Question

Most radioactive processes can only happen once for a given radioactive nucleus, changing the nucleus into another state (possibly a different radioactive state or even a different element) in the process. This means that for a given sample of radioactive material, the original radioactive substance is constantly being depleted. Since the rate of these decay events is directly proportional to the number of radioactive nuclei present, the decay is governed by the differential equation:

dN(t)/dt = -(lamda)N(t),

where N(t) is the number of nuclei of the original substance at time t, and the decay constant? ?, is positive because the amount of the substance is decreasing. The solution to this equation is an exponential, namely,

N(t) = No(e^(-lamda*t)) or N(t) = No(e^(t/Tao))

1. If the average count rate for a radioactive decay were 10 counts per minute, how long would you need to count to measure it to a precision of 5% of its value with 68% confidence (one standard deviation)?

2. The half-life of a decaying substance is the time it takes for 1/2 of that substance to disappear. Given an initial sample size of 1000 particles, how many particles are left after one half-life? Two?

Answer 1

SOLUTION

Part (1)

A 68% confidence interval means that the counts must fall with one standard deviation of the mean (1), therefore:

If

Then:

Where:

is the number of counts

is the standard deviation and is defined as:

Replacing (2) into (1) we get:

or

Isolating the number of counts we have:

Solving we obtain:

Now, the count rate is defined as:

Isolating :

Replacing (5) into equation (7) for data given we get:

Solving we obtain:

Part (2)

After one half life the number of particles is:

Similarly, after two half life we get: