Question

The radioactive substance polonium-210, with a half-life of 138.39 days, is present in tobacco. Suppose that the soot in an ex-smoker's lungs contains 100 milligrams of polonium-210. How much radioactive polonium will remain after one year of not smoking

Answer 1

Answer:

16.07 milligrams

Step-by-step explanation:

From the question,

Using the formula of radioactive decay,

R/R' = 2ᵃ/ᵇ..................... Equation 1

Where R = Original mass of the polonium-210, R' = Final mass of polonium-210 after decaying, a = Total disintegration time, b = half life of polonium-210.

Make R' the subject of the equation

R' = R/(2ᵃ/ᵇ)................ Equation 2

Given: R = 100 milligrams, a = 1 years = 365 days, b = 138.39 days

Substitute these values into equation 2

R' = 100/(2³⁶⁵⁰⁰/¹³⁸³⁹)

R' = 100/(6.222)

R' = 16.07 milligrams of plonium-210