Suppose 32000 radioactive nuclei are in a sample. About how many remain after two days if the half-life is 22 hrs? What is the initial activity of the sample in decays per minute?
Two days is 48 hours or 2.18 half-lives. The original number of radioactive nuclei, 2 days earlier, is higher by a factor 2^2.18 = 4.53, so there were N0 = 145,000 radioactive nuclei then.
There is a formula that relates activity per molecule (k) to half life (T). It is k = ln2/T = 0.693/T
T = 22 hours = 1320 min
The initial activity is
A = N0*k = (145,000)(0.693)/1320
= 76 decays/min
To find out how many radioactive nuclei remain after two days with a half-life of 22 hours, we can use the formula:
N = N₀ * (1/2)^(t / T₁/₂)
Where:
N is the final number of radioactive nuclei
N₀ is the initial number of radioactive nuclei
t is the time elapsed
T₁/₂ is the half-life of the radioactive substance
In this case, the initial number of radioactive nuclei (N₀) is 32,000, the time elapsed (t) is 2 days = 48 hours), and the half-life (T₁/₂) is 22 hours.
Let's start by calculating the number of radioactive nuclei that remain after 48 hours:
N = 32,000 * (1/2)^(48 / 22)
Now, let's calculate the initial activity of the sample in decays per minute. Activity is defined as the rate of decay, which is given by the formula:
Activity = lambda * N
Where:
lambda is the decay constant, which is ln(2)/T₁/₂
N is the initial number of radioactive nuclei
In this case, lambda = ln(2)/22, and N = 32,000.
Now we can calculate the initial activity:
Initial Activity = lambda * N
Now, let's plug in the values and calculate the initial activity in decays per minute.