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Radioactivity and Nuclear Reactions Radioactivity Radioactivity was first discovered in 1896 by Henri Becquerel when a photographic plate wrapped in black paper was exposed when placed in close proximity to a uranium salt. Later, experiments by Marie and Pierre Curie uncovered other radioactive substances and eventually it was shown that the radiation from radioactive materials could be classified in three fundamentally different groups:1 alpha rays or particles (α) — 2

4 He nuclei beta rays or particles (β) — electrons (created in and emitted from the nucleus) gamma rays (γ) — high-energy photons

(From: Six Ideas that Shook Physics, by Thomas Moore)

Afterwards, it was discovered that some radioactive substances emitted positrons, which are essentially positively charged electrons (antiparticle to the electron). Hence, there are both negative and positive beta rays: β– – electron; β+ – positron. Emission of β– radiation is far more common in natural radioactivity. The phenomenon of radioactivity is due to unstable nuclei. In radioactive processes involving α or β rays, the radioactive nucleus (parent) emits an α or β particle and is transformed into the nucleus of a different element (offspring or daughter nucleus). If the offspring is also radioactive, then it may emit an α particle or a β particle and transform into a different nucleus. This process of radioactive disintegration or decay will continue until a stable nucleus is obtained. In radioactive decay processes involving gamma-ray emission, a nucleus that is in an excited state can lower its energy (and thereby increase its stability) by emitting a gamma photon and in the process, falling to a lower energy state. Eventually, such transitions will take the nucleus to its ground state, at which time gamma emissions will cease for that nucleus. Note that in γ decay, no new nucleus is formed. The energies of γ photons are usually in the range 0.1–10 MeV, which is significantly higher than the energies of x-ray photons (emitted during atomic- electron transitions) whose energies are usually in the range 1–100 keV. 1 There are other emissions, e.g., protons, but these are far less common. (See Krane, Mod. Phys.)

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Radioactive Decay Law Consistent with the nature of quantum mechanics, one cannot predict precisely when a radioactive nucleus will decay. One can, however, calculate precisely the probability that any one nucleus will decay within a given time interval. Consider a sample containing a large number of identical radioactive nuclei. (For example, an ordinary macroscopic sample with mass 10–100 g will contain ~ 1023 nuclei.) Let the probability per unit time that any one nucleus will decay be r. This probability depends only on the properties of the nucleus. It does not depend on the number of nuclei present in the sample.2 Thus, r is a constant for a given radioactive nucleus for all time. The quantity r is called the decay constant. The decay constant can be precisely calculated using quantum mechanics. In the following, we assume the decay constant has a definite (usually different) value for each radioactive nucleus. Let the number of undecayed nuclei at time ′t be ′N . Let Δ ′N nuclei decay in a short time Δ ′t . Then from the definition of r,

r =

Δ ′N

′N

Δ ′t=

Δ ′N

Δ ′t

′N. (10.1)

Hence,

Δ ′NΔ ′t

= r ′N . (10.2)

Since ′N is decreasing, Δ ′N is negative. So we can write

−Δ ′NΔ ′t

= r ′N , (10.3)

or,

Δ ′NΔ ′t

= −r ′N , (10.4)

Taking the limit Δ ′t → 0 , we have

d ′Nd ′t

= −r ′N . (10.5)

To solve this differential equation, we separate the variables and integrate:

d ′N′N∫ = −r d ′t .∫

The limits on the integrals are the initial and final values. At ′t = 0, ′N = N0.At ′t = t, ′N = N . Hence,

d ′N′NN0

N

∫ = −r d ′t0

t

∫ .

This gives ln N − ln N0 = −rt,

2 This is true for any spontaneous transition process, e.g., atomic transitions involving photon emission, according to quantum field theory.

3

i.e., N = N0e

−rt . (Radioactive decay law) (10.6) [Show graph of radioactive decay.] The activity of a sample is the rate at which the radioactive nuclei decay, i.e.,

Activity A =

dNdt

= rN0e−rt . (10.7)

We can write Eq. (10.7) as A = A0e

−rt , (10.8) where A0 = rN0 is the initial activity. Note that the activity also follows the exponential decay law. Activity is measured in becquerels (Bq) or curies (Ci). 1 Bq = 1 decay/s. 1 Ci = 3.7× 1010 Bq. Half-Life The half-life of a radioactive nucleus is the time period taken for half the nuclei originally present to decay (or for the activity to drop to one-half of its original value). Thus,

N =

N0

2= N0e

−rt1/2 . (10.9)

Solving for the half-life gives

t1/ 2 =

ln2r

=0.693

r. (10.10)

The known half-lives of certain radionuclides can be used to estimate the ages of objects. This is called radioactive dating.3 Carbon-14 ( 6

14C ), which has a half-life of 5730 yrs, is usually used to estimate the ages of dead organisms (or objects made from such organisms). This is useful for ages up to about 50,000 years. Uranium-238 and other long-lived radionuclides can be used for estimating geological ages. Example: Problem 17.11 from Taylor et al. textbook, 2nd edition Solution:

N = N0e−rt = N0e

− ln 2t1/2

⎛⎝⎜

⎞⎠⎟ t= N0e

−(ln 2)(t / t1/2 )

= N0 eln 2( )(− t / t1/2 )= N0 2− t / t1/2 .

Thus, N = N0 2− t / t1/2 . (10.11)

3 See Physics Today, Oct. 2001, p.32; April 2002, p.13.

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Example: The half-life of 235U is 7.04× 108 y. A sample of rock that solidified with the Earth 4.55× 109 years ago, contains N atoms of 235U. How many 235U atoms did the same rock have at the time of solidification? Solution:

N = N0 2− t / t1/2 .

This gives

N0 = N 2t / t1/2( ) = N 2(4.55×109 y/7.04×108 y)⎡⎣

⎤⎦

= 26.46 N = 88.2N .

Beta Decay We mentioned previously that radioactivity is due to unstable nuclei. More specifically, a given nucleus will be unstable if there is a mechanism for it to increase its binding energy and therefore decrease its rest energy. Radioactive decay represents such a mechanism. In general, the excess energy in unstable nuclei is due to three conditions: (1) Neutron number/proton number ratio is not optimal; (2) Neutron number/proton number ratio is fine, but the nucleus is so large that electrostatic

repulsion between protons makes it energetically favorable for the nucleus to break up; (3) The nucleus is in an excited state, i.e., in a higher energy state than its ground state. Condition 1 results in beta decay, condition 2 results in alpha decay, and condition 3 produces gamma decay. We now describe the beta decay process. In beta decay, the ratio of neutron number to proton number changes, but the total number of nucleons remains the same. There are 3 types of beta decay: 4 Z

A X → Z +1A Y + e− + νe. (β

− decay) (10.12)

ZA X → Z −1

A Y + e+ + νe. (β+ decay) (10.13)

e− + Z

AX → Z −1A Y + νe. (electron capture) (10.14)

In β− decay, νe symbolizes an antineutrino. (The subscript "e" indicates that the antineutrino is

associated with the electron family of particles.) In β + decay, νe symbolizes a neutrino. The neutrino and antineutrino are antiparticles of each other. In electron capture (EC), an inner atomic electron is captured by the nucleus, which then undergoes beta decay. Examples:

614C → 7

14N + e− + νe (β− ); 1122Na → 10

22Ne + e+ + νe (β + ); e− + 2757Co → 26

57Fe + νe (EC).

4 Picture, Table 9-3, p. 251, Krane

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In β– decay, a neutron is converted to a proton and the N/Z ratio is lowered. In β+ decay, a proton is converted to a neutron and N/Z is increased. In electron capture, a proton is also converted into a neutron, thereby raising N/Z. By conservation of energy, the spontaneous beta decay process is only possible if the rest energy of the initial (parent) nucleus is greater than the sum of the rest energies of the decay products. Consider β– decay, with parent and offspring in their ground states. Energy conservation requires that

m

ZA X,nuc

c2 = mZ+1

A Y,nucc2 + mec

2 + mνec2 + K , (10.15)

where K is the kinetic energy of the decay products.5 In fact, historically, it was the firm belief in energy conservation that led to the discovery of the neutrino. Neutrinos are neutral and do not interact electromagnetically. They also do not interact via the strong interaction. Therefore, neutrinos are extremely difficult to detect – they can pass through light years of solid material without interacting! The neutrino was not detected in early beta decay experiments, but it was found that the kinetic energy of the electrons emitted did not have a fixed value ≈ K as would be expected in a reaction in which only the offspring nucleus and the much lighter electrons are produced. In fact, the KE of the electrons had a distribution (see figure). Also, momentum conservation would be violated if there were only the offspring nucleus and the electron after the decay. Pauli solved this problem in 1931 by proposing the existence of the neutrino, which was found twenty-five years later, in 1956. The distribution of K for the electron is due to the sharing of the total KE by the electron and the neutrino in different ratios. Modern experiments have shown that the neutrino has a rest mass, with an upper limit of approximately 0.1 eV/c2. [Sloan Digital Sky Survey, 2003; Super Kamiokande Collaboration, 1998.] A tritium (3H) beta decay experiment yielded the best current upper limit of 2.3 eV/c2 (Mainz group, 2001, 2005). These values are much smaller than the electron's rest mass, which is the next smallest rest mass in the beta-decay process. For simplicity, we shall therefore take the rest mass of the neutrino to be zero. With the neutrino having approximately zero rest mass, we can rewrite Eq. (10.15):

m

ZA X,nuc

c2 = mZ+1

A Y,nucc2 + mec

2 + K . (10.16)

If we add the rest energy of Z electrons to both sides (to get atomic masses) and solve for K, we get

K = (m

ZA X,atom

− mZ+1

A Y,atom)c2. (β− decay) (10.17)

The first term in parentheses on the right hand side is the rest mass of the parent atom; the second term is the rest mass of the offspring atom. Example: Derive an expression for K in β+ decay. Solution:

5 K is usually called Q.

No.

of e

lect

rons

energy K

6

For β+ decay,

m

ZA X,nuc

c2 = mZ−1

A Y,nucc2 + mec

2 + K .

Note that the rest mass of the positron is equal to that of the electron. If we add the rest energy of Z electrons to both sides and solve for K, we get

K = (m

ZA X,atom

− mZ−1

A Y,atom− 2me )c2. (β + decay) (10.18)

For electron capture, one finds that

K = (m

ZA X,atom

− mZ−1

A Y,atom)c2. (electron capture) (10.19)

[The student should prove this.] We note that a free neutron should be unstable since

mnc

2 > mpc2 + mec

2. In fact, the free neutron is unstable and decays with a half-life of about 15 minutes according to the reaction: n→ p + e− + νe . Note that the decay of a free proton to form a neutron, positron and neutrino is energetically impossible, since

mpc

2 < mnc2.

Though a free neutron is unstable, a neutron within a nucleus may or may not be stable. Similarly, though a free proton is stable, a proton in the nucleus can be unstable (resulting in β+ decay). Note that β decay does not change the value of A. The energetics of β decay can therefore be understood by considering the rest energies of a group of isobars (same A). In a group of isobars, the stable nucleus (or nuclei) has the lowest rest energy. If one considers a group of isobars with neighboring Z values, if N/Z is too large for stability, then β– decay will occur (neutron changes to a proton). If N/Z is too small for stability, then β+ decay or electron capture will occur (proton is converted to a neutron). One can actually predict which member of a group of isobars will be stable by using the expression for the binding energy Eb in the semi-empirical binding energy formula and minimizing the rest energy as a function of Z. [Rest energy

m

ZA X,atom

c2 = (Nmn + Zm11 H

)c2 − Eb = f(A,Z); df/dZ = 0 gives the value of Z for which the rest

energy is minimum, i.e., for a stable nucleus.] (cf. Problem in the homework for previous chapter.) Alpha Decay In alpha decay, a heavy unstable nucleus decays into a less massive offspring nucleus and an alpha particle: Z

A X → Z −2A−4 Y + 2

4 He. (10.20) Using the Greek-letter symbol for the alpha particle, Eq. (10.20) can be rewritten as

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ZA X → Z −2

A−4 Y +α. (10.21) An example: 92

238 U → 90234 Th + 2

4 He. Alpha decay is a spontaneous process and energy is liberated:

K = (m

ZA X,nuc

− mZ−2A−4 Y,nuc

− m24 He,nuc

)c2. (10.22)

If we add the masses of the appropriate number of electrons, we find that

K = (m

ZA X,atom

− mZ−2A−4 Y,atom

− m24 He,atom

)c2. (10.23)

If we choose a reference frame in which parent X is at rest before the decay, then it is clear that Y 0,p pα + = i.e., pα = pY. (10.24) Also, Kα + KY = K . (10.25) Alpha decay energies are typically a few MeV, which means that the kinetic energies of the α particle and the offspring nucleus are much smaller than their rest energies. Thus, the motion can be treated non-relativistically. Hence,

pα2

2mα

+pY

2

2mY

= K .

Since pα = pY , we have

pα2

2mα

1+mα

mY

⎛

⎝⎜⎞

⎠⎟= K .

⇒ Kα =

pα2

2mα

=K

1+ (mα mY ).

But

mα

mY

≈4

A− 4,where A is the mass number of the parent. Thus, we find

Kα ≈

A− 4A

⎛⎝⎜

⎞⎠⎟

K . (10.26)

Perhaps the most striking characteristic of alpha decay is that the half-lives of alpha-emitters vary over 24 orders of magnitude while the α-particle energies vary from about 4 MeV to 9

8

MeV. This rather amazing result is due to the fact that α-emission is due to tunneling or barrier penetration. [Show picture of tunneling barrier for alpha particle.] As you recall, the wave function for a tunneling particle is exponentially decaying:

ψ ~ e−ax , where x is the penetration distance and a is given by2 ( )m U E

a α −=

.

Thus, the tunneling probability P ∝ e−2ax . Now, the barrier height is just the Coulomb potential energy when the alpha particle just escapes nucleus. If R is the nuclear radius, then6

Umax = k 2(Z − 2)e2

R

We take E = Kα, since the total energy of the α particle is Kα when it is far away from the nucleus after escape. Note that the energy of the α particle is also Kα inside the nucleus due to energy conservation. Since the height of the barrier varies with tunneling distance, the tunneling probability has to be found by integration. However, we can do a rough calculation by using a barrier of fixed height with half the true maximum barrier height. With this approximation, P ∼ e−2a( ′R −R) , where

max

122

,m U K

aα α⎡ ⎤−⎢ ⎥⎣ ⎦=

and ′R is the distance from the center of the parent nucleus to where U = Kα. To estimate the decay constant, we recall that r = probability per unit time. Now when the α particle is trapped in the nucleus, the period of its wave function, which corresponds to the period of its back and forth

motion inside the nucleus, is given by T = v

2R, where v is the speed of the α particle inside the

nucleus. Using typical values of v and R, one finds T ~ 10-22 s. Thus,

r ~ P/T =

v2R

e−2a( ′R −R) .

Since the parameters a and ′R − R are in the exponent, r is extremely sensitive to changes in these parameters (note that a depends on Kα ) and this leads to the observed immense range of

values for r and t1/ 2. Gamma Decay [Show Fig. 9.15, Krane] 6 α-particle radius should be added.

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After α or β decay, the offspring nucleus may be left in an excited state (a nucleon is in an excited state). The nucleus can reach its ground state by emitting one or more gamma photons. Gamma energies are usually in the range 100 keV to 10 MeV. Typical half-lives are 10-12 s to 10-9 s. Occasionally, the properties of the two quantum states involved in the transition (e.g., large differences in angular momentum) lead to a very long half-life (e.g., Ba-137m that we saw in the lab). Such states are called isomeric states. Example 9.10, Krane

712 N beta-decays to an excited state of 6

12C , which subsequently decays to the ground state with the emission of a 4.43-MeV gamma ray. What is the maximum kinetic energy of the emitted beta particle? The decay type is β + decay. To determine the K value for this decay, we first need to find the mass of the product nucleus 12C in its excited state. In the ground state, 12C has a mass of

12.000000 u, so its mass in the excited state is 12.000000 u + 4.43 MeV

931.5 MeV/u= 12.004756 u.

Using Eq. (10.18), the K value is therefore K = (12.018613 u – 12.004756 u – 2 × 0.000549 u)× (931.5 MeV/u) = 11.89 MeV Notice that we could have just as easily found the K value by first finding the K value for decay to the ground state, 16.32 MeV, and then subtracting the excitation energy of 4.43 MeV, since the decay to the excited state has that much less available energy. Neglecting the small correction for the recoil kinetic energy of the 12C nucleus, the maximum electron kinetic energy is 11.89 MeV. Natural Radioactivity All elements except for the least massive (H and He) were produced by nuclear reactions taking place in the interior of stars. The radioactive ones with relatively short half-lives have long since decayed to stable elements. However, there are a few elements with half-lives of the same order of magnitude as the age of the Earth, and so these elements are still present in nature and result in natural radioactivity. The three natural radioactive decay series are for 90

232 Th, 92235 U, and 92

238 U. These elements decay by α emission producing radioactive offspring, which in turn decay by α and/or β emission.7 Thus, one has a sequence or series of decays until a stable element is reached. Note that alpha particles are emitted by the heavy nuclei rather than single protons or single neutrons because of the relatively high binding energy per nucleon of the 2

4 He nucleus, or in other words, its relatively small rest mass. (Recall that for spontaneous decay to be possible, the total rest mass of offspring plus emitted particles must be less than the rest mass of the parent.) Example (From Krane): Compute the total energy released for the 238U → 206Pb chain. Solution: This decay chain includes 8 α and 6 β– decays. Thus, for the entire chain: 7 Gamma decay also occurs when nuclei are formed in excited state.

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K∑ = [m92238 U,atom

− m82206 Pb,atom

− 8m24 H,atom

]c2 .

(The masses of the β– particles have been combined with the nuclear masses so that we can use atomic masses.) K∑ = [238.050786 u – 205.974455 u – 8× 4.002603 u]× 931.5 MeV/u = 51.7 MeV The energy released in natural radioactive decay is partially responsible for the internal high temperatures of the planets.