Half life count rates problem?

Question

In 6.00 hours, the activity of a radioactive compound falls from 999 counts per hour to 634 counts per hour. What is the half-life of the isotope?

Please explain how to do this; I don't understand.

Answer 1

Use N = N0e^-Lt

N = number of remaining nuclei
N0 = number of original nuclei
e = exponetial function
Lamda = L which is the decay constant
t = time taken for this amount to decay

Our first goal is to find lamda, the decay constant:

Rearrange to solve for lamda (L)

Ln(N/N0) = -Lt
-Ln(N/N0)/t = L

Note that Ln is the inverse of the e function (look up natural log laws)

L = -Ln (634/999)/6
L = 0.0758 hours^-1

Now we use the equation Ln(2) = LT
Where:
L = decay constant (from above)
T = half life.

Rearrange to solve for T

T = Ln(2)/L
T = 9.15 hours

Answer 2

A little more clarification ...

the 'counts' you're talking about here are the number of decay events detected by the counter.

Each atom in the material will decay in a random way and it is not possible to know when the decay will occur. This uncertainty is a result of the quantum-mechanical nature of the decay process. It is possible, however, to measure the probability of a decay occuring within a give time...

The usual way of phrasing this probability is to say that after a time T, 50% of the atoms in a sample will have decayed. This time, T, is the half-life. T is different for each different kind of radioactive isotope.

In any particular small time interval, the number of decay events will be proportional to the total number of atoms present - the constant of proportionality (lambda) being different for different kinds of isotopes (like the half-life)

dN/N = - lambda dt

Given M atoms (all of the same isotope) to start with, then after a time t there will be N atoms left. The above differential equation can be integrated to give:

N = M * e ^ ( - lambda t)

lambda can be shown to be related to the half-life (T):

T = ln(2) / lambda

lambda = ln(2)/T

(see wikipage)

In your problem, t = 6hrs, M = 999, N = 634.

So 634 = 999 e ^ ( - ln(2) t / T )

Hopefully, you can follow some of the solutions give above to finish the calculation :)

Answer 3

Nobody wants to answer your question. Let me try....

"half-life" means the amount of time it takes for 1/2 of the radioactivity to go away. (Well, not really, but it works for this explanation - sometimes a radioactive substance changes into something with a longer half-life, but never mind).

It works like this. If a substance has a count of 1,000 and then one month later it has a count of 500, then it has a half like of one month. The time it takes for half of it to go away.

BUT, you can see that over the next month the count will go from 500 to 250 (half the radioactivity). And on, and on.

There are equations that the others gave you that allow you to calculate it exactly, but here is a more simple example (if natural logarithms are not "natural" to you),,,

You can say that in about 10 hours it will go from 1,000 to 500 counts If 100 counts is leathal, how long must you stay inside? See how it works?

In 10 hours it is down to about 500, and in 10 more hours it is down to 250, and in 10 more hours it is down to 125 - - - SO, I would stay inside for more than 35 to 40 hours.

That's what it means, and a rough idea of how to figure it. Hope I helped...

Ron.

Answer 4

use N=Noexp(-l*t)..>634=999*exp(-6*l) t=6hrs N= N(t), No=initial pop of atoms,l=decay const
>ln634=ln999*exp(-6*l)
>ln634=ln999-6*l
>6*l=ln(999/634)
>l=0.07578.
To get T1/2 you subst N=No/2 in the same eqn with l=0.07578
No/2=No*exp(-0.07578*T)
ln 1/2= exp(-0.07578*T)
or T=ln2/l (gen equn)=9.1468hrs.
Hope you corrected for background count and calculated the dead time of your GM tube!

Answer 5

lnNo/lnN = lambda x time.

Work out the nat logs of 634 (N) and 999 (No), and then work out lambda.

Half life = 0.693/lambda.

Answer 6

I dont understand it either