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.

My answer: 1.77x10^7

please explain steps, thank you!

Answer 1

No. of moles of NaClO4 = given weight/molecular weight

                                  = 54.3 *10^-3g/ 122.5 g/mol

                                  = 0.443 moles

In one mole of NaClO4 ----------------------- 1 mole of chlorine atoms are there

in 0.443 moles --------------------------- 0.443 moles of Cl atoms will be there.

no.of Cl atoms in 0.443 moles = 0.443 8 6.023*10^23

                                            = 2.67 * 10^23 atoms

Multiply by 0.296 to find the number of atoms that are radioactive.

= 7.9*10^22 atoms are radioactive.

Rate of disintegration = K (rate constant) * no.of Cl atoms

Rate constant K = 0.693/t_1/2

Given t_1/2 = 3.0×105 yr.

Therefore k = 0.693/3.0×10⁵ yr.

                 = 0.231 *10^-5 years

Threfore rate = 7.9*10^22 Cl-atoms * 0.231 *10^-5 years-1

                    = 1.825 * 10^17 atoms/year

Therefore, the rate we calculate will be the decays per year. Convert that to dps:

                   = 1.825 *10^17 atoms/365 * 24*60*60

                   = 5.787 * 10^9 dps