Chapter 21, Problem 21.58QA

Interpretation Introduction

To:

Find the radioactivity of the Radium in the watch.

Answer to Problem 21.58QA

Solution:

$0.945 \mu Ci$

$3.50\times {10}^4 Bq$

Explanation of Solution

1) Concept:

The radioactivity is related to the number of atoms decaying by the relation $A=kN$; where A is the radioactivity, $k$ is the decay rate constant and $N$ is the number of atoms decaying. The half-life of the Radium-226 is provided through which we can calculate the decay rate constant. The number of atoms in the sample can be calculated from its mass given.

2) Formulae:

i) $k= \frac{0.693}{t_{1/2}}$

ii) $ln\frac{N_t}{N_o}= -0.693 \frac{t}{t_{1/2}}$

iii) $A=kN$

iv) $1 Ci=3.7 \times {10}^{10} Bq$

3) Given:

i) Half-life of Radium-226 $t_{1/2}=1.60 \times {10}^3 years$

ii) Initial quantity of Radium-226 used $=1.00 \mu g$

iii) Initial time $=year 1914$

iv) Final time $=year 2018$

4) Calculations:

Half-life of the Radium-226 is given in years, which we have to convert it into seconds because the radioactivity is measured in seconds. Therefore, the half-life in seconds is

$1.60 \times {10}^3 years \times \frac{365 days}{1 year} \times \frac{24 hours}{1 day} \times \frac{60 min}{1 hour} \times \frac{60 s}{1 min}$

$=5.0457 \times {10}^{10} s$

The decay rate constant for the decay of Radium-226 is:

$k= \frac{0.693}{t_{1/2}}= \frac{0.693}{5.04 \times {10}^{10} s}=1.3734 \times {10}^{-11} \frac{decay events}{atom.s}$

The number of atoms in $1.00 \mu g$ of Radium-226 are calculated as:

$1.00 \mu g \times \frac{1.00 g }{1.00 \times {10}^6 g} \times \frac{1 mole of Ra}{226 g of Ra} \times \frac{6.022 \times {10}^{23} atoms of Ra}{1 mole of Ra}$

$=2.664 \times {10}^{15} atoms of Ra$

The time interval is:

$Final time-Initial time=2018-1914=104 years$

We have to convert this time in seconds, which is done as:

$104 years \times \frac{365 days}{1 year} \times \frac{24 hours}{1 day} \times \frac{60 min}{1 hour} \times \frac{60 s}{1 min}=3.2797 \times {10}^9 s$

The number of atoms remaining after decay can be found out by the following relation:

$ln\frac{N_t}{N_o}= -0.693 \frac{t}{t_{1/2}}$

$ln\frac{N_t}{N_0}= -0.693 \frac{3.2797 \times {10}^9 s}{5.0457 \times {10}^{10} s}= -0.693 \times 0.06499=-0.04504$

$\frac{N_t}{2.664 \times {10}^{15}}= e^{- 0.04504}$

$\frac{N_t}{2.664 \times {10}^{15}}= 0.9559$

${ N}_t=2.5466 \times {10}^{15} atoms of Ra$

Now, we know the atoms remaining, we can find out the radioactivity remaining in the watch.

$A=kN= 1.3734 \times {10}^{-11} decay events/(atom.s) \times 2.5466 \times {10}^{15} atoms of Ra$

$A=3.4975 \times {10}^4\frac{ decay events}{s}=3.4975\mathrm{ }\times {10}^4\mathrm{B}\mathrm{q}$

Thus, the radioactivity of the watch today, is $3.50 \times {10}^4 Bq$.

Now, we know that $1 Ci=3.7 \times {10}^{10} Bq$.

Therefore, the radioactivity in Ci is:

$3.4975\mathrm{ }\times {10}^4\mathrm{B}\mathrm{q}\mathrm{ }\times \mathrm{ }\frac{1 Ci }{3.7 \times {10}^{10} Bq } \times \frac{{10}^6\mu Ci }{1 Ci}=0.9452 \mu Ci$

Thus, the radioactivity of the watch today is $0.945 \mu Ci$

Conclusion:

The radioactivity left after a certain time period can be found out if we have the initial quantity and the half-life of the radioactive element.