A radioactive isotope sits unused in the lab for 10 years at which it decays to 80% of original amount. Find the half life and the addition years it takes to reduce to 15%of original amount
you want a function like
y = a(1/2)^(t/k)
where a is the original amount and the half-life is k years. You have
a(1/2)^(10/k) = 0.8a
(1/2)^(10/k) = 0.8
10/k ln0.5 = ln0.8
k = 10ln0.5/ln0.8 = 31.06
so now you want to solve
(1/2)^(t/31.06) = 0.15
and just finish it off (expect t to be about 2.5 half-lives, since 1/8 < 0.15 < 1/4)
To find the half-life of the radioactive isotope, we can use the following formula:
N(t) = N0 * 0.5^(t / T),
where N(t) is the amount of the isotope remaining at time t, N0 is the initial amount of the isotope, T is the half-life of the isotope, and ^ denotes exponentiation.
Given that the isotope decays to 80% of its original amount after 10 years, we can set up the following equation:
0.8 = 0.5^(10 / T).
To solve for T, we need to isolate the exponent. Taking the logarithm (base 0.5) of both sides, we get:
log(0.8) = (10 / T) * log(0.5).
Simplifying, we have:
log(0.8) / log(0.5) = 10 / T.
Using a calculator, we find that log(0.8) / log(0.5) ≈ 0.3219.
Substituting this value back into the equation, we get:
0.3219 = 10 / T.
Simplifying further, we find:
T ≈ 10 / 0.3219 ≈ 31.01 years.
Therefore, the half-life of the radioactive isotope is approximately 31.01 years.
To find the additional years it takes for the isotope to reduce to 15% of its original amount, we can set up a similar equation:
0.15 = 0.5^(t / 31.01).
Taking the logarithm (base 0.5) of both sides, we have:
log(0.15) = (t / 31.01) * log(0.5).
Using a calculator, we find that log(0.15) / log(0.5) ≈ 2.7369.
Substituting this value back into the equation, we get:
2.7369 = t / 31.01.
Simplifying further, we find:
t ≈ 2.7369 * 31.01 ≈ 84.95 years.
Therefore, it takes approximately an additional 84.95 years for the isotope to reduce to 15% of its original amount.
To find the half-life of a radioactive isotope, we can use the formula:
t1/2 = (ln 2) / λ
Where:
t1/2 = half-life of the isotope
ln = natural logarithm
2 = constant representing half
λ = decay constant
In this case, we are given that after 10 years, the isotope decays to 80% of its original amount. This means that 20% of the isotope has decayed.
So, to find the half-life, we need to solve for λ first. We can use the following equation:
N = N₀ * (e^(-λt))
Where:
N = final amount (80% of original)
N₀ = initial amount (100% of original)
t = time (10 years)
0.8N₀ = N₀ * (e^(-λ * 10))
Simplifying the equation gives:
0.8 = e^(-λ * 10)
Next, we'll take the natural logarithm of both sides:
ln 0.8 = -λ * 10
Now, we can solve for λ:
λ = (ln 0.8) / -10
Using this value, we can now find the half-life:
t1/2 = (ln 2) / λ
t1/2 = (ln 2) / [(ln 0.8) / -10]
Calculating this gives the value of the half-life.
To determine the additional years it takes to reduce to 15% of the original amount, we can use the same approach as above, but change the final amount to 15% and solve for the time (t).
0.15N₀ = N₀ * (e^(-λt))
0.15 = e^(-λ * t)
Taking the natural logarithm of both sides:
ln 0.15 = -λ * t
Then solving for t:
t = (ln 0.15) / -λ
Calculating this will determine the additional years it takes to reduce to 15% of the original amount.