Question I don't understand about Half-Life? thanks for you help ;)?

Question

Iodine-131 has a half-life of 8 days. How do I calculate the number of days it takes for the activity of a sample of iodine to fall to 1/64 of its original value? this is an example which I don't understand. Would really love some help on this so I can carry on with other equations like this, thanks so much!

Do I use the formula: N = No (1/2) ^n

Answer 1

Half-Life (as you probably know) means that over a period of time (8 days in this case), the substance will have decayed and there will only be half of it left. So if you started with say 100kg of it, then after 8 days, you'd have 50kg. After 16 days, you'd have half of that, or 25kg. After 24 days, you'd have half of that, or 12.5kg.



So after ...

8 days, 1/2

16 days, 1/4

24 days, 1/8

32 days, 1/16

40 days, 1/32

48 days, 1/64



Okay, so I didn't really need a formula, but this is a simple case. With something more complicated, you're going to need a formula. Let me do this again:



8x1 (8) days, 1/2

8x2 (16) days, 1/2 * 1/2 [ or (1/2)^2), which equals 1/4]

8x3 (24) days, 1/2 * 1/2 * 1/2 [(1/2)^3 = 1/8]

8x4 (32) days, (1/2)^4 [1/16]

8x5 (40) days, (1/2)^5 [1/32]

8x6 (48) days, (1/2)^6 [1/64]



Note the 1,2,3,4,5,6 on both sides of the equation (times the half-life and as an exponent of the 1/2. If "H" is your half-life, and "n" is this number - the numer of half-lives that it is going to take, then your formula is:



H * N = (1/2)^N



In this case, you have H, you're solving for N.

Answer 2

You're missing some steps. N=NO(1/2)^(t)/(n/2) Put all your values in. N is the remaining value (1/64)

NO is the starting amount (1 in this case). t is the time overall (unknown). n is the halflife (8 days)

So we have (1/64)=1(1/2)^t/(8/2)

(1/64)=(1/2)^(t/4)

and solve from there....

There is a reason I stopped there. Here's an easier way to solve. Just work it out in your hear(or on paper). You start with 1, after 1 halflife cycle you have 1/2, then after two: 1/4, three: 1/8 four: 1/16 five:1/32 six: 1/64 SO it takes six cycles to get 1/64. Now, multiply 6 by the halflife (8) to get 48 days.

Answer 3

I would do it by writing 1/2, 1/4, 1/8, 1/16, 1/32, 1/64 and then counting that there are 6 'halvings'.

Then you know that 6 half lives (= 6 x 8 days = 48 days) have passed.



You don't have to write it out in full, but this is what you are doing:



After 1 half life, fraction of original amount remaining = 1/2

After 2 half lives, fraction of original amount remaining = 1/2 x 1/2 = 1/4

After 3 half lives, fraction of original amount remaining = 1/2 x 1/2 x 1/2 = 1/8

After 4 half lives, fraction of original amount remaining = 12 x 1/2 x 1/2 x 1/2 = 1/16

After 5 half lives, fraction of original amount remaining = 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32

After 6 half lives, fraction of original amount remaining = 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/64



So 6 half-lives have passed = 6 x 8 days = 48days.

Answer 4

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Answer 5

Your formula is fine if you use (no. of days) = n*(half-life). In other words, n is normalized to the half-life of the substance.



Using your formula, No/64 = No(1/2)^n,

1/64 = (1/2)^n



Take logs of both sides to get n. Use any base you want. But as a physicist, get used to natural logs:



ln(1/64) = n ln 0.5 = -ln(64)

n= -ln(64)/ln(0.5) = 6

So no. of days = 6*8 = 48

Answer 6

n = 8 x log2(64) = 8 x 5 = 40