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Questions: 1 2 3 4 5 6

Radioactivity Solutions

  1. The alpha-decay of 238U releases 4.25 MeV of energy. Virtually all this energy is in the form of the kinetic energy of the alpha-particle. How fast is the alpha particle travelling?

    We are given that

    Ereleased = ½mv2 ,

    where Ereleased = 4.25 MeV. We need to know the mass of the alpha particle. According to the text, the mass is 4.00260 u. Neither of these quantities are in SI units, so we first convert

    E = 4.25 × 106 eV × 1.609 × 10-19 J/eV = 6.808 × 10-13 J,

    and

    m = 4.00260 u × 1.661 × 10-27 kg = 6.6483 × 10-27 kg .

    Hence

    v = [2E/m]½ = [2(6.808 × 10-13 J)/( 6.6483 × 10-27 kg) ]½ = 1.43 × 107 m/s.

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  2. A neutron outside the nucleus decays into a proton, an electron, and a neutrino. Note that n = 1.008665 u, p = 1.007285 u, and e = 5.48578 × 10-4 u. Assume that the neutrino is massless.

    (a) Assuming the neutrino is massless, how much energy is released?

    (b) Assuming that all this energy is converted into the kinetic energy of the electron, how fast is the electron moving?

    (a) The decay reaction is n -> p + e + n. The difference in mass between the left and right-hand sides of the reaction is converted into energy according Einstein's formula, E = (Δm)c2. The mass difference is

    Δm = 1.008665 u - 1.007285 u - 0.000549 u = 0.000831 u .

    Converting to kg, this is

    Δm = 0.000831 u × 1.661 × 10-27 kg/u = 1.381 × 10-30 kg.

    Hence the released energy is

    E = (1.381 × 10-30 kg)(2.998 × 108 m/s)2 = 1.241 × 10-13 J.

    (b) This energy is converted to kinetic energy, E = ½mv2. Solving for v, we get

    v = [2E/m]½ = [2(1.241 × 10-13 J)/(9.11 × 10-31 kg) ]½ = 5.22 × 108 m/s .

    This value exceeds the speed of light. It doesn't. At such relativistic speeds, our formula for kinetic energy breaks down. The correct result is v = 0.99 c.

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  3. Consider 31P which has 15 protons and 16 neutrons. Find the binding energy per nucleon. Note that n = 1.008665 u, 1H = 1.007825 u, and 31P = 30.973762 u.

    The binding energy is the energy difference between the atom 31P and its nucleons separately. It appears as a mass difference between the two. The binding energy is then given by Einstein's formula, E = (Δm)c2. The mass difference is

    Δm = 15(1.007825 u) + 16(1.008665 u) - 30.973762 u = 0.282253 u .

    Converting to kg, this is

    Δm = 0.282253 u × 1.661 × 10-27 kg/u = 4.6882 × 10-28 kg.

    Hence the released energy is

    E = (4.6882 × 10-28 kg)(2.998 × 108 m/s)2 = 4.2138 × 10-11 J.

    Converting the energy to MeV,

    E = 4.2138 × 10-11 J (1 MeV /1.602 10-13 J) = 263.0 MeV .

    Dividing this by the number of nucleons (15 + 16 = 31), we find

    Eper nucleon = 263.0 MeV / 31 = 8.48 MeV

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  4. Graph the data below and determine the decay constant and the half-life using LineFit.

    The decay rate formula is R = R0e-λt. This is linearized by taking the natural logarithm of both sides,

    ln(R) = ln(R0) - λt.

    To plot the data we will need to convert the data and find the uncertainty in ln(R). Recall that ln(R ± R) = ln(R) ± ΔR/R,

    Time t
    (minutes)
    Counting Rate R

    Counts/min
    ln(R)
    60
    3100 ± 110
    8.039 ± 0.035
    120
    2400 ± 100
    7.783 ± 0.042
    240
    1500 ± 80
    7.313 ± 0.053
    360
    985 ± 60
    6.893 ± 0.061
    480
    525 ± 50
    6.263 ± 0.095
    600
    310 ± 40
    5.737 ± 0.129
    720
    220 ± 30
    5.393 ± 0.136

    From the slope of the plotted data, we get the decay constant

    λ = (4.11 ± 0.31) × 10-3 min-1 .

    The half-life is

    τ = ln2 / λ = 169 ± 13 minutes.

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  5. A Geiger counter, held 15.0 cm away from a small piece of radioactive ore, records a constant 192 counts per second. The Geiger counter is only recording the number of radioactive particle through the end of the counter. The end of the counter has an area of 2.00 cm2.
    (a) What is the total activity of the ore sample?
    (b) Chemical analysis indicates that there is 6.256 × 1021 radioactive nuclei present in the sample. What is the decay constant λ and half-life τ?

    (a) One would expect that the radioactivity spreads out equally in all directions, and that the Geiger counter is only intercepting a small portion of the total signal. The surface area of a sphere of radius 15.0 cm is A = 4πr2 = 2827.4 cm2. Thus we expect the total count to be

    R = (2827.4 cm2 / 2.00 cm2)(192 counts/s) = 2.714 × 105 counts/s .

    (b) The decay rate is constant indicating a long half-life. We have the formula that R = λN, so

    λ = R/N = 2.714 × 105 counts/s / 6.256 × 1021 = 4.338 × 10-17 s-1 .

    The half-life is

    τ = ln2 / λ = 1.598 × 1016 s = 5.07 × 108 y .

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  6. In a sample of wood found at an archeological dig, the ratio 14C to 12C is found to be only 23.4 ± 0.2% of that found in living organism. The half-life of 14C is 5370 years. Determine how long ago the tree died.

    The decay of 14C to 12C is governed by N = N0e-λt. We can rearrange this to find t,

    t = -ln(N/N0)/-λ .

    Since τ = ln2 / λ , this becomes

    t = -τln(N/N0)/ln2

    = -τ ln(.234 ± 0.002) / (ln2)

    = -(5370 y) [ ln(.234) ± 0.002/0.234] / ln2

    = 11252 ± 662 y

    = 11300 ± 700 y

    So the artifact is between 10,600 and 12,000 years old.

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