Question

A sample of phosphorus-32 has a half-life of 14.28 days. If 55 g of this radioisotope remain (1 point) unchanged after approximately 57 days, what was the mass of the original sample? 

Answer 1

875 gram

Step-by-step explanation:

Let h be the half life

Let t be the time

Radioactive decay formula (A)

A = Ao * 2^(-t/h)

A = amount after time t

Ao = initial amount (t = 0)

A = 55

h = 14.28

t = 57

55 = Ao * 2^(-57/14.28)

55 = Ao * 2^(-3.9916)

Ao = 55 / 0.062865

Ao = 874.889

Ao = 875 grams

Answer 2

For this problem, we use the formula for radioactive decay which is expressed as follows:

An = Aoe^-kt

where An is the amount left after time t, Ao is the initial amount and k is a constant.

We calculate as follows:

An = Aoe^-kt
0.5 = e^-k(14.28)
k = 0.05

An = Aoe^-kt
An = 55e^-0.55(57)
An = 1.33x10^-12 g