10. If 90 kg of Cobalt 60 decays to 3.9 kg in 24 years, what is the half life?

The correct answer can be found using the formula for radioactive decay:<br /><br />N(t) = N₀ * (1/2)^(t/T)<br /><br />Where:<br /><br />* N(t) is the amount remaining after time t (3.9 kg)<br />* N₀ is the initial amount (90 kg)<br />* t is the elapsed time (24 years)<br />* T is the half-life (what we want to find)<br /><br />Let's plug in the values and solve for T:<br /><br />3.9 = 90 * (1/2)^(24/T)<br /><br />Divide both sides by 90:<br /><br />3.9/90 = (1/2)^(24/T)<br /><br />0.04333 = (1/2)^(24/T)<br /><br />Now, take the logarithm base (1/2) of both sides:<br /><br />log_(1/2)(0.04333) = 24/T<br /><br />Or, using the property of logarithms, we can use log base 10 or the natural logarithm (ln):<br /><br />ln(0.04333) / ln(1/2) = 24/T<br /><br />-3.140 = 24/T<br /><br />Now, solve for T:<br /><br />T = 24 / -3.140 * -1<br /><br />T ≈ 4.59 years<br /><br />Therefore, the half-life of Cobalt-60 is approximately 4.59 years.<br />