What is the half-life of an isotope that decays to 50% of its original activity in 55.6 hours?
To find the half-life of an isotope, we can use the formula:
t1/2 = ln(2) / λ
Where:
t1/2 is the half-life of the isotope,
ln(2) is the natural logarithm of 2 (approximately 0.693),
and λ is the decay constant.
In this case, we know that the isotope decays to 50% of its original activity in 55.6 hours. This means that after the half-life, only half of the isotope remains.
Substituting the values into the formula:
t1/2 = ln(2) / λ
t1/2 = 0.693 / λ
We need to find the value of λ to determine the half-life. We know that after the half-life, the isotope decays to 50% of its original activity. In other words, the remaining activity is half of the original activity.
So, we can write the decay equation as:
A(t) = A(0) * (1/2)^(t / t1/2)
Where:
A(t) is the activity at time t,
A(0) is the initial activity,
t is the time, and
t1/2 is the half-life.
Since the initial activity A(0) decays to 50% (1/2), we rewrite the equation as:
1/2 = (1/2)^(t / t1/2)
Simplifying, we get:
(1/2)^(1) = (1/2)^(t / t1/2)
Here, we can see that the exponent of the left side is 1 because it represents the half-life.
Therefore, we have:
1 = t / t1/2
t1/2 = t
Since we know that t is equal to 55.6 hours, we can conclude that the half-life of the isotope is 55.6 hours.