Answer : The age of the mineral is, [tex]8.89\times 10^9\text{ years}[/tex]
Explanation :
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}{4.5\times 10^9\text{ years}}[/tex]
[tex]k=1.54\times 10^{-10}\text{ years}^{-1}[/tex]
Now we have to calculate the time.
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
t = time passed by the sample = ?
a = initial amount of the reactant = 61.0 mg
a - x = amount left after decay process = 15.5 mg
Now put all the given values in above equation, we get
[tex]t=\frac{2.303}{1.54\times 10^{-10}\text{ years}}\log \frac{61.0mg}{15.5mg}[/tex]
[tex]t=8.89\times 10^9\text{ years}[/tex]
Therefore, the age of the mineral is, [tex]8.89\times 10^9\text{ years}[/tex]
The age of the mineral is 8.9×10⁹ years.
We'll begin by calculating the rate constant.
Half-life (t½) = 4.5×10⁹ years
[tex]K = \frac{0.693}{t_½} \\ \\ K = \frac{0.693}{4 \times {10}^{9} } \\ \\ K = 1.54 \times {10}^{ - 10} /year \\ \\ [/tex]
Original amount (N₀) = 61.0 mg
Remain remaining (N) = 15.5 mg
Rate constant (K) = 1.54×10¯¹⁰ / year
[tex]t = \frac{2.303}{k} log( \frac{N_0}{N} ) \\ \\ t = \frac{2.303}{1.54 \times {10}^{ - 10} } log( \frac{61}{15.5}) \\ \\ t = 8.9 \times {10}^{9} years[/tex]
Therefore, the mineral is 8.9×10⁹ years old.
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