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Gwar [14]
3 years ago
14

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 a

mount (in grams). What percent of the fermium-253 decays each hour? Round your answer to the nearest hundredth of a percent.
Mathematics
1 answer:
almond37 [142]3 years ago
7 0

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%.

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