A sample of phosphorus-32 has a half-life of 14.28 days. If 55 g of this radioisotope remain unchanged after approximately 57 days, what was the mass of the original sample?
help please, im confused.
boomer burger
First, use the half life to calculate k.
k = 0.693/half life = ?
Then ln (No/N) = kt
No = solve for this
N = 55 g
you know k from above
t = 57 days (the problem says APPROXIMATELY 57 days. What kind of double talk is this? How many days is it really.
Post your work if you get stuck.
To solve this problem, we can use the concept of exponential decay and the formula for calculating the remaining amount of a radioactive substance after a certain time.
The formula for calculating the remaining amount of a radioactive substance is given by:
Remaining_amount = Initial_amount * (1/2)^(time / half-life)
In this case, the remaining amount is given as 55 g, and the half-life is 14.28 days. The time, in this case, is 57 days.
So, we can substitute the known values into the formula and solve for the initial amount.
55 g = Initial_amount * (1/2)^(57 / 14.28)
To solve this equation, we can rearrange it to isolate the initial amount:
Initial_amount = 55 g / (1/2)^(57 / 14.28)
Now let's calculate it step by step:
Step 1: Calculate the exponent:
57 / 14.28 ≈ 3.994
Step 2: Calculate the fraction:
1/2 ≈ 0.5
Step 3: Raise the fraction to the power of the calculated exponent:
(0.5)^(3.994) ≈ 0.064
Step 4: Substitute the calculated results into the equation:
Initial_amount ≈ 55 g / 0.064
Step 5: Calculate:
Initial_amount ≈ 859.375 g
Therefore, the mass of the original sample was approximately 859.375 grams.