# Half Life

1. Sep 11, 2005

how can we fnd out the half life of sum element?

2. Sep 11, 2005

### H_man

Well firstly you need something to measure the activity of the sample such as a Geiger Counter.

Record this figure.

Then wait till the number of counts has dropped to half the value you measured and hey presto you have the half life. As the half life is the amount of time it takes for the activity to drop by half.

Remember the half life is just the time for half the number of radioactive particles to decay.

3. Sep 11, 2005

### mathmike

the half life equation is given by ( - 1 / k * ln 2)

4. Sep 11, 2005

### H_man

Mathmike.... I think you made an error. The equation for Half Life is actually given by...

$$T_\frac{1}{2} = \frac {ln2}{\lambda}$$

5. Sep 11, 2005

### mathmike

no i am right you forgot the negative which represents the fact that it is
decreasing

6. Sep 11, 2005

### H_man

How can this time be negative?

Also in my humble opinion the ln2 goes on the top as in the equation in my post above.

I am sure a third party can either determine who is correct or where the source of confusion is.

7. Sep 11, 2005

### mathmike

suppose y has an exponintial growth model so

y = y_0 * e^(kt)

at any time t_1 let

y_1 = y_0 * e^(kt) be the value of y.

and let T denote the amount of time required for y to double in size. thus at time t_1 + T the value of y will be 2y_1 so

2y_1 = y_0 * e^(kt) = y_0 * e^(kt_1) * e ^(kT)

or

2y_1 = y_1 * e^(kT)

thus

2 = e^(kT)

and

ln 2 = kT

therfore doubling time is

T = 1/k * ln 2

halving time is easily derived from this as

T = - 1/k * ln 2

8. Sep 11, 2005

### Renge Ishyo

I have studied your work a bit Mathmike and believe that the source of "error" in your calculations is due to your treatment of the value of "k" as a negative quantity. The "correct" relation for the T (half life) is (ln 2)/(k) = T (the equation H-man provided). Work:

yfinal = 1/2(y-initial)

1. 1/2(y-initial) = y-initial(e^-kt) (note the minus sign attached to k, your equation lacks this key component)

2. Divide out y initial. Moved e^-kt to the denominator on the right side.

3. 1/2 = 1/e^kt

4. Flipped equations.

5. 2 = e^kt

6. ln (2) = kt

7. t half life = ln (2)/k

K by convention is reported as a *positive* quantity (even if the rate is decreasing for the reaction, the value of k is still reported as being positive). To see why, consider that half life reactions are first order reactions and the integrated rate law for that is:

Ln [Xfinal] = - kt + Ln[Xinitial]

Note that if k was negative and you substituted a negative k into this equation, you would have that value times a negative (if k was negative that equals a *positive* slope ultimately which would not do). So k is reported as a positive number (since they decided by convention to indicate the negative slope in the equation rather than have k be a negative number. They probably did this so that people who just see the algebra without substituting any numbers will recognize that the slope for the function is decreasing).

Your thinking is NOT incorrect though I (finally) recognize. If you substitute a negative k into your equation:

T = - 1/(-k) * ln 2

Then combine.

T = - (ln 2)/(-k)

Cancel the negatives.

T = (ln 2)/(k)

Same thing.

Last edited: Sep 11, 2005