Question

The barium isotope 133ba has a half-life of 10.5 years. a sample begins with 1.1×1010 133ba atoms. how many are left after (a) 5 years, (b) 30 years, and (c) 190 years?

Answer 1

Answer: Atoms of $_{56}^{133}\textrm{Ba}$ left in

a) $N=78.608\times 10^8atoms$

b) $N=14.650\times 10^8atoms$

c) $N=31.35\times 10^3atoms$

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:

$k=\frac{0.693}{t_{1/2}}$

$t_{1/2}=10.5years$

$k=\frac{0.693}{10.3years}\\k=0.0672years^{-1}$

To calculate how much amount of sample is left, we use the rate law expression for first order kinetics, which is:

$N=N_oe^{-kt}$    ....(1)

where,

k = rate constant  = $0.0672years^{-1}$

t = time taken for decay process

$N_o$ = initial amount of the reactant  = $1.1\times 10^{10}g$

N = amount left after decay process

• For a)

t = 5 years

Putting values in equation 1, we get:

$N=(1.1\times 10^{10})\times e^{(-0.0672years^{-1}\times 5years)}\\N=78.608\times 10^8atoms$

• For b)

t = 30 years

Putting values in equation 1, we get:

$N=(1.1\times 10^{10})\times e^{(-0.0672years^{-1}\times 30years)}\\N=14.650\times 10^8atoms$

• For c)

t = 190 years

Putting values in equation 1, we get:

$N=(1.1\times 10^{10})\times e^{(-0.0672years^{-1}\times 190years)}\\N=31.35\times 10^3atoms$

Answer 2

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