Question

An unidentified corpse was discovered on April 21 at 7:00 a.m. The pathologist discovered that there were 1.57×10¹⁷ atoms of 3215P remaining in the victim's bones and placed the time of death sometime on March 15. The half-life of 3215P is 14.28 days. How much 3215P was present in the bones at the time of death?   Please explain how the answer was recieved.

Answer 1

Given data ,

The amount of ³²₁₅P remaining in the victim's bones = 1.57 x 10¹⁷ atoms

Time of death of victim = March 9

Time of discovery of corpse = April 21 , 7: 00 am

Half - life of ³²₁₅P = 14.28 days.

Time elapsed from time of death to discovery of corpse is about 43 days.

The number of half lives of ³²₁₅P after 43 days = total time of decay / half life

= 43 / 14.28

= 3 (three half life)

Since , Fraction remaining = 1 / 2ⁿ , where n is number of half-lives

= 1 / 2³

= 1 / 8

In this solution , we require total amount of ³²₁₅P present in the bones at the time of death.

So , we need to multiply the amount of after death in the bones with fraction of 8 we get ,

8 x 1.57×10¹⁷atoms = 12.56 x 10¹⁷ atoms

Therefore , 12.56 x 10¹⁷³²₁₅P atoms were present in the victim's bones at the time of death