Question

A scientist is studying a \(2.234\) g sample of thorium-229 \(\left( {{t_{1/2}} = 7340{\rm{y}}} \right)\) in a laboratory.

(a) What is its activity in Bq?

(b) What is its activity in Ci?

Short answer

We get the required answer by converting\(Bq\)into\(Ci\).

a)\(1\;{\rm{Bq}} = 2.702 \times {10^{ - 8}}{\rm{Ci}}1.\)

b) \(757 \times {10^{10}}\;{\rm{Bq}} \times 2.702 \times {10^8} = 474.86{\rm{C}}\).

Step-by-step solution

Definition of Thorium

Thorium is a weakly radioactive metallic chemical element with the symbol Th and atomic number 90.

Using the formula for decay constant

There is a formula from which we can calculate the activity:

Activity\( = \lambda \cdot N\)

We need this activity in\(Bq\):

\(1\;{\rm{Bq}} = {{\rm{s}}^{ - 1}}\)

So, we can first convert years into seconds:

\(1y = 365\)Days

1 day\( = 24\)hours\(\begin{array}{l} = 1440\;{\rm{min}}\\ = 86400\;{\rm{s}}\end{array}\)

This means 1 year, which has 365 days\(31536000\;{\rm{s}}\).

\(7340{\rm{y}} = 2.3147 \cdot {10^{11}}\;{\rm{s}}\).

There is also a formula for the decay constant:

\(\begin{array}{l}\lambda = \frac{{\ln (2)}}{{{t_{1/2}}}}\\\lambda = \frac{{\ln (2)}}{{2.3147 \cdot {{10}^{11}}\;{\rm{s}}}}\\\lambda = 2.99 \cdot {10^{ - 12}}\;{{\rm{s}}^{ - 1}}\end{array}\)

Therefore,\(\lambda = 2.99 \cdot {10^{ - 12}}\;{{\rm{s}}^{ - 1}}\).

Step 3: Calculate the activity

We also need\(N\)so we can calculate the activity:

c\(n = \frac{m}{M}\)

And also:

\(n = \frac{N}{{{N_a}}}\)

Where\({N_a}\)is the Avogadro constant.

\(\begin{array}{l}N = n \cdot {N_a}\\N = \frac{m}{M} \cdot {N_a}\\N = \frac{{2.234\;{\rm{g}}}}{{229{\rm{gmol}}}} \cdot 6.022 \times {10^{23}}\;{\rm{mol}}\\N = 5.87 \times {10^{21}}\end{array}\)

Now we can calculate the activity:

Activity=\(2.99 \times {10^{12}}\;{{\rm{s}}^{ - 1}} \times 5.87 \times {10^{21}}\).

\(\begin{array}{c}Activity = 1.757 \times {10^{10}}\;{{\rm{s}}^1}\\ = 1.757 \times {10^{10}}\;{\rm{Bq}}\end{array}\).

Therefore, we need to convert\(Bq\)into\(Ci\).

\(\begin{array}{c}1\;{\rm{Bq}} = 2.702 \times {10^{ - 8}}{\rm{Ci}}1.\\757 \cdot {10^{10}}\;{\rm{Bq}} \times 2.702 \times {10^8} = 474.86{\rm{C}}\end{array}\)