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23.3Nuclear properties and the radioactive decay law

Nuclear radii

When an alpha particle of kinetic energy E K is directed at a nucleus of proton number Z, the distance of closest approach, d, is found by equating the kinetic energy far from the nucleus to the electric potential energy at the point of closest approach (Figure 23.8). That is:

E K=k(2 e)(Z e)d â‡’d=2 k Z e 2 E K

Figure 23.8: The alpha particle momentarily stops at the distance of closest approach d.

So for an alpha particle of kinetic energy 4.5 MeV that is directed at a nucleus of nickel of proton number 28 we will find a distance of:

d=2 kZ e 2 E K=2 Ã—8.99 Ã—10 9 Ã—28 Ã—(1.6 Ã—10 âˆ’19)2 4.5 Ã—10 6 Ã—1.6 Ã—10 âˆ’19=1.8 Ã—10 âˆ’14 Â m

As the energy of the incoming alpha particles increases, the distance of closest approach decreases and will eventually approach the nuclear surface. When the alpha particle is near the nuclear surface deviations from Rutherford scattering will be observed (see next section) because the nuclear force will start acting on the alpha particle. The number of scattered alpha particles will not be what is expected based on Rutherford scattering. Thus, the distance of closest approach measured in this case will essentially be the nuclear radius. In this way it was discovered that the radius of a nucleus is given by the empirical formula

R=R 0 A 1 3

where R 0 = 1.2 Ã— 10âˆ’15 m and A is the nucleon number of the nucleus, i.e. the total number of protons and neutrons in the nucleus. A nucleus has a typical radius of order of magnitude 10âˆ’15 m whereas the atom has a radius of order 10âˆ’10 m.

Worked example 23.9

Explain why the empirical formula for the nuclear formula implies that all nuclei have the same density. Estimate this density.

Answer

The density of a nucleus is given in terms of mass and volume by Â Ï�=M V. A nucleus with nucleon number A has a mass of approximately Au. The volume is given by V=4 Ï€3 R 3=4 Ï€3(R 0 A 1 3)3=4 Ï€3 R 0 3 A. Both mass and volume are proportional to the nucleon number so in the ratio the dependence on A disappears. Specifically,
Ï�=A u 4 Ï€3 R 0 3 A=3 u 4 Ï€R 0 3=3 Ã—1.66 Ã—10 âˆ’27 4 Ï€(1.2 Ã—10 âˆ’15)3 â‰ˆ 2 Ã—10 17 kgm âˆ’3.

(The only macroscopic body with a density of this order of magnitude is a neutron star.)

Deviations from Rutherford scatteringâ€”evidence for the strong nuclear force

Rutherford derived a theoretical formula for the scattering of alpha particles from nuclei. The Rutherford formula states that as the scattering angle Î¸ increases, the number of alpha particles scattered at that angle decreases very sharply. This is shown in Figure 23.9a. In fact, Rutherfordâ€™s formula states that the number N of alpha particles scattering at an angle Î¸ is proportional to 1 sin 4 â�¡(Î¸2). If this is the case, the product N sin 4 â�¡(Î¸2) should be constant.

Table 23.3 contains some of the original data in the Geigerâ€“Marsden experiment with a gold foil. The last column in the table shows that the product N sin 4 â�¡(Î¸2) is indeed fairly constant, which is strong evidence in support of the Rutherford formula.

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The derivation of the Rutherford formula is based on a number of assumptions. The most important is that the only force in play during the scattering process is the electric force. As the energy of the alpha particles increases, the alpha particles can get closer to the nucleus. When the distance of closest approach gets to be about 10âˆ’15 m or less, deviations from the Rutherford formula are observed (Figure 23.9b). This is due to the fact that the alpha particles are so close to the nucleus that the strong nuclear force begins to act on the alpha particles. Therefore, the presence of these deviations from perfect Rutherford scattering is additional evidence for the existence of the strong nuclear force.

Scattering angle Î¸ / Â° | N | N sin 4 â�¡(Î¸2)
15 | 132 000 | 38.4
30 | 7 800 | 35.0
45 | 1 435 | 30.8
60 | 477 | 29.8
75 | 211 | 29.1
105 | 69 | 27.5
120 | 52 | 29.0
150 | 33 | 28.8

Table 23.3: Data from the Geigerâ€“Marsden experiment, reproduced in the book by E. Rutherford, J. Chadwick and C. D. Ellis, Radiations from Radioactive Substances, Cambridge University Press, 1930.

Figure 23.9: a The logarithm of the number of alpha particles scattered at some angle ğ�œ½ as a function of ğ�œ½. b The logarithm of the number of alpha particles scattered at an angle of 60Â° as a function of the alpha particle energy. The dotted curve is based on Rutherford scattering. The blue curve is the observed curve. We see deviations when the energy exceeds about 28 MeV. The energy at which deviations start may be used to estimate the nuclear radius.

Worked example 23.10

Suggest how the results of the scattering of alpha particles would change if the gold (Au Â Â 79 197) foil was replaced by an aluminium (Al 13 30) foil of the same thickness.

Answer

Aluminium has a smaller nuclear charge and so the alpha particles would approach closer to the nucleus. This means that the alpha particles would start feeling the effects of the nuclear force and deviations from perfect Rutherford scattering would be observed at lower alpha particle energies compared to those in the case of gold.

Approximate constancy of the binding energy per nucleon

We have already noted that a feature of the binding energy curve, Figure 23.1, is that the binding energy per nucleon is approximately constant for nuclei with A > 20. How do we explain this observation? The short range of the force implies that a given nucleon can interact with its immediate neighbours only and not with all of the nucleons in the nucleus. So, for large nuclei (roughly A > 20) any one nucleon is surrounded by the same number of immediate neighbours, Figure 23.10, and so the energy needed to remove that nucleon from the nucleus is the same. Thus, the short-range nature of the nuclear force explains why the binding energy per nucleon is roughly constant above a certain value of A. Conversely, this feature of the curve is evidence for the short-range nature of the nuclear force.

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Figure 23.10: For nuclei that are large enough, any one nucleon has the same number of closest neighbours that fall within the range of the strong nuclear force (dotted circle). The nucleon painted yellow has about 12 closest neighbours in both nuclei (in this unrealistic two-dimensional representation).

Check yourself 9

State and explain the evidence for

athe existence of the strong nuclear force

bthe short range of the nuclear force.

Nuclear stabilityâ€”the role of neutrons

Figure 23.11 shows that very few nuclides are stable; there are about 250 of them and are shown as points in black. Most nuclides (about 2500) are unstable.

Figure 23.11: A plot of neutron number versus proton number for nuclides. The stable nuclides are shown in black. Most nuclides are unstable.

Notice that stable nuclei with small values of Z have equal numbers of neutrons and protons. As Z increases, stable nuclei have more neutrons than protons. How do we understand this observation? Recall that two of the forces acting within the nucleus are the electromagnetic force and the strong nuclear force. The electromagnetic force acts on protons and is a repulsive force. It has infinite range so all the protons in a nucleus repel each other, contributing to the tendency for the nucleus to break apart. After a certain nuclear size, adding more protons only contributes to the tendency to break the nucleus apart through the electric force. The strong nuclear force does not help because for large enough nuclei the number of nearest neighbours of a given proton is the same and because of the short range of the force only these act on a given proton. So, to keep a larger nucleus stable, more neutrons are needed, both because the extra neutrons make the distance between protons larger and because the neutrons contribute to binding.

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Nuclear energy levels

The nucleus, like the atom, exists in discrete energy levels. The main evidence for the existence of nuclear energy levels comes from the fact that the energies of the alpha particles and gamma ray photons that are emitted by nuclei in alpha and gamma decays are discrete. This is to be contrasted with beta decays, in which the electron has a continuous range of energies, a fact that we will explain shortly.

Figure 23.12 shows the lowest nuclear energy levels of the magnesium nucleus Mg 12 24. Also shown is a gamma decay from the level with energy 5.24 MeV to the first excited state. The emitted photon has energy 5.24 âˆ’ 1.37 = 3.87 MeV.

Figure 23.12: Nuclear energy levels of magnesium, Mg 12 24. Notice the difference in scale between these levels and atomic energy levels.

Figure 23.13 shows an energy level of plutonium (Pu Â Â 94 242) and a few of the energy levels of uranium (U Â Â 92 238). Also shown are two transitions from plutonium to uranium energy levels. These are alpha decays:

Pu Â Â 94 242 â†’U Â Â 92 238+Î±2 4

The energies of the emitted alpha particles are 4.983 âˆ’ 0.148 = 4.835 MeV and 4.983 âˆ’ 0.307 = 4.676 MeV.

Figure 23.13: Energy levels for plutonium and uranium. Transitions from plutonium to uranium energy levels explain the discrete nature of the emitted alpha particle in the alpha decay of plutonium. Photons may emitted as uranium makes transitions from excited states to lower energy states.

Worked example 23.11

The nucleus of bismuth (Bi Â Â 83 211) decays into lead (Pb Â Â 82 207) in a two-stage process. In the first stage, bismuth decays into polonium (Po Â Â 84 211). Polonium then decays into lead. The nuclear energy levels that are involved in these decays are shown in Figure 23.14.

Figure 23.14: For worked example 23.11.

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aWrite down the reaction equations for each decay.

bCalculate the energy released in the beta decay.

cExplain why the electron does not always have this energy.

Answer

a Bi Â Â 83 211 â†’Po Â Â 84 211+e âˆ’1 Â Â Â 0+Î½ Â¯0 0 and Po Â Â 84 211 â†’Pb Â Â 82 207+Î±2 4

bThe energy released is the difference in the energy levels involved in the transition, i.e. 0.57 MeV.

cThe energy of 0.57 MeV must be shared between the electron, the anti-neutrino and the polonium nucleus. So the electron does not always have the maximum energy of 0.57 MeV. Depending on the angles (between the electron, the anti-neutrino and the polonium nucleus), the electron energy can be anything from zero up to the maximum value found in b.

international mindedness

Research into neutrino physics is a great example of very large international research teams that work together and collaborate widely, including the Kamiokande and the super-Kamiokande Japaneseâ€“American collaboration, the GALLEX and SAGE groups in Italy and Russia and the Solar Neutrino Observatory (SNO) in Canada with Canadian, American and British participation.

Check yourself 10

How are the photons emitted in transitions between nuclear energy levels different from the photons in atomic energy level transitions?

The neutrino

In the 1930s it was thought that beta minus decay was described by:

n 0 1 â†’p 1 1+e âˆ’1 Â Â Â 0

The mass difference for this decay is:

1.008665 Â u âˆ’(1.007276+0.0005486)Â u=0.00084 Â u

and corresponds to an energy of:

Q=0.00084 Ã—931.5 Â MeV=0.783 Me Â V

If only the electron and the proton are produced, then the electron, being the lighter of the two, will carry most of this energy away as kinetic energy. To see this, assume that the neutron is at rest when it decays. Then the total momentum before the decay is zero. After the decay the electron and the proton will have equal and opposite momenta, each of magnitude p. Equating the total kinetic energy after the decay to the energy released, Q:

p 2 2 m e+p 2 2 m p=Q p 2=2 Q m e m p m e+m p

And so:

E e=p 2 2 m e=Q m p(m e+m p)

E e=0.783 Ã—1.007 5.49 Ã—10 âˆ’4+1.007

E e=0.78257 â‰ˆ 0.783 Â MeV

Thus, we should observe electrons with kinetic energies of about 0.783 MeV. In experiments, however, the electron has a range of energies, from zero up to 0.783 MeV (Figure 23.15). If the electron is not carrying 0.783 MeV of energy, where is the missing energy?

Figure 23.15: The number of electrons that carry a given energy as a function of energy.

Wolfgang Pauli (1900â€“1958) hypothesised the existence of a third particle in the products of a beta decay in 1933. Since the energy of the electron in beta decay has a range of possible values, it means that a third very light particle must also be produced so that it carries the remainder of the available energy.

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Enrico Fermi coined the word neutrino for the â€˜little neutral oneâ€™ (Fermi is shown with Heisenberg and Pauli in Figure 23.16).

Figure 23.16: E. Fermi (left), W. Pauli (right) and W. Heisenberg in happy times, before they were separated by World War 2.

As the neutrino is electrically neutral, it has no electromagnetic interactions. Its mass is negligibly small and so gravitational interactions are irrelevant. It also does not have strong nuclear interactions. This leaves the weak interaction as the only interaction with which the neutrino can interact. This means that the neutrino can go through matter with very few interactions. In fact, about 10 billion neutrinos pass through your thumbnail every second, yet you do not feel a thing. For every 100 billion neutrinos that go through the earth only one interacts with an atom in the earth! Most of the neutrinos that arrive at earth are produced in the sun in the fusion reaction p+p â†’H 1 2+e++Î½ e. Read the fascinating story of the solar neutrino problem in the Nature of Science section at the end of this topic.

Links

Pauli had an additional argument suggesting the existence of a third particle in beta decay in addition to energy conservation that we talked about above. A nucleus with zero angular momentum can decay by beta minus decay into a nucleus with zero angular momentum. But the electron produced in the decay has a property called spin, which is a kind of angular momentum. So to conserve angular momentum another particle had to be produced with an angular momentum equal and opposite to that of the electron.

The radioactive decay law

The law of radioactive decay states that the rate of decay is proportional to the number of nuclei present that have not yet decayed:

d N d t=âˆ’Î»N

The constant of proportionality is denoted by Î» and is called the decay constant probability of decay per unit time.

To see the meaning of the decay constant we argue as follows: in a short time interval dt the number of nuclei that will decay is dN = Î»N dt (we ignore the minus sign). The probability that any one nucleus will decay is therefore:

probabilityÂ =d N N

probabilityÂ =N Î»d t N

probabilityÂ =Î»d t (this is valid provided Î»d t is smaller than one)

and finally, the probability of decay per unit time is:

probabilityÂ d t=Î»

KEY POINT

The decay constant Î» is the probability of decay per unit time.

Check yourself 11

The decay constant of a radioactive isotope is Î» = 6 hrâˆ’1. What is the probability of decay in 1 hour?

The decay law is a differential equation, which when integrated gives:

N=N 0 e âˆ’Î»t

This is the number of nuclei present at time t given that the initial number (at t = 0) is N 0.

As expected, the number of nuclei of the decaying element decreases exponentially as time goes on (Figure 23.17).

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Figure 23.17: Radioactive decay follows an exponential decay law.

The (negative) rate of decay (i.e. the number of decays per second) is called activity, A:

A=âˆ’d N d t

It follows from the exponential decay law that activity also satisfies an exponential law:

A=Î»N 0 e âˆ’Î»t

Thus, the initial activity of a sample is given by the product of the decay constant and the number of atoms initially present, A 0 = Î»N 0. Notice also that A = Î»N.

After one half-life, Â T 1 2, half of the nuclei present have decayed and the activity has been reduced to half its initial value. So using either the formula for N or A (here we use the N formula):

N 0 2=N 0 e âˆ’Î»t

Taking logarithms we find:

Î»T 1 2=ln â�¡2

Î»T 1 2=0.693

This is the relationship between the decay constant and the half-life.

Worked example 23.12

Carbon-14 has a half-life of 5730 yr, and in living organisms it has a decay rate of 0.25 Bq gâˆ’1. A quantity of 20 g of carbon-14 was extracted from an ancient bone, and its activity was found to be 1.81 Bq. What is the age of the bone?

Answer

Using the relationship between decay constant and half-life:

Î»=ln â�¡2 T 1 2

Î»=ln â�¡2 5730 yr âˆ’1

Î»=1.21 Ã—10 âˆ’4 yr âˆ’1

When the bone was part of the living body the 20 g would have had an activity of 20 Ã— 0.25 = 5.0 Bq. If the activity now is 1.81 Bq, then:

A=A 0 Â e âˆ’Î»t

1.81=5.0 Â e âˆ’1.21 Ã—10 âˆ’t t

e âˆ’1.21 Ã—10 âˆ’t t=0.362

Taking logarithms of both sides: âˆ’1.21 Ã— 10âˆ’4 t = âˆ’1.016

t=1.016 1.21 Ã—10 âˆ’4

t â‰ˆ 8400 Â yr

Exam tip

It is important to know that the initial activity A 0 is Î»N 0.

A graph of A against N gives a straight line whose slope is the decay constant.

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Worked example 23.13

aA container is filled with a quantity of a pure radioactive element X whose half-life is 5.0 minutes. Element X decays into a stable element Y. At time zero no quantity of element Y is present. Determine the time at which the ratio of atoms of Y to atoms of X is 5.

bPolonium-208 decays by alpha decay into a stable isotope of lead. The half-life is 2.90 years. Initially there were 8.00 mg of Po in a sample and none of lead. What is the mass of lead produced after 1.30 years?

Answer

aAfter time t the number of atoms of element X is given by N = N 0 eâˆ’Î»t. The number of atoms of element Y is given by N = N 0 âˆ’ N 0 eâˆ’Î»t. The decay constant is Î»=ln â�¡2 5.0=0.1386 Â min âˆ’1 and so we have that:

N 0 âˆ’N 0 e âˆ’0.1386 t N 0 e âˆ’0.1386 t=5 1 âˆ’e âˆ’0.1386 t=5 e âˆ’0.1386 t e âˆ’0.1386 t=1 6 0.1386 Ã—t=1.7918 t=12.9 Â min

bThe number of lead nuclei produced in 1.30 years is N 0 âˆ’N 0 e âˆ’ln 2 200 Ã—1.30=0.267 N 0, where N 0 is the initial number of Po nuclei. N 0=8.00 208 Ã—N A so the mass M of lead is M=0.267 Ã—8.00 208 Ã—N A â�Ÿ number ofliea nuclaÂ Ã—204 N A â�Ÿ kaad utÃ´mmasÂ =0.267 Ã—8.00 Ã—204 208=2.09 Â mg.

nature of science

The solar neutrino problem

In 1968, Ray Davis announced results of an experiment that tried to determine the number of neutrinos arriving at earth from the sun. The idea was that the very rare interaction of neutrinos with ordinary chlorine would produce radioactive chlorine atoms that could then be detected, and hence the number of neutrinos determined. The results showed that the number of neutrinos was about one-third of what the theoretical calculation predicted. This created the â€˜solar neutrino problemâ€™. There were three ways out: either the Davis experiment was wrong or the theory was wrong, or there was new physics involved.

Happily, it turned out that it was the last possibility that actually was in play. The number of neutrinos predicted by theory was based on the assumption that the neutrino was massless. If the neutrino had mass, then the theory would have to be modified because in that case â€˜neutrino oscillationsâ€™ would take place. This is a rare quantum phenomenon, in which the three types of neutrinos could turn into each other. The Davis and subsequent experiments all measured electron neutrinos. Much later, when advances in instrumentation and computing power allowed experiments to detect all three types of neutrinos, the number was in agreement with the theory.

But the neutrinos produced in the sun were only electron neutrinos!

By this time, experiments in Japan and elsewhere provided convincing evidence that neutrinos had a tiny mass. So, because of neutrino oscillations, by the time the electron neutrinos reached earth some of them had turned into muon neutrinos and some into tau neutrinos. On the average, about one-third would be electron neutrinos, in agreement with Davisâ€™ results! Ray Davis shared the 2002 Nobel Prize in Physics.

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TEST YOUR UNDERSTANDING

48Large, stable nuclei have more neutrons than protons. Explain this observation by reference to the properties of the strong nuclear force.

49An alpha particle of energy 4.10 MeV is fired at a nucleus of iron 56 (proton number 26). Calculate the distance of closest approach assuming the iron nucleus does not move.

50An alpha particle directed towards a nucleus of platinum (Pt Â Â 78 195) stops at a distance d from the centre of the nucleus. If the nucleus is replaced by the isotope Pt Â Â 78 198, how does d change, if at all?

51 aEstimate the nuclear radii of an alpha particle and a nucleus of platinum (Pt Â Â 78 195).

bDetermine the initial energy of an alpha particle directed towards the platinum nucleus such that the alpha particle stops just as the nuclear surfaces touch.

52An alpha particle is fired head-on at a stationary gold nucleus from far away. Calculate the initial speed of the particle so that the distance of closest approach is 8.5 Ã— 10âˆ’15 m. (Take the mass of the alpha particle to be 6.64 Ã— 10âˆ’27 kg.)

53A particle of mass m and charge e is directed from very far away toward a massive (M >> m) nucleus of charge +Ze with a velocity v, as shown in Figure 23.18. The distance of closest approach is d. Sketch (on the same axes) a graph to show the variation with separation of:

athe particleâ€™s kinetic energy

bthe particleâ€™s electric potential energy.

cThe incoming particle is a proton and Z = 28; v = 3.0 Ã— 107 m sâˆ’1 and d = 2.0 Ã— 10âˆ’14 m. Using energy and angular momentum conservation find b, the distance by which the proton misses the nucleus.

Figure 23.18: For question 53.

54Show that the nuclear density is the same for all nuclei.

55 aDeviations from Rutherford scattering are expected when the alpha particles reach large energies. Suggest an explanation for this observation.

bSome alpha particles are directed at a thin foil of gold (Z = 79) and some others at a thin foil of aluminium (Z = 13). Initially, all alpha particles have the same energy. This energy is gradually increased. Predict in which case deviations from Rutherford scattering will first be observed.

56Plutonium (Pu Â Â 94 242) decays into uranium (U Â Â 92 238) by alpha decay. The energy of the alpha particles takes four distinct values: 4.90 MeV, 4.86 MeV, 4.76 MeV and 4.60 MeV. In all cases a gamma ray photon is also emitted except when the alpha energy is 4.90 MeV. Use this information to suggest a possible nuclear energy level diagram for uranium.

57The first excited state of the nucleus of uranium-235 is 0.051 MeV above the ground state.

aWhat is the wavelength of the photon emitted when the nucleus makes a transition to the ground state?

bWhat part of the spectrum does this photon belong to?

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58Figure 23.19 shows a few nuclear energy levels for Ar 18 40, K 19 40 and Ca 20 40.

Figure 23.19: For question 58.

Identify the four indicated transitions.

59 aState the evidence in support of nuclear energy levels.

Radiumâ€™s first excited nuclear level is 0.0678 MeV above the ground state.

bWrite down the reaction that takes place when radium decays from the first excited state to the ground state.

cCalculate the wavelength of the photon emitted in b.

60 aFind the decay constant for krypton-92, whose half-life is 3.00 s.

bSuppose that you start with 1 100 mol of krypton. Estimate how many undecayed atoms of krypton there are after

i1 s

ii2 s

iii3 s.

61 aState the probability that a radioactive nucleus will decay during a time interval equal to a half-life.

bCalculate the probability that it will have decayed after the passage of three half-lives.

cA nucleus has not decayed after the passage of four half-lives. State the probability it will decay during the next half-life.

62 aEstimate the initial activity of 1.00 g of polonium-210. The half-life of polonium-210 is 138 days.

bPolonium-210 decays into lead-206 by alpha decay. Estimate the mass of lead produced after 31 days.

63The half-life of an unstable element is 12 days. Find the activity of a given sample of this element after 20 days, given that the initial activity was 3.5 MBq.

64The half-life of a radioactive isotope is 6.0 days. The activity of the sample now is 0.50 MBq. What was the activity one day before?

65The age of very old rocks can be found from uranium dating. Uranium is suitable because of its very long half-life: 4.5 Ã— 109 yr. The final stable product in the decay series of uranium-238 is lead-206. Find the age of rocks that are measured to have a ratio of lead to uranium atoms of 0.80. You must assume that no lead was present in the rocks when the rocks were formed.

66A piece of an ancient bone has a mass of 150 g. The ratio of carbon-14 to carbon-12 atoms in a living bone is constant at 1.3 Ã— 10âˆ’12. The activity of the bone is measured to be 11 Bq. The half-life of carbon 14 is 5730 years. How old is the bone?

67Two unstable isotopes are present in equal numbers (initially). Isotope A has a half-life of 4 min and isotope B has a half-life of 3 min. Calculate the ratio of the activity of A to that of B after:

a0 min

b4 min

c12 min.

68The decay constant of an isotope is 1.2 hrâˆ’1.

aUsing the radioactive decay law, calculate the fraction of nuclei that will decay in 1 hr.

bBy expressing the decay constant in minâˆ’1 calculate the probability of decay within 1 hr.

cHow do the answers to a and b compare?

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69The half-life of an isotope with a very long half-life cannot be measured by observing its activity as a function of time, since the variation in activity over any reasonable time interval would be too small to be observed. Let m be the mass in grams of a given isotope of long half-life.

aShow that the number of nuclei present in this quantity is N 0=m Î¼ N A where Âµ is the molar mass of the isotope in g molâˆ’1 and N A is the Avogadro constant.

bFrom A = N 0 Î»e âˆ’Î»t show that the initial activity is A 0=m N A Î¼ Î» and hence that the half-life can be determined by measuring the initial activity (in Bq) and the mass of the sample (in grams).

70In a pure sample of a radioactive material the number of nuclei that have not yet decayed is 2.0 Ã— 1020. The half-life is 5.0 min. What is the activity?