Question

The amount y (in grams) of the radioactive isotope fermium-253 remaining after t hours is y=a(0.5)t/72, where a is the initial amount (in grams). What percent of the fermium-253 decays each hour? Round your answer to the nearest hundredth of a percent.

Answer 1

Answer:

Per hour decay of the isotope is 0.96%.

Step-by-step explanation:

Amount of radioactive element remaining after t hours is represented by

$y=a(0.5)^{\frac{t}{72}}$

where a = initial amount

t = duration of decay (in hours)

Amount remaining after 1 hour will be,

$y=a(0.5)^{\frac{1}{72} }$

y = 0.9904a

So amount of decay in one hour = a - 0.9904a

                                                      = 0.0096a gms

Percentage decay every hour = $\frac{\text{Amount of decay}}{\text{Initial amount}}\times 100$

                                                  = $\frac{0.0096a}{a}\times 100$

                                                  = 0.958 %

                                                  ≈ 0.96 %

Therefore, per hour decay of the radioactive isotope is 0.96%.