for 10 points on word problem for any 1 that can solve it?

Question

The half-life of palladium-100 is 4 days. after 12 days a sample of palladium-100 has been reduced to a mass of 1 mg. what was the initial mass (in mg) of the sample?
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what is the mass 7 weeks after the start?________________

this is wrong

Answer 1

Every 4 days it loses half its mass.

Start with n.
After 4 days --> n/2
After 8 days --> ½(n/2) = n/4
After 12 days --> ½(n/4) = n/8

n/8 = 1 mg.
n = 8 * 1 mg.
n = 8 mg.

Essentially every 4 days, you take 1/2 of the value. This would be ½^(n/4). Multiply this by the initial value to get the final value.

f(n) = 8 * (½)^(n/4)

Double-checking:
f(12) = 8 * (½)^(12/4)
f(12) = 8 * (½)^3
f(12) = 8 * 1/8 = 1

Now try this formula for 7 weeks (49 days)
f(49) = 8 * (½)^(49/4)
f(49) = 0.00164237581 mg (or 1.64 micrograms)

Answer 2

after 12 days, it has gone through 12/4 half lives, i.e. 3 half lives. So the mass has been halved three times. To be exact, the final mass is the original mass* 2^-3

so, 1 = M*2^-3

1/2^-3 = M

2^3 = M

8 = M, so the original mass was 8mg.

7 Weeks is 49 days. so there were 49/4 half lives; or 12.25 half lives, so the final mass is the original mass * 2^-12.25

Mass = 8 * 2^-12.25

Mass = 8 * 2.05296...*10^-4
Mass = 1.64237... * 10^-3 mg

Mass = 1.64 * 10^-3 mg (to 3 significant figures).


A quick note here, if you actually measure the mass of the sample, it will not change much, because the palladium-100 is being converted into a different element, you'd have to extract the palladium-100 from the products of it's decay to actually see this result.

Remember, each half life multiplies the original mass by 1/2, so after 3 half lives, the factor is (1/2)*(1/2)*(1/2) = 1/(2^3) = 2^-3

Answer 3

Since the half life is 4 days, then you know that every four days before the sample was weighed it would be twice as much. So 12 days earlier is 3 half-life periods, so:

1mg * 2 * 2 * 2 = 8mg

Or you could use the half-life equation (see link below) to determine what λ is based on the 4-day half-life.

λ = 0.173/day

Then plug this into the N(t) equation with N(t) equal to 1mg and calculate No.

Now use the N(t) equation with the 7 weeks (49 days) and the initial mass of 8mg to calculate the mass at 7 weeks:

N(7wk) = 0.00164mg

Answer 4

12 days = 3 half-lives
Initial sample Pd₁₀₀/2³ = 1mg ⇒ Pd₁₀₀ = 8mg
7 wks = 49 days = 12.25 half-lives
Pd₁₀₀/2^(12.25) = 0.00164mg

Answer 5

m = 1/2m e^-4k
2 = e^-4k
ln(2) = -4k
k = ln(2)/-4 = -.1733
m = m1e^-.1733t
1 = m1 e^-.1733t
m1 = 1/e^-(12*.1733) = 8 mg <-- Original amount
m = 8e^-.1733t
m = 8e^(-.1733*49) = .0016 mg <-- after 49 days

Answer 6

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Answer 7

1st blank 16 mg.

2nd blank, 5 mg.