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

Per hour decay of the isotope is 0.96%.

Explanation:

Amount of radioactive element remaining after t hours is represented by

`y=a(0.5)^{(t)/(72)}`

where a = initial amount

t = duration of decay (in hours)

Amount remaining after 1 hour will be,

`y=a(0.5)^{(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}}* 100`

=
`(0.0096a)/(a)* 100`

= 0.958 %

≈ 0.96 %

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