quesition dealing with half life?

Question

A chemist observes that a particular compound decays continuously, and notes that 90% of the initial amount of the compound remains after 1 hour, 10 minutes. Estimate the half-life of the compound to the nearest minute.



could someone please show me step by step on how to do this?

Answer 1

This formula will work with every instance of halflife problems that you will come across. (They can be expressed in any units, you just need to ensure that N and n are expressed in the same units and that t and k are expressed in the same units)

N = n (1/2)^(t/k)

Where N = final sample amount (%, kg, g... any unit)
n = initial sample amount (same units as final sample)
t = time passed (hours, minutes, seconds)
k = halflife (same units as time passed)

90% = 100%(1/2)^(70minutes/k)
We need to isolate k.
90%/100% = 100%/100% * (1/2)^(70minutes/k)
0.9 = (1/2)^(70minutes/k)
We take the log of both sides.
log(0.9) = log((1/2)^(70minutes/k))
You have to remember a basic property of logs to be able to do this next part. Log[a^b] = b*Log[a] Using this, we can literally "take down the exponent" if you will.
log(0.9) = (70minutes/k)*log(1/2)
log(0.9)/log(1/2) = 70minutes/k
Rearrange to get
k = 70minutes*log(1/2)/log(0.9)
k = 460.517 minutes
Estimated to the nearest minute this gives k = 461 minutes.
We can verify by putting it back into the equation ( always a good idea).
N = 100% (1/2)^(70/461)
We get
N = 90%
This is the correct value of n according to the data we were given, therefore we know that our halflife value is correct.
Hope this helped.
-The Mad Scientist

Answer 2

The compound is very likely a radioactive compound, which are the only known substances that spontaneouly decay into more stable chemical elelments. Radioactive decay is governed by the following, well known formula:

Q = Qo exp(-kt)

Qo is the initial amount (quantity) of radioactive substance, and k a constant yet to determine.

Since 90% of the compound remains after 70 min, you can write:

0.9 Qo = Qo exp(-70 k).

Cancelling Qo and taking logarithms (natural, I mean),

ln 0.9 = - 70 k,

so

k = - ln 0.9/70 = 0.0015

Our formula now remains as

Q / Qo = exp (-0.0015 t)

To estimate half-life, put Q / Qo = ½ in the expression above, then take logarithms again

ln ½ = - 0.0015 t ∴ t = - ln ½ / 0.0015 = 462 min

The formula Q = Qo exp(-kt) is easy to derive, since it is the result of the simplest type of ordinary differential equation, namely,

dQ/dt = - kQ.

Right side is set to negative because the rate of change of a radioactive substance is negative (i.e., amount of substance decreases.) Variables are readily separated, yielding

dQ/Q = -k dt.

Integrating,

ln Q - ln C = - k t,

or

ln Q/C = - k t.

To determine integration constant C, the antilogarithm of both sides is taken,

Q/C = exp (- k t);

When t = 0, exp (- k t) =1, and C = Qo.

Answer 3

This is more a physics problem rather than math one :-)
You have to know the equation for the number of nuclei at radioactive decay:
N(t) = No x 2^(-t / T), where we wrote:
N = quantity of the compound after time t
No = initial quantity of the compound
T = half-time
t = 1 h 10 min = 70 min
In our problem, 90% of the initial amount remains, so we know the ratio N/No = 90% = 0.90.
We wrote the equation as:
N/No = 2^(-t/T)
To get T, we would have to use the logarithm in base 2, but it is more common to use the natural logarithm ln (in base e). Also, we can only use this logarithm with our scientific calculator. Doing so, we get:
ln N/No = -t/T x ln2
(we used the identity: ln (a^b) = b x ln a)
We solve for T:
T = -t x ln2 / ln(N/No) = -70 min x ln2 / ln0.90 = 460.5 min

Answer 4

Half-life - time required to have 50% of original material.

70 minutes = 90% of original remains.

Now, if 70 minutes reduces by 90%, then 140 minutes is 81% ( 90% of 90%) and so forth.

So .5 = (.9) ^x, where x is the number of 70 minute periods.

Take logs of both sides to solve for x.

I don't recall my logarithms - haven't used them in... well, too long.

Answer 5

Half-life is the time required for the quantity to decay to half of its initial value. The formula you need is this:

N = exp( - lamda * t)

We have N= 0.90 at time = 70 minutes.
0.90 = exp( - lamda * 70)
ln(0.90) = - lamda * 70
ln(0.90) / 70 = - lamda = -0.001505

Then it can be shown that the half-life time is
t[1/2] = ln(2) / lamda = 0.693147181 / 0.001505
t[1/2] = 460.517 minutes, or about 461 minutes

You could build this little table (time is in minutes):
Time... amount left
70.........90%
140.......81%
210.......73%
280.......66%
350.......59%
420.......53%
490.......48%

Answer 6

It reduces to 90% of its previous value every 70 minutes, and so after 70t minutes it has reduced to
(0.9)^t) of its original value.

We need t such that
(0.9)^t = 0.5
Take logs of both sides:
t log(0.9)= log(0.5)
t = (log(0.5))/log(0.9)
= 6.5788 approx

So the half-life is
6.5788*70 minutes
= 460.516,
which is about 7 hours 40 minutes and 30 seconds.

Notice Brian D used a similar approach, but as the half-life must be much more than 1 hour and 10 minutes I think my answer is correct.
OK, I see he's now cancelled. Must have realised.

Answer 7

Haha, I believe lots of the different solutions, what if your IQ already is severe? human beings could bore me two times as much as they do already, i do no longer understand how i ought to stay like that, yet howdy, a minimum of in line with threat i ought to construct an quite flux capacitor, that is superb. nonetheless in spite of the undeniable fact that, no deal, existence is staggering.

Answer 8

it seems to me this is basic math:

10% -------- 70 minutes
50% --------- x minutes


50% * 70m / 10% = 350 minutes