Two Questions:
1) A certain radioactive isotope has leaked into a small stream. 100 days later after the leak 10% of the original amount of the substance. Determine the half-life of this radioactive isotope.
2) During a research experiment, it was found that the number of bacteria in a culture at a rate proportional to its size. At 5am there were 3,000 bacteria present in the culture. At noon, the number of bacteria grew to 6,200. How many bacteria will be at midnight.
I have no idea how to do this! please help.
1. amountremaing=original(e^(.692*time/hslf)
.1=1 e^ ( ) take ln of each side
ln(.1)=.692*100/halflife
halflife= you do it. I get about 30 days.
2. find half life from 5am, noon counts as done above. Then, figure amount remaining from the same formula knowing noon count, and halflife.
im sorry but im still confuse i have 8 other problems that are similar to this. if you just show me how to do these, i can figure out the rest of my hw. please.how to find the halflife?
Sure! I can help you with both of these questions. Let's start with the first one.
1) To determine the half-life of a radioactive isotope, we need to use the formula:
N = N₀ * (1/2)^(t / T)
where:
N = amount of the substance remaining after time t
N₀ = initial amount of the substance
t = time elapsed
T = half-life of the substance
In this case, we are given that 10% of the substance remains after 100 days, and we want to find the half-life. Let's consider the amount remaining as N/N₀ = 0.1, and t = 100 days. Plugging these values into the formula, we get:
0.1 = (1/2)^(100 / T)
To solve for T, we need to isolate it. Taking the logarithm (base 2) of both sides, we have:
log₂(0.1) = log₂((1/2)^(100 / T))
Using the logarithmic property, we can bring the exponent down:
log₂(0.1) = (100 / T) * log₂(1/2)
Now, we can solve for T. Divide both sides by log₂(1/2):
log₂(0.1) / log₂(1/2) = 100 / T
Finally, rearrange the equation to solve for T:
T = 100 / (log₂(0.1) / log₂(1/2))
Using a calculator, you can find the value of T.
Moving on to the second question:
2) The given problem states that the rate of bacterial growth is proportional to the current size of the culture. This is a common scenario modeled by exponential growth. The general formula for exponential growth is:
N = N₀ * e^(kt)
where:
N = number of bacteria at a given time t
N₀ = initial number of bacteria
e = Euler's number (approximately 2.71828)
k = constant of proportionality
t = time in hours
Now, let's use the given information to find the constant k. From 5 am to noon, 7 hours passed, and the number of bacteria increased from 3,000 to 6,200. Plugging in these values, we get:
6,200 = 3,000 * e^(k * 7)
To solve for k, we divide both sides by 3,000 and take the natural logarithm (ln) of both sides:
ln(6,200/3,000) = 7k
Calculate the left side using a calculator, and then divide by 7 to find the value of k.
Once we have the value of k, we can use the formula to find the number of bacteria at midnight. Calculate the time difference from noon to midnight (12 hours), and plug it into the equation along with the given initial amount of bacteria (6,200 in this case).
N = N₀ * e^(k * t)
where:
N₀ = 6,200 (number of bacteria at noon)
k = calculated constant of proportionality
t = 12 (time from noon to midnight)
Solve for N using the equation above, and you'll get the number of bacteria at midnight.
Both of these problems involve mathematical formulas that require some calculations, but by following the steps I've explained, you should be able to find the answers.