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Found in: Page 1304

### Fundamentals Of Physics

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..

See the step by step solution

## Step 1: The given data

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, ${A}_{He}=4\frac{\mathrm{g}}{\mathrm{mol}}$

## Step 2: Understanding the concept of decay

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)

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

## Step 3: Calculate the milligrams of helium produced by the Plutonium isotope

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:

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

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:

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

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