Posted by Derrick on Monday, July 19, 2010 at 8:17pm.
Sorry I wasn't finished posting, I accidentally pressed enter and it ended up posting. I don't get how to solve the half-life qquestion.
My other questions are...
The half-life of a particular radioactibe isotope is 6 hours. What percent of the daughter isotope would be present after 1 day?
A 50%
B 75%
C 87.5%
D 93.75%
Your assistance will be longed-for.
The half life of Ni-28 is six days. What fraction of a sample of this isotope will remain after 18 days?
A 1/2
B 1/4
C 1/8
D 1/16
You can do this two or three ways.
a. 18 days/6 days = 3 half lives; the sample will be 1/2 after 1 half life, 1/4 after two half lives, and 1/8 after 3 half lives. This works ok for a small number of half lives as long as they are even (no "part of a half live"). But it wouldn't work very well for say 20 half lives--too long to do it one by one.
b. No/N = where No = initial number of atoms, N = final number of atoms, n = # half lives.
No/N = 2^{n} = 2^{3} = 8. Therefore N = No/8 or 1/8 of the initial number. This works even for fractional half lives.
c. The long way but it is just a variation of the second method above.
k = 0.69315/t_{1/2},
k = 0.693/6 hrs = 0.1155, then substitute into
ln(No/N) = kt
ln(1/N) = 0.1155(18)
(1/N) = 8.00
N = (1/8)No.
The half-life of a particular radioactibe isotope is 6 hours. What percent of the daughter isotope would be present after 1 day?
A 50%
B 75%
C 87.5%
D 93.75%
24 hours/6 hours 4 half lives.
(No/N) = 2^{4} = 16.
N = (1/16)No
If we started with 100 for No, then we would be left with 100/16 = 6.25 which means that 100-6.25 = 93.75 have decayed and that is 93.75%
how many hours are in a day? 12! So... just take the 6 as the numerator and the 12 as the denominator (6/12) and if you simplify so it will be (1/2) and if you turn that into s percentage it will be %50. THE ANSWER IS AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAaa