a radiotactive isotope lose 1/5 of its atoms in a sample decay after 14 years. find the half life.
if k is the half-life time, then
.8 = 1 (1/2)^(14/k)
log .8 = (14/k)(log .5)
14/k = log.8/log.5
k/14 = log.5/log.8
k = 14(log.5)/log.8 = 43.48
thanks :D
To find the half-life of a radioactive isotope, we need to use the formula:
Half-life = (t * ln(2)) / ln(N₀/N)
Where:
- Half-life is the time it takes for half of the atoms to decay,
- t is the given time period (14 years in this case),
- ln is the natural logarithm function,
- N₀ is the initial number of atoms in the sample,
- N is the final number of atoms in the sample.
In this problem, the isotope loses 1/5 of its atoms during the decay, which means the final number of atoms is 4/5 of the initial number of atoms. Therefore, we have N = (4/5)N₀.
Now, let's substitute the values and solve for the half-life:
Half-life = (14 * ln(2)) / ln(N₀ / ((4/5)N₀))
= (14 * ln(2)) / ln(5/4)
Using a calculator, we can evaluate this expression:
Half-life ≈ (14 * 0.693) / 0.223
≈ 43.3 years
Therefore, the half-life of the radioactive isotope is approximately 43.3 years.