To understand decay in terms of half-life and to solve radioactive dating proble

Question

To understand decay in terms of half-life and to solve radioactive dating problems.

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=N0et,

where is known as the decay constant. Note that at t=0, N(t)=N0, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part C. If a sample shows only one-fourth of its estimated original quantity of 14C, how old is it? Think about half-lives; don't try to just plug into the decay equation. (You shouldn't need to plug into it at all for this problem.)

Explanation / Answer

PART A:

N = No e - t

N = No / 2 when t = thalf

plug in to get     No / 2 = No e - thalf

The factors of No cancel out in the above step to give    1/2 = e - thalf

now take ln of each side to get     ln (1/2) = ln e - thalf

using the rule for logs on the left side, and recognizing the ln and e are inverses of each other, we get:

ln 1 - ln 2 = - thalf   

but ln 1 = 0

so,   = ln 2 = - thalf

thus,   thalf = ln 2 / exact final answer.     This is approximately thalf = 0.693 / . .   final answer.

easy way:   after 1 half-life, No / 2 nuclei remain.

After another half-life, only 1/2 of (No /2) or No / 4 nuclei remain.  

hard way:   

plug t = 2 thalf into the original eqn, and then use part A result that thalf = ln 2 / . .   final answer.

PART B:

N = No e - t

N = No e- (2 ln 2/)     the factors of cancel out in the exponent on the right

N = No e - 2 ln 2 using a law of logs, the exponent on the right can be written as ln 2-2 which is ln (1/22) or ln (1/4).

thus ,   N = No eln (1/4)   the e and ln are inverses and thus undo each other

N = No (1/4).    we get the same result as by the easy way !!    N = No / 4.   final answer.

PART C:

reduced to 1/4 the original number of C-14 means 2 half-lives have passed.

thus, age of the C-14 sample is 2 (5730 years) = 11,460 years.

Rounded to 3 sig figs, this time is   11,500 years. = 1.15*10^4 years    final answer.

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