Question

Consider an arbitrary radionuclide X (τ = 30 min), which decays to the stable nuclide Y through β− decay. If a sample contains 2 pmol1 of X, what is the activity of this sample? If the sample were allowed to stand for a period of 7 hours, what would be its activity?

Answer 1

The equation of radioactive decay and half-life time allows to find the answers are:

a) The activity is valid is   λ = 3.85 10-4 s⁺¹ and is constant for any time

b) the amount of matter in atom X is  1.2 10⁻¹⁶ mol

Given parameters

• The half-life time tau = 30 min = 1800 s

• The amount of material n = 2 10⁻¹² mol

To find

• Activity

• The particles after a time t = 7 h

Radioactivity is a process by which heavy and unstable atoms can decay into other lighter and more stable atoms. There are three main emission processes: alpha particle, beta particle and gamma rays.

All processes can be characterized by a probabilistic decay described by the expressions

            $N = N_0 \ e^{- \lan \ t}$

don N and N₀ is the current and initial particle number, lamba the decay constant and t the time.

The half-life time is a very useful quantity defined as the time that half of the radioactive material decays

           $T_ \frac{1}{2} = \frac{ln 2}{\lambda}$

          λ = $\frac{ln 2}{ T_\frac{1}{2} }$

          λ = $\frac{ln \ 2}{1800}$

          λ = 3.85 10⁻⁴ s⁻¹

This value of the activity does not depend on the time that passes, it is a property of the material

b) Let's find the particle number that remains after t = 7 h

           t = 7 h ($\frac{3600s }{1 h}$) = 2.52 10⁴ s

We look for the initial quantity of matter with a direct rule of proportions. If 1 mole of matter is avogadro's number 6.022 1023, how much matter is 210-12 mole

           # _particles = 2 10⁻¹² mol (6.022 10²³ particles /1 mol)

           # _particles= 1.2 10¹² particles

We use the initial expression

           N = $1.2 \ 10^{12} \ e^{- 3.85 \ 10^{-4} \ 2.52 \ 10^4 }$

           N = 7,339 10⁷ particles

Let's reduce this number to quantity of materia

             n = 7.339 10⁷ particles ( $\frac{1 mol}{6.022 \ 10^{23} \ particles }$ )

             n = 1.219 10⁻¹⁶ mol

In conclusion, using the radioactive decay and half-life time equations we can find the answers are:

a) The activity is valid and is constant for any time   λ = 3.85 10⁻⁴ s⁻¹

b) the amount of matter in atom X is  1.2 10⁻¹⁶ mol

Learn more about radioactivity here:

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