Respuesta :
a) 7.9x10^9
b) 1.5x10^9
c) 3.9x10^4
To determine what percentage of an isotope remains after a given length of time, you can use the formula
p = 2^(-x)
where
p = percentage remaining
x = number of half lives expired.
The number of half lives expired is simply
x = t/h
where
x = number of half lives expired
t = time spent
h = length of half life.
So the overall formula becomes
p = 2^(-t/h)
And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are:
a) 1.1x10^10 * 2^(-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9
b) 1.1x10^10 * 2^(-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9
c) 1.1x10^10 * 2^(-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4
Answer: Atoms of [tex]_{56}^{133}\textrm{Ba}[/tex] left in
a) [tex]N=78.608\times 10^8atoms[/tex]
b) [tex]N=14.650\times 10^8atoms[/tex]
c) [tex]N=31.35\times 10^3atoms[/tex]
Explanation: The given reaction is a type of radioactive decay and all the radioactive decay follows first order reactions. Hence, to calculate the rate constant, we use the formula:
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]t_{1/2}=10.5years[/tex]
[tex]k=\frac{0.693}{10.3years}\\k=0.0672years^{-1}[/tex]
To calculate how much amount of sample is left, we use the rate law expression for first order kinetics, which is:
[tex]N=N_oe^{-kt}[/tex] ....(1)
where,
k = rate constant = [tex]0.0672years^{-1}[/tex]
t = time taken for decay process
[tex]N_o[/tex] = initial amount of the reactant = [tex]1.1\times 10^{10}g[/tex]
N = amount left after decay process
- For a)
t = 5 years
Putting values in equation 1, we get:
[tex]N=(1.1\times 10^{10})\times e^{(-0.0672years^{-1}\times 5years)}\\N=78.608\times 10^8atoms[/tex]
- For b)
t = 30 years
Putting values in equation 1, we get:
[tex]N=(1.1\times 10^{10})\times e^{(-0.0672years^{-1}\times 30years)}\\N=14.650\times 10^8atoms[/tex]
- For c)
t = 190 years
Putting values in equation 1, we get:
[tex]N=(1.1\times 10^{10})\times e^{(-0.0672years^{-1}\times 190years)}\\N=31.35\times 10^3atoms[/tex]
