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Fundamentals Of Physics
Found in: Page 1304

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Short Answer

Plutonium isotope Pu239 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 Pu239is 87.9 m.g..

See the step by step solution

Step by Step Solution

Step 1: The given data

a) The Plutonium isotope Pu239decays by alpha decay.

b) Half-life of Plutonium isotope Pu239,

c) Time of decay, t = 20000y

d) Mass of the sample, m = 12g

e) Molar mass of the Plutonium sample, APu=239gmol

f) Molar mass of the helium sample, AHe=4gmol

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.


The undecayed sample remaining after a given time as follows:

N=N0e-λt …… (i)

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

N=mANA (ii)

Here, NA=6.022×1023atomsmol

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:

N0-N=N01-e-tIn 2/T1/2

Substitute the values as follows:

N0-N=6.022×1023articlesmol12g239g/mol1-e-20000In 2/24100 =1.32×1022 alpha 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:

mHe=1.32×10226.022×1023 mol-14gmol =87.9×1023 mol

Hence, the mass of the sample is 87.9×10-3 g.


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