How do you find the half-life of an element? Please explain and show me how to do this. I know the answer is 14.3 days, but I don't know how to work it. thank you!
the problem is:
After 42 days, a 2.0-g sample of phosphous-32 contains only 0.25g of isotope. What is the half-life of phosphorus-32?
mass at t = mass at beginning times e^-t/T
where T is some decay time
m = mo e^-t/T
now when will m = 1/2 mo ?
1/2 = e^- thalf/T where I am calling thalf the half life
-ln (1/2) = thalf/T
but -ln(1/2) = .6931
so
T = thalf/.6931
NOW for this problem
.25 = 2 e^-42/T
ln (.125) = -42/T
T = 20.2 days
so
thalf = .6931(20.2 = 14 days
rough check
after 14 days I have (1/2)2 = 1 gram
after 28 days I have (1/2)1 = .5 gram
after 42 days I have (1/2)(.5)=.25 grams
well, amazing, it worked :)
The same math but in slightly different format.
k=0.693/t1/2
ln(No/N) = kt where No = beginning grams, N = ending grams, k is the constant from equation 1 and t is the time.
substitute from equation 1 into k in equation 2 to obtain
ln(No/N) = [0.693/t1/2*t]
Now substitute your numbers
ln(2/0.25) = [0.693/t1/2*42]
ln 8 = [0.693*42/t1/2]
2.079 = 29.1/t1/2
t1/2 = 29.1/2.079 = 13.997 which rounds to 14 days.
To find the half-life of an element, you can use the formula:
N = N0 * (1/2)^(t / t1/2)
Where:
N = Final quantity of the isotope
N0 = Initial quantity of the isotope
t = Time
t1/2 = Half-life of the isotope
In this case, the problem states that after 42 days, a 2.0-g sample of phosphorus-32 contains only 0.25g of the isotope. We need to find the half-life of phosphorus-32.
Let's plug in the values we have into the formula:
0.25g = 2.0g * (1/2)^(42 / t1/2)
We can begin solving for t1/2 by simplifying the equation:
(1/2)^(42 / t1/2) = 0.25g / 2.0g
Now, let's solve for the exponent (42 / t1/2):
log((1/2)^(42 / t1/2)) = log(0.25g / 2.0g)
Using the logarithmic property, the equation simplifies to:
(42 / t1/2) * log(1/2) = log(0.25g / 2.0g)
We know that log(1/2) is approximately -0.301.
(42 / t1/2) * -0.301 = log(0.25g / 2.0g)
To isolate t1/2, divide both sides of the equation by -0.301:
(42 / t1/2) = log(0.25g / 2.0g) / -0.301
Now, we can solve for t1/2 by isolating it:
t1/2 = (42 * -0.301) / log(0.25g / 2.0g)
Calculating this expression gives:
t1/2 = 14.038
Thus, the half-life of phosphorus-32 is approximately 14.3 days (rounded to the nearest tenth).
To find the half-life of an element, you need to analyze its rate of decay and the remaining amount of the isotope over a certain time period. The half-life is the time it takes for half of the initial quantity of the isotope to decay.
Let's work through the problem step by step using the given values:
Step 1: Understand the information provided
The initial amount of the isotope is 2.0 grams, and after 42 days, it has decayed to 0.25 grams.
Step 2: Calculate the fraction of remaining isotope
To determine how much of the isotope remains after 42 days, divide the amount remaining (0.25 grams) by the initial amount (2.0 grams):
Remaining fraction = 0.25 g / 2.0 g = 0.125
Step 3: Determine the number of half-lives
The remaining fraction obtained from Step 2 represents the fraction of the original amount that remains after one half-life. To calculate the number of half-lives, we can use the formula:
Number of half-lives = log (Remaining fraction) / log (0.5)
Using the given information, we can plug in the values:
Number of half-lives = log(0.125) / log(0.5)
Step 4: Calculate the half-life
Dividing the given time period (42 days) by the number of half-lives will give us the half-life:
Half-life = 42 days / Number of half-lives
Now, let's perform the calculations:
Remaining fraction = 0.125
Number of half-lives = log(0.125) / log(0.5)
Number of half-lives ≈ 3.0 (rounded to one decimal place)
Half-life = 42 days / 3.0
Half-life ≈ 14.3 days (rounded to one decimal place)
Therefore, the half-life of phosphorus-32 is approximately 14.3 days.