An unknown radioactive element decays into non-radioactive substances. In 700 days the radioactivity of a sample decreases by 60 percent.

Question

(a) What is the half-life of the element? half-life: (days)

(b) How long will it take for a sample of 100 mg to decay to 54 mg? time needed: (days)

Answer 1

Assuming you still want an answer to this question, here it is.

We know that in 700 days, the radioactivity of the sample decreases by 60 percent. In other words, the sample now has 40 percent of its original radiation or 0.4 * original radiation. When something decreases exponentially, it can be modeled by the function e^-kx, for some value k.

For us to find the half-life, in other words, when e^-kx = 0.5, we first need to find out our k value.

Given what we know in the problem, x = 700 days, and we have 0.4 * our original radiation. This means our equation is e^-700k = 0.4, for some k. To solve this, we need to take the ln (natural log, log_e) of both sides, so we get ln(e^-700k) = ln(0.4), which implies -700k = ln(0.4), so k = -(ln(0.4)/700).

Since we know our k value, we can solve our first equation.

e^{--(ln(0.4)/700)x} = e ^{(ln(0.4)/700)x} = 0.5

Taking the ln of both sides, we get:

(ln(0.4)/700)x = ln(0.5)

Cross multiplying, we get:

x = 700 ln(0.5)/ln(0.4) ≈ 529.52 days

Since we wanna know when we are down to 54 percent of the original radiation, we just use the same equation and substitute 0.54 for 0.5 and solve in the same manner as the above problem.

e ^{(ln(0.4)/700)x} = 0.54

(ln(0.4)/700)x = ln(0.54)

x = 700 ln(0.54)/ln(0.4) ≈ 470.73 days