Question

Phosphorus-32 is radioactive and has a half life of 14.3 days. Calculate the activity of a 3.5mg sample of phosphorus-32. Give your answer in becquerels and in curies. Round your answer to 2 significant digits.

Answer 1

Answer:

The activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.

Explanation:

The activity of P-32 can be calculated with the following equation:

$A = \lambda N$   (1)

Where:

N: is the number of atoms of P-32

λ: is the decay constant

We can find the number of atoms of P-32 as follows:

$N = \frac{N_{A}*m}{M}$  (2)

Where:

$N_{A}$: is the Avogadro's number = 6.022x10²³ atoms/mol

m: is the mass of P-32 = 3.5x10⁻³ g

M: is the molar mass of the radionuclide (P-32) = 32 g/mol    

Now, the decay constant is given by:

$\lambda = \frac{ln(2)}{t_{1/2}}$   (3)

Where:

${t_{1/2}}$: is the half-life of P-32 = 14.3 days

Finally, we can find the activity of P-32 by entering equations (2) and (3) into (1):

$A = \lambda N = \frac{ln(2)}{t_{1/2}}*\frac{N_{A}*m}{M} = \frac{ln(2)}{14.3 d*\frac{24 h}{1 d}*\frac{3600 s}{1 h}}*\frac{6.022 \cdot 10^{23} mol^{-1}*3.5 \cdot 10^{-3} g}{32 g/mol} = 3.7 \cdot 10^{13} dis/s$      

Since a becquerel (Bq) is defined as a disintegration (dis) per second, the activity in Bq is:

$A = 3.7 \cdot 10^{13} Bq$

And, since a Curie (Ci) is 3.7x10¹⁰ Bq, the activity in Ci is:

$A = 3.7 \cdot 10^{13} Bq*\frac{1 Ci}{3.7 \cdot 10^{10} Bq} = 1.0 \cdot 10^{3} Ci$

Therefore, the activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.  

               

I hope it helps you!