Question

A 54.3 mg sample of sodium perchlorate contains radioactive chlorine-36 (whose atomic mass is 36.0 amu).

If 29.6% of the chlorine atoms in the sample are chlorine-36 and the remainder is naturally occurring nonradioactive chlorine atoms, how many disintegrations per second are produced by this sample? The half-life of chlorine-36 is 3.0×105 yr.

Answer 1

`5.79* 10^(6)\ Bq`

Step-by-step explanation:

Calculation of the moles of sodium perchlorate as:-

Mass = 54.3 mg

Also, 1 mg = 0.001 g

So, Mass =
`0.0543\ g`

Molar mass of sodium perchlorate = 122.44 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

`Moles= (0.0543\ g)/(122.44\ g/mol)`

`Moles= 0.0004435\ mol`

Also, given that it contains 29.6 % of the radioactive Chlorine

So, Moles of radioactive chlorine in the sample =
`(29.6)/(100)* 0.0004435\ mol=0.000131276\ mol`

1 mole of Chlorine contains
`6.023* 10^(23)` atoms of chlorine

So,

0.000131276 mole of Chlorine contains
`0.000131276* 6.023* 10^(23)` atoms of chlorine

Atoms of radioactive chlorine in the sample =
`7.9* 10^(19)`

Given that:

Half life =
`3.0* 10^5` year

1 year =
`3.154* 10^7` s

Half life =
`3.0* 10^5* 3.154* 10^7` s = 9462000000000 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=(ln\ 2)/(9462000000000)\ s^(-1)`

The rate constant, k =
`7.33* 10^(-14)` s⁻¹

Disintegration is:-

Disintegrations per second = Rate constant*Number of atoms =
`7.33* 10^(-14)* 7.9* 10^(19)\ Bq` =
`5.79* 10^(6)\ Bq`