If we start with 1.000 g of strontium-90, 0.908 g will remain after 4.00 yr. This means that the half-life of strontium-90 is ________ yr.

Answer 1

Radioactive decay obeys 1st order kinetics, therefore we will use the applicable integrated rate law.

In this question, what is given is the initial mass of Sr-90 of 1 g, final mass of 0.908 g, and time (t) of 4 years

Integrated rate law for 1st order reaction:

ln[Sr-90] = ln[Sr-90]_initial - kt

ln(0.908) = ln(1) - 4k

-9.651 x 10^-2 = 0- 4k

4k = -9.651 x 10^-2

k = 2.41 x 10^-2 yr^-1

To find the half life, utilize formula t_1/2= 0.693/k

Since k is calculated, plug this value into half-life formula:

t_1/2= 0.693/k = 0.693/ (2.41 x 10^-2 yr^-1) = 28.8 yr

Answer 2

The half life formula is N(t) = N₀(0.5)^t/H

N(t) is the quantity after time t, N₀ is the quantity at time zero, and H is the half-life.

We know N ₀=1.000 g, t = 4 years, and N (t) = 0.908. We need to find H.

0.908 = 1.000 (0.5)^4/H

0.908 = (0.5)^4/H

We take logs here (base 10).

Log 0.908 = log (0.5) ^4/H=(4/H) log (0.5). (properties of logs).

(4/H) = (log 0.908 / log 0.5)

4 = H log (0.908) / log (0.5)

H = 4 log (0.5) / log (0.908) = 28.7 years (to 3 significant figures). Our original data is given to 3 significant figures so that is why my answer is stated to the same.

I recommend your check your answer by substituting back. Putting H = 28.7 into our original equation gives 0.90791 on the left hand side which is very close to 0.908. There will be some error as our calculation of H is only a good approximation.

Answer 3

A relatively easy way to approach this problem (as it would be 1st order decay) is to use FR = 0.5ⁿ

FR = fraction remaining = 0.908

n = number of half lives that have elapsed = ?

0.908 = 0.5n

log 0.908 = n log 0.5

n = 0.139 half lives have elapsed

Since 4 yrs have elapsed, we know that 4 yrs = 0.139 half lives. Setting up a simple ratio, we see that...

4 yr/x yr = 0.139/1

x = 28.8 yrs. = half life