Question

Use the worked example above to help you solve this problem. The half-life of the radioactive nucleus _(88)^(226)text(Ra) is 1.6 103 yr. If a sample initially contains 4.00 1016 such nuclei, determine the following:________.
(a) the initial activity in curies µCi
(b) the number of radium nuclei remaining after 4.4 103 yr nuclei
(c) the activity at this later time µCi

Answer 1

Answer:

Explanation:

From the information given:

The half-life $t_{1/2}$ = 1.6103 years

The no. of the initial nuclei $N_o$ = $4.00 \times 10^6$

Using the formula:

$N = N_o exp(-\lambda t)$

where;

decay constant $\lambda = \dfrac{In2}{1.6*10^3} y^{-1}$

∴

$N = N_o exp ( \dfrac{-In2}{1.6*10^3}\times 4.4 \times 10^3)$

$N = N_o exp (- 1.906154747)$

The number of radium nuclei N = 5.94 × 10¹⁵

The initial activity$A_o = \lambda N_o$

$A_o =(\dfrac{In (2)}{1.61\times 10^3 \times 365 \times 24 \times 3600}\times 4.00 \times 10^{16})$

$A_o =546075.8487 \ Bq$

Since;

1 curie = 3.7 × 10¹⁰ Bq

Then;

$A_o =\dfrac{546075.8487 }{3.7\times 10^{10}}$

$A_o = 1.47588 \times 10^{-5}Ci$

$A_o = 14.7588 \ \mu Ci$

c) The activity at a later time is:

$=5.94 \times 10^{15}( \dfrac{In (2)}{1.60 \times 10^3 \times 365\times 24 \times 3600})$

$= 81599.09018 \ Bq \\ \\ = \dfrac{81599.09018}{3.7\times 10^{10}} \ Ci \\ \\ = 2.20538 \times 10^6 \ Ci \\ \\ = 2.20538 \ \mu Ci$