A sample of a radioactive substance decayed to 94.5% of its original amount after a year. (Round your answers to two decimal places.)
(a) What is the half-life of the substance?
(b) How long would it take the sample to decay to 65% of its original amount?
A certain radioactive isotope decays at a rate of 2% per 100 years. If t represents time in years and y
represents the amount of the isotope left then the equation for the situation is y= y0e-0.0002t. In how many years will there be 93% of the isotope left?
(a) Well, the substance decayed to 94.5% of its original amount after a year. So, if we divide that by 2, we'll get the amount it would decay to in half a year. And if we keep doing that, we can find out how many half-lives it takes to reach 0%... or just about. So, let's do some quick math. After one half-life, we're at 47.25%, after two half-lives we're at 23.63%, after three half-lives we're at 11.81%, and after four half-lives we're at 5.90%. So, it looks like it takes about four half-lives to get close to 0%. Therefore, the half-life of the substance is approximately a quarter of a year or 3 months.
(b) If it took a year for the substance to decay to 94.5% of its original amount, we can take that as one complete decay cycle. So, if we take that time and divide it by two (because it's the half-life), we find that the half-life is 6 months or 0.5 years.
Now, to find out how long it would take for the substance to decay to 65% of its original amount, we can use the half-life we just found. If we divide 65% by 100%, we get 0.65. And since every half-life cuts the amount in half, we can set up a little equation like this:
0.5^(n) = 0.65
Solve for "n," and we'll get the number of half-lives it takes for the decay. So, grab a calculator and let's get cracking! After doing some math magic, we find that it takes approximately 2.28 half-lives (which we'll round up to 3).
Since we know the half-life is 0.5 years, we can multiply that by 3 to get the time it takes for the sample to decay to 65% of its original amount. Doing that, we find that it takes approximately 1.5 years.
So, to sum it up, it would take the sample approximately 1.5 years to decay to 65% of its original amount. But hey, don't worry, you'll still have plenty of time to plan all those radioactive-themed parties while you wait!
(a) To find the half-life of the substance, we can use the formula:
\[ \text{{Final amount}} = \text{{Initial amount}} \times \left(\frac{1}{2}\right)^\frac{{\text{{time elapsed}}}}{{\text{{half-life}}}} \]
Given that the final amount is 94.5% of the original amount, we can write the equation as:
\[ 0.945 = 1 \times \left(\frac{1}{2}\right)^\frac{1}{\text{{half-life}}} \]
Taking the logarithm of both sides of the equation, the equation becomes:
\[ \log(0.945) = \frac{1}{\text{{half-life}}} \log\left(\frac{1}{2}\right) \]
Now, we can solve for the half-life:
\[ \text{{half-life}} = \frac{\log(0.945)}{\log\left(\frac{1}{2}\right)} \approx \frac{-0.0247}{-0.3010} \approx 8.20 \text{{ years}} \]
Therefore, the half-life of the substance is approximately 8.20 years.
(b) To find the time it takes for the sample to decay to 65% of its original amount, we can use the same formula:
\[ \text{{Final amount}} = \text{{Initial amount}} \times \left(\frac{1}{2}\right)^\frac{{\text{{time elapsed}}}}{{\text{{half-life}}}} \]
Given that the final amount is 65% of the original amount, we can write the equation as:
\[ 0.65 = 1 \times \left(\frac{1}{2}\right)^\frac{{\text{{time elapsed}}}}{{8.20}} \]
Taking the logarithm of both sides of the equation, the equation becomes:
\[ \log(0.65) = \frac{{\text{{time elapsed}}}}{{8.20}} \log\left(\frac{1}{2}\right) \]
Now, we can solve for the time elapsed:
\[ \text{{time elapsed}} = \frac{\log(0.65)}{\log\left(\frac{1}{2}\right)} \times 8.20 \approx \frac{-0.1878}{-0.3010} \times 8.20 \approx 5.13 \text{{ years}} \]
Therefore, it would take the sample approximately 5.13 years to decay to 65% of its original amount.
To find the answers to these questions, we can use the formula for exponential decay:
N(t) = N0 * (1/2)^(t / T)
where:
N(t) is the amount of the radioactive substance at time t,
N0 is the initial amount of the substance,
T is the half-life of the substance,
and t is the time elapsed.
(a) What is the half-life of the substance?
We are given that the substance decayed to 94.5% of its original amount after a year. This means that N(t) / N0 = 0.945.
Substituting these values into the decay formula, we get:
0.945 = (1/2)^(1 / T)
To solve for T, take the logarithm (base 1/2) of both sides:
log(0.945) = log((1/2)^(1 / T))
Using properties of logarithms, we can bring down the exponent to the front:
log(0.945) = (1 / T) * log(1/2)
Now, solve for T by isolating the variable:
T = 1 / (log(0.945) / log(1/2))
Using a calculator, evaluate the right-hand side to get the half-life.
(b) How long would it take the sample to decay to 65% of its original amount?
We need to find the time t when N(t) / N0 = 0.65.
Substituting into the decay formula, we get:
0.65 = (1/2)^(t / T)
To solve for t, take the logarithm (base 1/2) of both sides:
log(0.65) = log((1/2)^(t / T))
Using properties of logarithms, we can bring down the exponent to the front:
log(0.65) = (t / T) * log(1/2)
Now, solve for t by isolating the variable:
t = T * log(0.65) / log(1/2)
Using a calculator, evaluate the right-hand side to find the time it would take for the sample to decay to 65% of its original amount. Round the answer to the desired decimal places.
general equation
amount = starting (1/2)^(t/k) , where t is the time and k is the half-life period
a)
.945 = 1(.5)^1/k
ln .945 = (1/k)ln.5
1/k = ln.945/ln.5
k = ln.5/ln.945 = 12.25 years
b)
.65 = (.5)^t/12.25
t/12.25 = ln.65/ln.5 = .621488..
t = 7.61 years