Question

Strotium−90, a radioactive isotope, is a major product of an atomic bomb explosion. It has a half-life of 28.1 yr. (a) Calculate the first-order rate constant for the nuclear decay. (b) Calculate the fraction of 90Sr that remains after 10 half-lives. (c) Calculate the number of years required for 95.7 percent of 90Sr to disappear.

Answer 1

Answer :

(a) The first-order rate constant for the nuclear decay is, $0.0247\text{ years}^{-1}$

(b) The fraction of 90-Sr that remains after 10 half-lives is, $\frac{a_o}{1024}$

(c) The time passed in years is, 127.4 years.

Explanation :

Part (a) :

Half-life = 28.1 years

First we have to calculate the rate constant, we use the formula :

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

$k=\frac{0.693}{28.1\text{ years}}$

$k=0.0247\text{ years}^{-1}$

Thus, the first-order rate constant for the nuclear decay is, $0.0247\text{ years}^{-1}$

Part (b) :

Now we have to calculate the fraction of 90-Sr that remains after 10 half-lives.

Formula used :

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives

$a_o$ = Initial amount of the reactant

n = number of half lives  = 10

Now put all the given values in the above formula, we get:

$a=\frac{a_o}{2^{10}}$

$a=\frac{a_o}{1024}$

Thus, the fraction of 90-Sr that remains after 10 half-lives is, $\frac{a_o}{1024}$

Part (c) :

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant  = $0.0247\text{ years}^{-1}$

t = time passed by the sample  = ?

a = let initial amount of the reactant  = 100

a - x = amount left after decay process = 100 - 95.7 = 4.3

Now put all the given values in above equation, we get

$t=\frac{2.303}{0.0247}\log\frac{100}{4.3}$

$t=127.4\text{ years}$

Therefore, the time passed in years is, 127.4 years.