Question

Question 3 (1 point) Co-60 has a half life of 5.3 years. If a pellet that has been in storage for 26.5 years contains 14.5g of Co-60, how much of this radioisotope was present when the pellet was put in storage? Answer = grams Blank 1:​

Answer 1

Answer:

464 grams.

Step-by-step explanation:

Amount of substance:

The amount of substance after t years is given by an equation in the following format:

$A(t) = A(0)(1-r)^t$

In which A(0) is the initial amount and r is the decay rate, as a decimal.

Co-60 has a half life of 5.3 years.

This means that:

$A(5.3) = 0.5A(0)$

We use this to find r. So

$A(t) = A(0)(1-r)^t$

$0.5A(0) = A(0)(1-r)^{5.3}$

$(1-r)^{5.3} = 0.5$

$\sqrt[5.3]{(1-r)^{5.3}} = \sqrt[5.3]{0.5}$

$1 - r = 0.5^{\frac{1}{5.3}}$

$1 - r = 0.8774$

So

$A(t) = A(0)(0.8774)^{t}$

If a pellet that has been in storage for 26.5 years contains 14.5g of Co-60, how much of this radioisotope was present when the pellet was put in storage?

We have that $A(26.5) = 14.5$, and use this to find A(0). So

$A(t) = A(0)(0.8774)^{t}$

$14.5 = A(0)(0.8774)^{26.5}$

$A(0) = \frac{14.5}{(0.8774)^{26.5}}$

[tex]A(0) = 464[tex]

464 grams.