Question

Bromine-88 is radioactive and has a half life of 16.3 seconds. Calculate the activity of a 3.2 mg sample of bromine-88. Give your answer in becquerels and in curies. Be sure your answer has the correct number of significant digits.

Answer 1

`9.31* 10^(17)\ Bq`

`2.52* 10^(7)\ Ci`

Step-by-step explanation:

Calculation of the moles of Bromine-88 as:-

Mass = 3.2 mg

Also, 1 mg = 0.001 g

So, Mass =
`0.0032\ g`

Molar mass of Bromine-88 = 88 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (0.0032\ g)/(88\ g/mol)`

1 mole of Bromine-88 contains
`6.023* 10^(23)` atoms of Bromine-88

So,

`(0.0032)/(88)` mole of Bromine-88 contains
`(0.0032)/(88)* 6.023* 10^(23)` atoms of Bromine-88

Atoms of Bromine-88 in the sample =
`2.19* 10^(19)`

Given that:

Half life =
`16.3` s

The expression for half-life is:-

`t_(1/2)=\frac {ln\ 2}{k}`

Where, k is rate constant

So,

`k=\frac {ln\ 2}{t_(1/2)}`

`k=(\ln2)/(16.3)\ s^(-1)`

The rate constant, k = 0.04252 s⁻¹

Disintegration is:-

Disintegrations per second = Rate constant*Number of atoms =
`0.04252* 2.19* 10^(19)\ Bq` =
`9.31* 10^(17)\ Bq`

Considering that:-

1 Bq =
`2.7* 10^(-11)` Ci

Thus, disintegrations =
`9.31* 10^(17)* 2.7* 10^(-11)\ Ci` =
`2.52* 10^(7)\ Ci`