Answer:
2.5 half-lives
Step-by-step explanation:
Find [tex]t_{h}[/tex] in [tex]N(t)=N_{0}e^{-kt}[/tex] given [tex]N(t) = 2, N_{0} =64,t=\frac{25}{2}[/tex]
[tex]N(t)[/tex] is the amount after the time [tex]t[/tex], [tex]N_0[/tex] `is the initial amount, [tex]t_{h}[/tex] is the half-life
We know that after half-life there will be twice less the initial quantity:
[tex]N(t_{h})=\frac{N_{0} }{2}=N_{0}e^{kt_{h} }[/tex].
Simplifying gives [tex]\frac{1}{2} =e^{-kt_h} or k = -\frac{In(\frac{1}{2}) }{t_{h} } .[/tex]
Plugging this into the initial equation, we obtain that [tex]N(t)=N_{0}e^\frac{In(\frac{1}{2}) }{t_{h} }[/tex]or [tex]N(t)= N_{0} (\frac{1}{2})^\frac{t}{t_h}.[/tex]
Finally, just plug in the given values and find the unknown one.
From [tex]2=64(\frac{1}{2})^\frac{\frac{25}{2} }{t_h}[/tex], we have that [tex]t_h[/tex] = [tex]\frac{25 In (2)}{2 In (32)\\}[/tex]
So you should get: [tex]t_{h} =\frac{25 In (2)}{2 In (32)\\}=2.5.[/tex]
The number of half-lives that transpired during this time period is equal to 2.5 hours.
Given the following data:
Time = 12.5 hours.
Initial mass = 64 grams.
Remaining mass = 2.0 grams.
Mathematically, the amount of a radiocative element that is remaining after time (t-hours) is given by this formula:
[tex]ln(\frac{A}{A_0} )=ln(\frac{1}{2} )\frac{t}{H}[/tex]
Where:
Substituting the given parameters into the formula, we have;
ln(2/64) = ln(1/2) × 12.5/H
-3.466 = -0.693 × 12.5/H
H = 12.5 × -0.693/-3.466
H = 12.5 × 0.1999
H = 2.5 hours.
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