Specific activity

Radioactivity is expressed as the decay rate of a particular radionuclide with decay constant λ and the number of atoms N:

${\displaystyle -{\frac {dN}{dt}}=\lambda N.}$

The integral solution is described by exponential decay:

${\displaystyle N=N_{0}e^{-\lambda t},}$

where N₀ is the initial quantity of atoms at time t = 0.

Half-life T_1/2 is defined as the length of time for half of a given quantity of radioactive atoms to undergo radioactive decay:

${\displaystyle {\frac {N_{0}}{2}}=N_{0}e^{-\lambda T_{1/2}}.}$

Taking the natural logarithm of both sides, the half-life is given by

${\displaystyle T_{1/2}={\frac {\ln 2}{\lambda }}.}$

Conversely, the decay constant λ can be derived from the half-life T_1/2 as

${\displaystyle \lambda ={\frac {\ln 2}{T_{1/2}}}.}$

The mass of the radionuclide is given by

${\displaystyle {m}={\frac {N}{N_{\text{A}}}}[{\text{mol}}]\times {M}[{\text{g/mol}}],}$

where M is molar mass of the radionuclide, and N_A is the Avogadro constant. Practically, the mass number A of the radionuclide is within a fraction of 1% of the molar mass expressed in g/mol and can be used as an approximation.

Specific radioactivity a is defined as radioactivity per unit mass of the radionuclide:

${\displaystyle a[{\text{Bq/g}}]={\frac {\lambda N}{MN/N_{\text{A}}}}={\frac {\lambda N_{\text{A}}}{M}}.}$

Thus, specific radioactivity can also be described by

${\displaystyle a={\frac {N_{\text{A}}\ln 2}{T_{1/2}\times M}}.}$

This equation is simplified to

${\displaystyle a[{\text{Bq/g}}]\approx {\frac {4.17\times 10^{23}[{\text{mol}}^{-1}]}{T_{1/2}[s]\times M[{\text{g/mol}}]}}.}$

When the unit of half-life is in years instead of seconds:

${\displaystyle a[{\text{Bq/g}}]={\frac {4.17\times 10^{23}[{\text{mol}}^{-1}]}{T_{1/2}[{\text{year}}]\times 365\times 24\times 60\times 60[{\text{s/year}}]\times M}}\approx {\frac {1.32\times 10^{16}[{\text{mol}}^{-1}{\cdot }{\text{s}}^{-1}{\cdot }{\text{year}}]}{T_{1/2}[{\text{year}}]\times M[{\text{g/mol}}]}}.}$

For example, specific radioactivity of radium-226 with a half-life of 1600 years is obtained as

${\displaystyle a_{\text{Ra-226}}[{\text{Bq/g}}]={\frac {1.32\times 10^{16}}{1600\times 226}}\approx 3.7\times 10^{10}[{\text{Bq/g}}].}$

This value derived from radium-226 was defined as unit of radioactivity known as the curie (Ci).

Experimentally measured specific activity can be used to calculate the half-life of a radionuclide.

Where decay constant λ is related to specific radioactivity a by the following equation:

${\displaystyle \lambda ={\frac {a\times M}{N_{\text{A}}}}.}$

Therefore, the half-life can also be described by

${\displaystyle T_{1/2}={\frac {N_{\text{A}}\ln 2}{a\times M}}.}$

One gram of rubidium-87 and a radioactivity count rate that, after taking solid angle effects into account, is consistent with a decay rate of 3200 decays per second corresponds to a specific activity of 3.2×10⁶ Bq/kg. Rubidium atomic mass is 87 g/mol, so one gram is 1/87 of a mole. Plugging in the numbers:

${\displaystyle T_{1/2}={\frac {N_{\text{A}}\times \ln 2}{a\times M}}\approx {\frac {6.022\times 10^{23}{\text{ mol}}^{-1}\times 0.693}{3200{\text{ s}}^{-1}{\cdot }{\text{g}}^{-1}\times 87{\text{ g/mol}}}}\approx 1.5\times 10^{18}{\text{ s}}\approx 47{\text{ billion years}}.}$

For a given mass ${\displaystyle m}$ (in grams) of an isotope with atomic mass ${\displaystyle m_{\text{a}}}$ (in g/mol) and a half-life of ${\displaystyle t_{1/2}}$ (in s), the radioactivity can be calculated using:

${\displaystyle A_{\text{Bq}}={\frac {m}{m_{\text{a}}}}N_{\text{A}}{\frac {\ln 2}{t_{1/2}}}}$

With ${\displaystyle N_{\text{A}}}$ = 6.02214076×10²³ mol⁻¹, the Avogadro constant.

Since ${\displaystyle m/m_{\text{a}}}$ is the number of moles (${\displaystyle n}$), the amount of radioactivity ${\displaystyle A}$ can be calculated by:

${\displaystyle A_{\text{Bq}}=nN_{\text{A}}{\frac {\ln 2}{t_{1/2}}}}$

For instance, on average each gram of potassium contains 117 micrograms of ⁴⁰K (all other naturally occurring isotopes are stable) that has a ${\displaystyle t_{1/2}}$ of 1.277×10⁹ years = 4.030×10¹⁶ s,[4] and has an atomic mass of 39.964 g/mol,[5] so the amount of radioactivity associated with a gram of potassium is 30 Bq.