A sample of a radioactive substance decayed to 93.5% of its original amount after a year.
a) What is the half-life of the substance?
? years
(b) How long would it take the sample to decay to 10% of its original amount?
? years
a) Let the half life in years be H
2^-(1/H) = 0.935
Solve that for H.
-1/H log 2 = log 0.935
-1/H = -0.09696
H = 10.31 years
b) Solve 2^-(t/10.31) = 0.10
t will be the time required to decasy to 10% activity
You could also solve
(0.935)^t = 0.1
The answer should be the same either way you do it: about 30 years.
Thank you!
(a) What is the half-life of the substance? Well, if it decayed to 93.5% of its original amount after a year, that means it still has about 6.5% of itself left. Let's call it the resilience of the substance. So, since half of the substance has decayed, we can say that the resilience is equal to half of the original amount. Therefore, the half-life of the substance is... drumroll, please... it's the time it takes for my patience to run out while waiting for this answer. Just kidding! The half-life of the substance is one year.
(b) How long would it take the sample to decay to 10% of its original amount? Ah, it's time for some radioactive math! Since the substance decayed to 93.5% after one year, that means it lost 6.5% of itself. So, let's just do a little bit of division here. If it loses 6.5% in one year, then it should lose 1% every... umm... carry the one... ah yes, it should lose 1% every 6.5 years! So, if it started with 100% and it loses 1% every 6.5 years, it would take about... you ready for this... 65 years for it to decay to 10% of its original amount! Just don't ask it for a loan, it might just disintegrate on you.
To determine the half-life of the radioactive substance, we can use the decay formula:
N = N0 * (1/2)^(t / T)
Where:
N is the final amount (93.5% of the original amount)
N0 is the initial amount (100%)
t is the time passed (1 year)
T is the half-life we are trying to find.
For part (a), we can solve for T by substituting the values into the formula:
0.935 = 1 * (1/2)^(1 / T)
To solve for T, we can take the logarithm of both sides of the equation:
log(0.935) = log((1/2)^(1 / T))
Using the logarithm identity log(a^b) = b*log(a), we simplify the equation further:
log(0.935) = (1 / T) * log(1/2)
We can isolate T by dividing both sides by log(1/2):
log(0.935) / log(1/2) = 1 / T
To find T, we can take the reciprocal of both sides of the equation:
T = 1 / (log(0.935) / log(1/2))
Evaluating the equation, we get:
T ≈ 2.823 years
Therefore, the half-life of the substance is approximately 2.823 years.
For part (b), we can use the same decay formula with a different value for N (10% of the original amount). Let's substitute the values into the formula:
0.10 = 1 * (1/2)^(t / T)
We want to solve for t, so we can rearrange the equation:
(1/2)^(t / T) = 0.10
Taking the logarithm of both sides:
log((1/2)^(t / T)) = log(0.10)
Using the logarithm property log(a^b) = b*log(a):
(t / T) * log(1/2) = log(0.10)
Simplifying the equation:
t / T = log(0.10) / log(1/2)
To solve for t, we multiply both sides by T:
t = T * (log(0.10) / log(1/2))
Plugging in the value we found for T (2.823 years):
t = 2.823 * (log(0.10) / log(1/2))
Using a calculator, evaluate the equation:
t ≈ 9.327 years
Therefore, it would take approximately 9.327 years for the sample to decay to 10% of its original amount.