a sample of16.0 mg of nickle-57, half life 36.0h, is produced in a nuclear reactor. How much of the nickle -57

Question

sample is remaining after 7.5 days?



Technetium-99 decays to ruthenium-99 with a half life of 210,000y. Starting with 1.00 mg of technetium-99, how long will it take for 0.75 mg of ruthenium -99 to form?

Answer 1

M = Mo (1/2)^(t/T)
so in this case
M = 16(1/2)^(t/36)
if t = 7.5 x 24 = 180h
M = 16(1/2)^(180/36) =16(1/2)^5 = 16/32 ≈ 0.5mg

since the nuclear masses are equal
forming 0.75 mg of ruthenium -99 is the same as
decaying 0.75mg of technetium-99
so you need to find how long it will take till 0.25mg of technetium-99 remains
which is 1/4 of the original
so it will take 2 half lives
t = 2 x 210000 = 420000yr
(note: you could also have let Mo=1, T= 210000 and M=0.25 in the formula above and then solved for t to get this answer)

the end
.

Answer 2

Ni-57 has a half life of 36 hours that is 1 and a half days. 7.5 days is 5 times the half-life and thus the sample remaining will be 1 divided by 2^5 of the original sample. 2^5 is 32 and so 16.0 mg becomes 0.5 mg

1.0 mg becomes 0.5 g of Technetium and 0.5 g of Ruthenium in one half life.

0.5 g of Technetium becomes 0.25 g in another half-life. Thus 2 half lives are needed.

That means 4,20,000 years.

Answer 3

7.5 days =7.5*24=180 heures
180/36 =5 so time elapsed is 5 half-times
the formula N = N0*2^-t/T where N0 mass at time 0, N mass at time t, T=half time
So = 16*2^-5 =16/32= 1/2
so 0.5g are remaining after 7.5 days

Same formula 0.75 = 1* 2^(-t/210)
2^(-t/210) =0.75
2^(t/210)= 1/0.75 =1.333
t/210 = ln base 2 of 1.333 =0.4147=87.1 days