xoexp(-λt)