he radioisotope cesium-137 has a half-life of 30 years. In 1985, a sample starts at 544 mg. In what year will there be 17 mg left of the original sample?

Question

he radioisotope cesium-137 has a half-life of 30 years. In 1985, a sample starts at 544 mg. In what year will there be 17 mg left of the original sample?

Answer 1

FR = 0.5ⁿ

FR = fraction remaining = 17 mg/544 mg = 0.03125

n = number of half lives that have elapsed = ?

0.03125 = 0.5ⁿ

log 0.03125 = n log 0.5

-1.505 = -0.301 n

n = 5.00 half lives have elapsed

each half life = 30 years

Time taken = 5 half lives x 30 yrs/half life = 150 yrs

1985 + 150 = 2135

So, in the year 2135 there will be 17 mg remaining.

Answer 2

Solving for half life means that you need to set up an exponential decay equation. The general equation is as follows:

A = A₀(1/2)^t/h

where:

1. A = final amount
2. A₀ = initial amount
3. t = time
4. h = half-life time

You can solve the question by plugging in the values from your question into the equation where...

1. A = 17 mg
2. A₀ = 544 mg
3. t = ???
4. h = 30 years

A = A₀(1/2)^t/h

17 = 544(1/2)^t/30

0.03125 = (1/2)^t/30

log_1/20.03125 = t/30

5 = t/30

t = 150

This means it would take 150 years for 17 mg of cesium-137 to be left by the original sample. By adding the total number of years it takes with the original year (1985) you can find the answer.

The final answer would be that in 2135, there will be 17 mg left of the original sample. Hope this helps!