Respuesta :
Answer : The half life of the radioisotope is 2.7 hours.
Explanation :
Radioactive disintegration is a first order reaction. The disintegration equation can be written as,
[tex]\frac{N}{N_{0}} = e^{- \lambda t}[/tex]
Here N is the amount of radioactive substance left = 8 counts
No is the initial amount of Radioactive substance = 64 counts
t is the time = 8 hours
λ is the disintegration constant
Let us rearrange the equation to solve for λ
[tex]\frac{N}{N_{0}} = e^{- \lambda t}[/tex]
Take natural logarithm "ln" on both sides
[tex]ln [\frac{N}{N_{0}}]= - \lambda t[/tex]
Divide both sides by t
[tex]\lambda = \frac{ln(N/N_{0})}{-t}[/tex]
Let us plug in the given values.
[tex]\lambda = \frac{ln(8/64)}{-8}[/tex]
[tex]\lambda = \frac{(-2.079)}{-8}[/tex]
[tex]\lambda = 0.26 [/tex]
The disintegration constant, λ is 0.26.
λ and half life ( t1/2) are related to each other by following equation.
[tex]t_{1/2} = \frac{0.693}{\lambda}[/tex]
Let us plug in the value of λ
[tex]t_{1/2} = \frac{0.693}{0.26} = 2.7 hours[/tex]
The half life of the radioisotope is 2.7 hours.
Radioactive decay follows first order kinetics
ln (No/ Nt) = kt
k = rate constant
t = time
N0 = initial count = 64
Nt = count at time ti = 8
So
K = ln (64/8) / time = 2.0794 /8 = 0.2599 hours-1
half life = 0.693 / K = 0.693 / 2.599 = 2.67 hours
