In April 1986, a flawed reactor design played a part in the Chernobyl nuclear meltdown.

Question

Approximately 14252 becqurels (Bqs), units of radioactivity, were initially released into the environment. Only areas with less than 800 Bqs are considered safe for human habitation.

The function f(x)=14252(0.5)^x/32 describes the amount, f(x) in becqurels of a radioactive element remaining in the area x years after 1988.

Find F(40) to one decimal place in order to determine the amount of becqurels in 2026.

The determine if the area is safe for human habitation in the year 2026.

When I did the work myself I did:

F(40)=14252(0.5)^40/32= 5992.227855 but since I don't know if what i did was right I am not able to say yes or no for the area being safe to live.

Answer 1

You should check exactly what the problem is asking. The issue that I see is that finding f(40) finds the radioactivity 40 years after 1988 which is 2028, but the problem also asks for the radioactivity levels in 2026. Double-check that, but otherwise, you are on the right track by solving that equation by substituting x for the given value of years. Once you do that, you can decide whether or not that number is less than or greater than the value of 800 that the question provides!