Question

Phosphorus-32 has a half-life of 14.3 days. How many milligrams of phosphorus-32 remain after 57.2 days of you start with 6.3 mg of the isotope

Answer 1

[A] = 0.39mg

Step-by-step explanation:

The radioactive decay of an atom follows the equation:

Ln[A] = -kt + ln[A]₀

Where [A] is actual amount of isotope after time t; k is decay constant and [A]₀ initial amount of isotope

Also, we can write k in therms of half-life:

k = ln2 / Half-life

Solving for k:

k = ln2 / 14.3days

k = 0.04847days⁻¹

Solving in the equation for [A]:

Ln[A] = -0.04847days⁻¹*57.2days + ln[6.3mg]

ln[A] = -0.93193

[A] = 0.39mg