The Lifetimes of 238U Nuclei, the Escaping
Attempts of Their Alpha Particles and the Geiger-Nuttall Law
Pavle I. Premović
Laboratory for Geochemistry,
Cosmochemistry&Astrochemistry,
University of Niš, pavleipremovic@yahoo.com, Niš, Serbia
Abstract. Here, we
consider the large populations of 238U nuclei and their emitted
alpha particles. The main points are (a) a
time interval for which alpha particle of 238U nucleus makes
one attempt to escape is approximately 10-21 sec; (b) the number of the mean escape
attempts is ~ 2×1038; (c) contribution of the tunneling time of 238U nucleus to the meantime and the
lifetime of each nucleus is negligible; (d) the sum of the lifetimes of 238U nuclei and the sum of the number of escape attempts
of their alpha particles are constant. Consequently, the ratio of these two
sums is also constant, about 10-21 sec; and, (e) a new
mathematical form of the Geiger-Nuttall law is derived using a simple theoretical
approach.
Keywords: Radioactive decay, mean lifetime, 238U, α-particle, escape, Geiger-Nuttall law.
Introduction. Radioactive decay is a purely statistical (random)
process. If the number of radioactive nuclei is large enough then the total
number of radioactive nuclei Nt, that have not yet been transformed
at a time t, can be obtained from the following equation
Nt = N0e-λt … (1)
where
N0 is the number of radioactive nuclei at a time t = 0 and λ is
the decay constant. This equation is the final mathematical
representation of the radioactive
decay law, and
it applies to all decays. The decay pattern of a large number of
radioactive nuclei is easy to predict using eqn. (1). For practical purposes,
we will be using Avogadro's
number N (= 6.022×1023) as a reasonable
approximation for a large number of radioactive nuclei.
In general,
there are two ways to measure the time for which a radioactive nucleus is
stable: half-life t1/2 and the mean lifetime τ. By the definition of the term
half-life, when Nt = 1/2N0
then t = t½. By the
definition of the term half-life, when Nt = 1/2N0 then t = t½. A
simple algebra of eqn. (1) shows that t1/2 = (ln2)/λ.
The quantity τ is the reciprocal of
the decay constant λ or τ = 1/λ; τ is usually estimated from the measured
t1/2 using the following relation τ = 1.44t1/2. The author has some reservations
about these equations because they are derived by combining the discrete nature
of the number of radioactive nuclei and the continuous exponential
approximation. He will discuss this question in one of his
next communications. Of course, λ, t1/2 and τ have only meaning for a large population of radioactive nuclei (of
the same type, of course).
We will
concentrate on the
lifetimes of a
large population of the decayed 238U nuclei, as well as on the
escaping times of their emitted α-particles.
Results and Discussion. If there is an initial population of N radioactive, the sum of their all-individual lifetimes divided by N is the mean lifetime τ. Mathematically
speaking, τ is the arithmetic mean of lifetimes of all N individual nuclei
τ = (τ1 + τ2 + ... + τN -1 + τN)/N
where i and τi denote, respectively,
individual radioactive nuclei and their lifetimes
Rearranging this equation gives
Nτ = (τ1 + τ2 + ... + τN -1 + τN) …
(2).
After a bit of algebra and estimation,
we found that about 62.5 % of τi is smaller than 1.44t1/2 or τ. Therefore, for a large population of radioactive nuclei about 37.5 % of τi
exceeds τ.
Consider alpha (α-) decay of (parent) 238U
(initially at rest) with τ of about 2×1017 sec to (daughter) 234Th.
It is assumed that the parent 238U nucleus before decay consists of
the daughter 234Th and α-particle. This particle is in a free
(non-bound) state, otherwise, the decay could not occur. The emitted α-particle whose mass mα is ca.
6.65×10-27
kg has a speed vα = 1.42×107 m sec-1. This speed is determined from its non-relativistic
kinetic energy Eα = 1/2 mαvα2, ca. 4.2 MeV. In fact, vα is its
speed when the α-particle escapes
from the 238U
nucleus.
According to classical mechanics,
α-particle with the Eα of ca. 4.2 MeV could never be able to overcome the
Coulomb potential barrier of 238U, VC, which is probably about
20 MeV {1}, and to escape the nucleus. Quantum
mechanics provides an explanation based on the concept of tunneling where this particle can be found in a classically
forbidden (outside) region of the nucleus.
The size (diameter) d
of the nucleus of the heaviest atoms, such as uranium (U), is about 15 × 10−15 m or 15 fm. An α-particle must
make many attempts back and forth the across 238U nucleus before it can escape; in fact, this particle oscillates along the diameter of the nucleus.
A time interval for
which the α-particle of 238U nucleus
makes one “try” to escape is approximately given by d/vα = 10-21 sec. In other
words, α-particle
effectively attempts to tunnel through the Coulomb barrier vα/d = 1021 times per sec. Denote now with n and ni,
respectively, the mean escape attempts and the individual escape attempts of
the α-particle before
leaving 238U nucleus. Let also t(e)
and ti(e) stand for the mean escaping time and the escaping
time, respectively. Simple reasoning indicates that t(e) ~ τ and ti(e) ~ τi because the tunneling
time T could contribute to these escaping times.
A question that goes along with
tunneling is: how long does it take for α-particle to tunnel the Coulomb barrier? The tunneling time problem is one of the long-standing and controversial
problems in quantum mechanics. There are numerous attempts to solve it but none
of them gives a flawless answer to this question {2}.
The width of the barrier w a through which
an α-particle must tunnel can be roughly calculated using the following equation: W = (VC/Eα)d – d {1}. Inserting in this equation the
values for VC, Eα and d we obtain W ~ 57 fm
which is probably
much smaller about 30 fm {1}. Of course, the speed of this particle during the tunneling process
cannot exceed the speed of light c, therefore, the lower limit for the
tunneling time T = W/c ~ 1 ×10-22 sec; its upper limit W/vα ~ 1.5×10-21 sec. Therefore, the contribution of the
tunneling time to τ is negligible. Consequently,
t(e) ~ τ = nd/vα then the number of the mean escape attempts
n = τvα/d … (3).
We know that τ is about 2 × 1017
sec and vα = 1.42×107
m sec-1 for 238U so n ~ 2 × 1038 is the constant characteristic for this
uranium isotope. In general, n of eqn. (3) is the constant characteristic for
any isotope α-emitter of the heaviest atom (including any uranium
isotope α-emitter, of course). Since T contributes to each then
ti(e)
= nid/vα + T = 10-21ni
+ T. Of course, the maximum contribution of T is
about 1.5 × 10-21 sec and the maximum
τi
= 10-21ni
+ 1.5 × 10-21 sec. For ni
≥ 100, we can practically take that τi = 10-21ni. By
inserting this τi
into eqn. (2) we obtain
Nτ = 10-21(n1 + n2 + ...+ nN -1 + nN) … (4).
Plugging the values for N and τ into the equations (2) and (4) we get
τ1 + τ2 + ... + τN -1 + τN = 1.2×1041 sec
and
(n1 + n2 + ...+ nN -1 + nN) = 1.2×1062.
Thus, the sums of lifetimes of a
large population of 238U nuclei and the number of escape attempts of their α-particles
are fixed. A
similar conclusion is true for other appropriate radioactive isotopes.
If the number of
decayed 238U nuclei
is k times greater or smaller (but still large) then the above sums would be k times higher or
lower. However, the ratio of eqns. (2) and (4) is constant and equals: τ/n = τi /ni
= 10-21 sec even if k → ∞, of course. This is of some importance.
Any single 238U nucleus which we observe is not all alone in the
Universe and it belongs to an infinitely large total number (k → ∞) of its 238U nuclei. Therefore, the ratio τ/n of 238U of about 10-21
sec is one
of the “magic” units of time in the Universe.
Geiger-Nuttall law {3, 4} is a semi-empirical law that expresses the half-life of a heavy α-emitter
in terms of the kinetic energy of its released α-particle. In its modern
natural logarithm (LN) form this law is: lnt1/2
= a + b/√Qα, lnt1/2 = a + b/√Qα where Qα is the α-decay energy a and b are the
constants that can be determined by fitting to experimental data for each isotopic series. The kinetic energy, Eα, of the emitted α-particle, is rather slightly less than Qα.. Therefore,
the above equation can be rewritten as lnt
1/2
= a + b/√E
α. The quantum tunneling enables one to obtain the Geiger-Nuttall law, including
coefficients,
via direct calculation.
We know that n = τvα/d and vα = √(2Eα/mα). Combining these two equations
and after a bit of algebra, we get τ = nd × √(mα/2)/√Eα. Substituting in this equation 1.44t1/2
instead τ, u instead
(nd/1.44) and w instead √(mα/2) we obtain
t1/2 = u × w × √(1/Eα)
… (5)
where u and w are the constants. Its
LN form is
lnt1/2 = lnu + lnw + ln(1/√Eα) … (6).
Evidently, u and w are also the
constants. This equation is a linear function of lnt1/2 vs. ln(1/√Eα). To illustrate this,
we will use some experimental data for uranium isotopes 228U – 238U
given in Table 1. The plot lnt1/2 vs. ln(1/√Eα)
for these isotopes gives a straight line, Fig. 1. Of course, it is before necessary to
convert seconds (sec) to years (yr) and joules (J) into MeV of this equation.
|
Uranium isotope
|
Eα [MeV]
|
t1/2
|
|
238
|
4.198
|
4.468× 109
yr
|
|
236
|
4.494
|
2.342 × 107
yr
|
|
235
|
4.395
|
7.037 × 108
yr
|
|
234
|
4.775
|
2.455 × 105
yr
|
|
232
|
5.320
|
69.8 yr
|
|
230
|
5.888
|
20.8 day
|
|
228
|
6.680
|
9.1 min
|
|
226
|
7.570
|
0.35 sec
|
Table 1. Some
experimental data of uranium isotopes 226U – 238U {5-8}.
It appears that the above new form
of the semi-empirical Geiger-Nuttall law represents until now the simplest
theoretical way to derive this law. It is also possibly applicable to other α-emitters of the heaviest atom isotopes.
We know that mα = 6.65 × 10-27, 1 yr = 3.15 × 107 sec and that 1 J =
6.24 × 1012 MeV. After a simple algebra the eqn. (5) can be written
as follows
t1/2 = nd/1.44 × (3.15 × 107) × √[(6.25
×1012) × mα/2](1/√Eα)
… (7).
In the case of 238U, for example,
plugging into this equation the values for n (= 2 × 10-38), d (= 15
× 10-15 m), mα (= 6.65 × 10-27kg) and Eα
(= 4.198 MeV) we calculate t1/2
~ 4.74 × 109 years. The experimental value for t1/2 is ~
4.47 × 109 years, Table 1. Our theoretical approach is
therefore not “too” bad. A detailed
consideration of eqn. (7) will be the subject of the next author’s study.
For the sake of clarity, let us
consider briefly u and w of eqn. (5). According to eqn. (7), the constant u is a product of two variable terms: n
and d and the constant w
consists of three fixed terms: mα and two conversion factors J into MeV (6.25 ×1012) and sec into yr (3.15 × 107). Therefore in this equation u (=
nd/1.44) is the constant which is characteristic for each of the isotope
α-emitters of the heaviest atoms and
w [= 1/(3.15 × 107] × √(6.25
×1012 × mα/2)] is the general constant, the same for all of these
α-emitters. In contrast, the a and b constants in the original mathematical
form of the Geiger-Nuttall are characteristic of each isotopic series.
Figure 1. A comparison of the eqn. (6) with
experimental data for 226U - 238U {5-8}.
The linear fit
to the values is shown by the solid line.
There are some intriguing details
related to α-particles inside 238U nucleus. Making the mean escape
attempts n (~ 2×1038) α-particle has to travel a distance:
2×1038×d ~ 2 × 1038 ×15× × 10-15 m ~ 3 × 1021
km. The light-year is about ≈ 1013 km. As a result, on
average, the α-particle has to travel a distance of about 300 million light-years
before escaping the 238U nucleus. Of
note, our galaxy Milky Way is just about 100
thousand light-years across.
It is suggested that a part of the Earth’s 238U
atoms was formed in
a supernova explosion about 6 billion years ago. Let us suppose that at that
time there were two “neighboring” nuclei ns and ne of 238U
nuclei and the nucleus ns disintegrated at that time emitting α-particle. Allow right
now that ne decays and emits α-particle. This nucleus has made a
“trip” of about 270 billion light-years after the disintegration of ns.
Of note, the
diameter of the observable universe is about 93 billion light-years. A question is why ns needs
so an incredibly long “trip” and corresponding incredible time to escape
238U
nucleus? What is the cause of such a 6 billion years delay? Moreover, if the
decay happens right now why it did not occur before since there was a very long time to occur. These are some of the intriguing questions
for which modern physics has not yet provided reasonable answers.
Conclusion. The main points are:
1. The time interval for which the α-particle of the 238U
nucleus makes one attempt to escape is
approximately 10-21 sec.
The number of the mean escape attempts n ~ 2×1038.
2.
The contribution of the tunneling time T to τ
and τi of 238U is negligible.
3. The sum of the lifetimes of the 238U nuclei and the sum
of the
number of escape attempts of their α-particles are constant. Consequently, the
ratio of these two sums is also constant and is about 10-21 sec.
4. A new expression for the Geiger-Nuttall law
is reproduced in a very simple way.
References
[2] N. L. Chuprikov, Tunneling time problem: At the intersection
of quantum mechanics, classical probability theory and special relativity. arXiv:1303.6181v4 [quant-ph] 24 Mar 2014.
[3] H. Geiger and
J.M. Nuttall, The ranges of the α particles from various
radioactive substances and a relation between range and period of
transformation. Phil. Mag. 22, 613-621 (1911).
[6]
. N. E. Holden, The uranium half-lives: a
critical review. National nuclear data center Brookhaven National
Laboratory (1981).
[7] J. Elkenberg, Radium Isotope Systematics in
Nature: Applications in Geochronology and Hydrogeochemistry. Habilitation
Thesis, Earth Science Department, Swiss Federal Institute of Technology, Zürich
(2002).
[8] Uranium
isotopes. Wikipedia, the free encyclopedia.
Acknowledgements. A thank goes to Miloš Stefanović (Department
of Engineering Sciences & Applied Mathematics, Faculty of Technology,
University of Niš, Niš, Serbia) for his technical help.
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