The half-life of krypton-91 (91Kr) is 10 s. At time
t = 0
a heavy canister contains 5 g of this radioactive gas.
(a) Find a function
m(t) = m02−t/h
that models the amount of 91Kr remaining in the canister after t seconds.
every 10s the amount drops by 1/2, so
m(t) = 5 * 2^(-t/10)
To find a function that models the amount of krypton-91 remaining in the canister after t seconds, we can use the formula for exponential decay:
m(t) = m₀ * (1/2)^(t / h)
Where:
- m(t) is the amount of krypton-91 remaining at time t
- m₀ is the initial amount of krypton-91 (5 g in this case)
- t is the time in seconds
- h is the half-life of krypton-91 (10 s in this case)
Substituting the values, we get:
m(t) = 5 * (1/2)^(t / 10)
To find a function that models the amount of 91Kr remaining in the canister after t seconds, we can use the formula for exponential decay:
m(t) = m0 * (1/2)^(t / h)
Where:
m(t) is the amount of 91Kr remaining after t seconds
m0 is the initial amount of 91Kr in the canister
t is the time in seconds
h is the half-life of the radioactive substance
In this case, the half-life of krypton-91 is given as 10 seconds, and the initial amount of 91Kr in the canister is 5 grams.
Plugging these values into the formula, we get:
m(t) = 5 * (1/2)^(t / 10)
So, the function that models the amount of 91Kr remaining in the canister after t seconds is m(t) = 5 * (1/2)^(t / 10).