Question

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

Answer 1

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

Answer 2

10/k = 4.67 * 10^4

Answer 3

Thank you very much .... Your a star