116In nuclei are produced at the rate of 10 s-1. Calculate the maximum number of 116In nuclei (number in equilibrium) in the sample if the half-life of 116In is 54 m.
So far I have come up with ...
The half life = 54m x 60 s-1 = 3.24 x 103 s-1
(In2) / λ = In2 / 3.24 x 10^3 s-1 = 2.14 x 10^-4 s-1,
10 s-1 So, N = 10 / 2.14 x 10^-4 s-1 = 4.67×10^4 nuclei
Thankks
number produced per second = 10
number lost per second = r = k n
so
dn/dt = 10 - k n
max when dn/dt = 0
or when
n = 10/k
now find k from half life experiment
dn/dt = -k n
dn/n = -k dt
ln n = -k t + c
n = Ni e^-kt
when t = 3.24 * 10^3
n/Ni = .5
ln .5 = -k (3.24*10^3)
-.693 = -k 3.24*10^3
k = .214 * 10^-3
n max = 10/k = 4.67 *10^3
10/k = 4.67 * 10^4
Thank you very much .... Your a star
To calculate the maximum number of 116In nuclei in the sample (number in equilibrium), you can use the equation for radioactive decay and the concept of equilibrium.
The equation for radioactive decay is given by:
N(t) = N0 * e^(-λt)
Where:
N(t) = Number of radioactive nuclei at time t
N0 = Initial number of radioactive nuclei
λ = Decay constant
t = Time
In this case, the initial number of 116In nuclei is not given, but we can calculate it using the given information about the decay rate.
The decay rate is given as 10 s^-1, meaning that 10 116In nuclei are produced per second. Since this is the same as the decay rate, it means that at equilibrium, 10 nuclei will be decaying and 10 will be produced per second. Therefore, the initial number of 116In nuclei is 10.
Now, we can calculate the decay constant (λ) using the half-life of 116In, which is given as 54 minutes.
λ = ln(2) / half-life
λ = ln(2) / (54 minutes * 60 seconds per minute)
λ ≈ 1.28 x 10^-5 s^-1
Now, we can calculate the maximum number of 116In nuclei at equilibrium.
N = N0 / (1 + λt)
N = 10 / (1 + (1.28 x 10^-5 s^-1) * t)
Since the system is in equilibrium, the value of N will not change with time, so we set t to infinity.
N(in equilibrium) = 10 / (1 + (1.28 x 10^-5 s^-1) * ∞)
N(in equilibrium) = 10 / 1
N(in equilibrium) = 10
Therefore, the maximum number of 116In nuclei in the sample, at equilibrium, is 10 nuclei.