Half Life

Practice

practice problem 1

Write something.

solution

practice problem 2

Ernest Rutherford suggested that the difference in the abundance of the naturally occurring isotopes of uranium was due to their different half lives. He assumed that the two most common isotopes were originally found in equal abundance on earth. The reason we see less 235U and more 238U today is because the former decays faster than the latter. Given this, he made the first radioisotopic calculation of the age of the earth in 1929.
 
Naturally Occurring Uranium Isotopes
mass number abundance (%) half life (years)
 235   0.7201  7.04 × 108
 238   99.275  4.46 × 109
 

Determine the age of the earth using Rutherford's method and contemporary data, then compare your results with the currently accepted value of 4.55 × 109 years and discuss.

solution

Solution …

Divide the half life equation by itself. Give the numerator to the lighter isotope and the denominator to the heavier isotope. Plug in the appropriate values.

N235   =  N02 −t/τ235
 		  ⇒  0.7201%  =   2 −t/7.04 × 108
 		  = 0.00725
N238  N02 −t/τ238
 		 99.275%  2 −t/4.46 × 109
 

Normally, I'd reduce this equation down until the unknown quantity (t) was on the left side of the equals sign and everything else was on the right side. I don't recommend that in this case. You'll wind up with a base 2 log function in the results. I don't own a calculator with that capability built in. However, I do own a calculator that can draw graphs and allows me to trace over the curve, testing the value at various spots.

Plot this function (or something similar looking).

equation

Set an appropriate range for the displayed window. I suspect the answer should be in the billions, so I'll set the horizontal range from 0 to 10 billion. Draw the graph and then probe it to find values around 0.00725.

0 to 10 billion x=5.85106e9 y=.00781644 x=5.95745e9 y=.00715649
Window Dimensions Left Side Right Side

Those values are pretty close. Let's zoom in more to see if we can land on a value closer to our goal of 0.00725.

5.85 to 5.95 billion x=5.94149e9 y=.00725181 x=5.94255e9 y=.0072454 value
Tighter Window Dimensions Left Side Right Side

Note how the curve looks like a straight line at this resolution. We've zoomed in quite a bit. The values to the left and right are converging towards our goal and are in agreement to within 3 significant digits. I think we've found our solution.

t = 5.94 × 109 years

This result is …

Δt  =  (5.94 −4.55) × 109 years  = 31% too large
taccepted 4.55 × 109 years

This method is only good enough to give an order of magnitude estimate of the age of the earth. The problem lies in the assumption that the two main isotopes of uranium ever had equal abundances. This is not the case.

The sun was formed from the residue of stars that went supernova. We know that it is a second generation star because of the presence of heavy elements. The only way to naturally produce nuclei heavier than iron is in a supernova. This process was first described in a 1957 paper that is usually known by the last names of the authors in alphabetical order — Burbidge, Burbidge, Fowler, and Hoyle. The abundance of all the elements heavier than hydrogen and helium can be calculated by the processes described in this paper. In the case of uranium, supernova always produce more 238U than 235U.

Rutherford devised another method for determining the age of the earth in the same year as the one described in this problem. This second method gave him a value of 3 billion years, which is 34% too small. The accepted value of 4.55 billion years quoted in this problem was determined by Claire Patterson (1922–1995) of the California Institute of Technology in 1956. It remains the definitive value to this day.

practice problem 3


[magnify
When the Apollo astronauts landed on the moon they left behind equipment to monitor such things as the moon's internal temperature, its magnetic and gravitational fields, seismic activity caused by moonquakes and meteor impacts, and the moon's extremely thin atmosphere. Known collectively as the ALSEP (an acronym for Apollo Lunar Surface Experiment Packages) these devices were powered by a small and very simple nuclear power plant called a SNAP (an acronym for Space Nuclear Auxiliary Power). A SNAP generator is basically a can of plutonium 238 dioxide surrounded by radiator fins with thermocouples in between. The difference in temperature between the decaying plutonium (600 ℃) and the the radiator fins (275 ℃) produces a voltage across the thermocouples that can be used to generate electric current. The whole contraption is about the size of an office watebasket and with no moving parts is reliable for very long periods of time.

The SNAP-27 activated by the Apollo XIV crew on 5 February 1971 used 3.8 kg of plutonium 238 dioxide and generated 73 W of power when first turned on. If 238PuO2 has a half life of 87.74 years and decays via the emission of 5.593 MeV alpha particles, determine …
  1. the initial power radiated by the plutonium fuel
  2. the initial efficiency of the generator
  3. the power of the generator on the anniversary of the Apollo XIV mission from the day it was turned on until its centennial.

solution

  1. Let's begin by determining the number of atoms in 3.8 kg of plutonium 238 dioxide. Recall that oxygen has an atomic mass of 16 and therefore that 238PuO2 has a molecular mass of 238 + 2(16) = 270.
     
    N =  (3800 g)(6.02 × 1023 atom/mol)  = 8.47 × 1024 plutonium atoms
    (270 g/mol)

    Half of this will have decayed after one half life giving an average activity of …

    A =  N/2  =  (8.47 × 1024 decay)/2  = 1.53 × 1015 Bq
    t (87.74 yr)(365.25 × 24 × 3600 s/yr)

    This is an average value, but we want the initial value. If the activity decreased linearly, the average value would be the average of the initial and final values. But this is radioactivity and radioactivity goes down exponentially not linearly. We will have to resort to calculus for this next step. The average value of a varying quantity is its definite integral divided by the interval.

    τ
    A =  1   ⌠
    ⌡    		 A dt =  1   ⌠
    ⌡    		 A02−t/τ dt =  1   A0τ  =  A0
    t τ τ 2 ln 2 2 ln 2
    0

    Rearrange to make the initial activity the subject of the equation and substitute numbers for variables.

    A0 = 2 ln 2 A = 2(ln 2)(1.53 × 1015 Bq) = 2.12 × 1015 Bq

    Multiply this by the energy per decay (in joules) to get the power (in watts).

    P0 = A0E = (2.12 × 1015 decay/s)(5.593 × 106 eV/decay)(1.602 × 10−19 J/eV) = 1900 W

  2. Efficiency is the ratio of energy output to energy input.
     
    η =  Eout  =  73 J/s  = 3.8%
    Ein 1900 J/s

    Not very efficient, is it?

  3. Assume the power output decays in the same manner as the activity, use years as the unit of time, and generate a formula for a spreadsheet or data analysis application.

    P = P02−t/τ = 73 W × 2(1971 − t[yr])/(87.74 yr)

    The results are listed in the text file snap.txt.

practice problem 4

Write something.

solution

The Physics Hypertextbook

  • No condition is permanent.