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Physics Lesson 20.3.5 - Half-Life

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Welcome to our Physics lesson on Half-Life, this is the fifth lesson of our suite of physics lessons covering the topic of Radioactivity and Half-Life, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Half-Life

So far, we have explained various radioactive decays but without giving any information about the amount and rate of their occurrence. For different reasons, at a certain moment a nucleus may be in a state where a decay is possible. The conditions in which a decay occurs are not determined, so the radioactive decay is a pure causal event. Therefore, we rely on theory of probability to explain the rate of radioactive decay processes.

Let's suppose we have a number N unstable atomic nuclei of type A, which can transform into other nuclei according the chain

A → B + C

This does not mean that all N nuclei of the type A immediately transform into products B and C, rather, here we are discussing only about the probability (possibility) for this process to occur. This probability obviously depends on the time interval we are interested to and a number of other internal factors, which cannot be controlled and manipulated. Since every event takes place inside the nuclei A, it is obvious that the decay process does not depend on external factors such as state of matter, temperature, pressure, chemical structure, etc. The only factor that may affect this process is any possible interaction of nucleons with particles coming from outside the nucleus. In this way, entire process will depend only on the nuclei properties and the instant in which the decay takes place is purely casual.

Let's denote by N0 the number of nuclei of A-type present in a radioactive sample at a given instant, which we consider as the initial stage of process (t = 0). Obviously, after some time t, a number of radioactive decays have taken place, so the number of nuclei N remained in the original sample is N < N0. The question that arises here is: what is the relationship between the number of decayed nuclei and the time interval involved? In other words, our duty is to find the form of the function

N(t) = f(N0,t)

that gives the relationship between the number of nuclei still not decayed and the time elapsed given the original number of particles present in the sample.

This relationship relies on the supposition that every individual decay process is purely casual and independent from the other decays occurring in the rest of radioactive nuclei. Obviously, there is a linear relationship between the number ΔN of decayed particles and the interval Δt of this event's occurrence, which represents the time elapsed since the beginning of process. It is obvious that the number of decayed nuclei is ΔN = N(t) - N0 where N(t) is the number of undecayed nuclei at the time instant t. Hence, we can write

∆N ∝ N0 ∙ ∆t

From mathematics, it is known that we must multiply the right side of a proportion by a constant in order to turn it into an equation, Here, this constant if given by (). Thus, we obtain

∆N = -λ ∙ N0 ∙ ∆t

The negative sign before λ is used to make the right side of the above equation negative since the left side is negative as well (ΔN = N(t) - N0 and N(t) < N0).

The constant λ known as the constant of radioactive decay is an intrinsic property of material itself (unit: s-1) and it includes all factors affecting the decay of a given radioactive nucleus.

The equation for the number N(t) of undecayed nuclei at a given instant t is obtained by rearranging the above equation for N. Thus, we have

∆N = -λ ∙ N0 ∙ ∆t
∆N/N0 = -λ ∙ ∆t

Integrating both sides of the above equation in order to include all nuclei involved, we obtain after reducing the intervals to infinitely small ones to increase the accuracy of result (the symbol "Δ" is replaced by "d" in such cases, as explained in Electromagnetism in Section 16):

∆N/N0 = -λ ∙ ∆t
dN/N0 = -λ ∙ dt
dN/N0 = -λ ∙ dt
dN/N0 = -λ ∙ dt
ln⁡N/N0 = -λ ∙ t

Hence, we obtain for the number of undecayed nuclei in a radioactive sample as a function of time:

N(t) = N0 e-λ ∙ t

Example 4

The radioactive decay constant for Bismuth-214 is 4330 s-1. How many Bismuth-214 nuclei are still present in a sample of Bismuth-214 after 500 microseconds if initially there were 100,000 nuclei in the sample?

Solution 4

Clues:

t = 500 μs = 5 × 10-4 s
λ = 4330 s-1 = 4.33 × 103 s-1
N0 = 100 000 = 105
N(t) = ?

Using the equation

N(t) = N0 e-λ ∙ t

we obtain for the number of undecayed nuclei in the sample after substituting the known values:

N(500 μs) = 105 ∙ e(-4.33 × 103 s-1 ) ∙ (5 × 10-4 s)
= 10000 ∙ e-2.165
= 105 ∙ 0.11475
= 11475 nuclei

An important quantity used in equations of radioactive decay is half-life period, T1/2 (often referred to as simply "half-life"). It represents the time needed for the decay of half of the original radioactive nuclei in a sample. Like radioactive decay constant λ, half-life period T1/2 is a property that depends on the type of radioactive material too. Half-life is very useful as it makes the calculations much easier. The following table gives the value of half-life period for some radioactive isotopes.

Radioactive isotopes half-life period table
IsotopeHalf-life period (T1/2)
Uranium-2384.5 billion years
Plutonium-23912.110 years
Carbon-145730 years
Radium-2261600 years
Radon-22051.5 seconds
Litium-80.838 seconds
Bismuth-21420 minutes
Hydrogen-3 (tritium)12.32 years

We can find the relationship between λ and T1/2 by substituting t = T1/2 and N = N0/2 in the equation of radioactive decay. In this way, we obtain

N(t) = N0 e-λ ∙ t
N0/2 = N0 e-λ ∙ T1/2

Simplifying N0 from both sides, we obtain

e-λ ∙ T1/2 = 1/2
e-λ ∙ T1/2 = 2-1
eλ ∙ T1/2 = 2
ln⁡(eλ ∙ T1/2 ) = ln⁡2
λ ∙ T1/2 = ln⁡2

Since ln 2 ≈ 0.693, we obtain

λ = 0.693/T1/2

Another important quantity commonly used in quantitative approach of situations involving radioactive decay is the decay rate R(t) = ΔN/Δt. It shows how fast a radioactive decay occurs and varies by time as the speed of radioactive decay decreases by time. Thus, since

∆N = -λ ∙ N0 ∙ ∆t

we obtain for the initial rate of radioactive decay:

R0 (t) = ∆N/∆t = -λ ∙ N0

and for the rate R(t) of radioactive decay at a given time t:

R(t) = R0 e-λ ∙ t

The unit of radioactive decay rate (if measured in number of decays/second) is known as Becquerel, Bq.

The relationship between undecayed nuclei and the time elapsed is given by an inverse function (an inverse function has the form y(t) = C/x where C is a constant). The graph of such functions is a hyperbola, as shown in the figure below, where the data regarding radioactive decay of Magnesium-28 are shown.

Physics Tutorials: This image provides visual information for the physics tutorial Radioactivity and Half-Life

Example 5

Based on the graph shown above for radioactive action of Magnesium-28, calculate:

  1. Half-life period of Mg-28
  2. The number of undecayed nuclei after four half-life cycles if initially there were 1200 nuclei in the sample
  3. How long does it take to be still 10 nuclei undecayed in the sample?
  4. What is the rate of radioactive decay for Mg-28 after 60 days from the beginning of process?

Solution 5

Clues:

N0 = 1200
T1/2 = ?
N(t = 4·T1/2) = ?
t(N = 10) = ?
R(t) = ?

  1. From the graph, we see that the radioactive nuclei in the vertical axis are not given as a number but as a part of whole. We see that half of the initial sample (which correspond to the value 0.5 in the vertical axis) correspond to 20 days in the (horizontal) time axis. Thus, T1/2 = 20 days.
  2. Four half-life cycles correspond to t = 4 · T1/2 = 4 · 20 days = 80 days. Since we cannot find the number of undecayed nuclei at this instant as the graph units are not divided so minutely, we find the number of undecayed nuclei at t = 80 days through calculations. Thus, since the number of undecayed nuclei halves by every half-life cycle, we have for the remaining nuclei after 4 half-cycles:
    N(80 days) = N0/24
    = 1200/16
    = 75 nuclei
  3. We must find the constant of radioactive decay λ first, and then calculate the number of undecayed nuclei at the given time. We have
    λ = ln2/T1/2
    = 0.693/20 days
    = 0.03465 day-1
    Hence, the time in which N(t) = 10, is calculated by
    ∆N = -λ ∙ N0 ∙ ∆t
    ∆t = -∆N/λ ∙ N0
    = -N(t) - N0/λ ∙ N0
    = -10 - 1200/(0.03465 day-1 ) ∙ 1200
    = --1190/41.58 days
    = 28.62 days
  4. The initial rate R(t) of radioactive decay for Mg-28 (we can skip the negative sign, as it only indicates that the number of nucleons dercreases), is
    R0 (t) = λ ∙ N0
    = (0.03465 day-1 ) ∙ 1200
    = 41.58 nuclei/day
    Hence, the rate of radioactive decay at 60 days after the beginning of process is
    R(t) = R0 e-λ ∙ t
    R(60) = 41.58 ∙ e(-0.03465) ∙ (60)
    = 41.58 ∙ e-2.079
    = 41.58 ∙ 0.125
    ≈ 5.2 nuclei/day

You have reached the end of Physics lesson 20.3.5 Half-Life. There are 5 lessons in this physics tutorial covering Radioactivity and Half-Life, you can access all the lessons from this tutorial below.

More Radioactivity and Half-Life Lessons and Learning Resources

Nuclear Physics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
20.3Radioactivity and Half-Life
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
20.3.1Natural Radioactivity. Becquerel's Experiment
20.3.2Alpha Decay
20.3.3Beta Decay
20.3.4Gamma Decay
20.3.5Half-Life

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