Respuesta :
Answer :
(a) The first-order rate constant for the nuclear decay is, [tex]0.0247\text{ years}^{-1}[/tex]
(b) The fraction of 90-Sr that remains after 10 half-lives is, [tex]\frac{a_o}{1024}[/tex]
(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 :
[tex]k=\frac{0.693}{t_{1/2}}[/tex]
[tex]k=\frac{0.693}{28.1\text{ years}}[/tex]
[tex]k=0.0247\text{ years}^{-1}[/tex]
Thus, the first-order rate constant for the nuclear decay is, [tex]0.0247\text{ years}^{-1}[/tex]
Part (b) :
Now we have to calculate the fraction of 90-Sr that remains after 10 half-lives.
Formula used :
[tex]a=\frac{a_o}{2^n}[/tex]
where,
a = amount of reactant left after n-half lives
[tex]a_o[/tex] = Initial amount of the reactant
n = number of half lives = 10
Now put all the given values in the above formula, we get:
[tex]a=\frac{a_o}{2^{10}}[/tex]
[tex]a=\frac{a_o}{1024}[/tex]
Thus, the fraction of 90-Sr that remains after 10 half-lives is, [tex]\frac{a_o}{1024}[/tex]
Part (c) :
Now we have to calculate the time passed.
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]0.0247\text{ years}^{-1}[/tex]
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
[tex]t=\frac{2.303}{0.0247}\log\frac{100}{4.3}[/tex]
[tex]t=127.4\text{ years}[/tex]
Therefore, the time passed in years is, 127.4 years.
