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Q36P

Expert-verifiedFound in: Page 1304

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Plutonium isotope**** ${}^{{\mathbf{239}}}{\mathbf{Pu}}$ decays by alpha decay with a half-life of 24100y****. How many milligrams of helium are produced by an initially pure 12.0g**** sample of**** at the end of20000y****? (Consider only the helium produced directly by the plutonium and not by any by-products of the decay process.)**

The milligram of helium that is produced by an initially pure 12g sample of ${}^{239}\mathrm{Pu}$is 87.9 m.g..

a) The Plutonium isotope ${}^{239}\mathrm{Pu}$decays by alpha decay.

b) Half-life of Plutonium isotope ${}^{239}\mathrm{Pu}$,

c) Time of decay, t = 20000y

d) Mass of the sample, m = 12g

e) Molar mass of the Plutonium sample, ${A}_{Pu}=239\frac{\mathrm{g}}{\mathrm{mol}}$

f) Molar mass of the helium sample, ${A}_{He}=4\frac{\mathrm{g}}{\mathrm{mol}}$

**The radioactive decay constant or the disintegration constant represents the fraction of radioactive atoms that disintegrates in a unit of time. Using the given condition of decay, we can get the number of alpha particles produced by the isotope using the number of nuclei formula in terms of mass and the disintegration equation. Now, using the same mass equation, determine the unknown mass quantity from the calculated number of particles.**

Formulae:

The undecayed sample remaining after a given time as follows:

$N={N}_{0}{e}^{-\lambda t}$ …… (i)

The number of atoms in a given mass of an atom as follows:

$N=\frac{m}{A}{N}_{A}$ (ii)

$\mathrm{Here},{\mathrm{N}}_{\mathrm{A}}=6.022\times {10}^{23}\frac{\mathrm{atoms}}{\mathrm{mol}}$

One alpha particle (helium nucleus) is produced for every plutonium nucleus that decays. The number of alpha particles that have decayed can be found using equations (i) and (ii) as follows:

${N}_{0}-N={N}_{0}\left(1-{e}^{-t\mathrm{In}2/{T}_{1/2}}\right)$

Substitute the values as follows:

${N}_{0}-N=\left(6.022\times {10}^{23}\frac{\mathrm{articles}}{\mathrm{mol}}\right)\left(\frac{12\mathrm{g}}{239\mathrm{g}/\mathrm{mol}}\right)\left(1-{e}^{-20000\mathrm{In}2/24100}\right)\phantom{\rule{0ex}{0ex}}=1.32\times {10}^{22}\mathrm{alpha}\mathrm{particles}$

In terms of the amount of helium gas produced (assuming the particles slow down and capture the appropriate number of electrons), this corresponds to the mass of the helium sample from the Plutonium isotope using equation (ii) as follows:

${m}_{He}=\left(\frac{1.32\times {10}^{22}}{6.022\times {10}^{23}{\mathrm{mol}}^{-1}}\right)\left(4\frac{\mathrm{g}}{\mathrm{mol}}\right)\phantom{\rule{0ex}{0ex}}=87.9\times {10}^{23}\mathrm{mol}$

Hence, the mass of the sample is $87.9\times {10}^{-3}\mathrm{g}$.

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