Answer:
r = 3.5 * .5^(t/65); around 117.5 days
Step-by-step explanation:
r = remaining substance
t = time in days
The half life is 65 day, meaning the .5 must be put to a power of t/65.
The initial substance was 3.5, which we must substitute in.
After using logarithms to solve our new equation, we are left with t=(about)117.5 days for there to be less than 1 gram of substance remaining.
The expression of the decay is required and the time taken to decay to 1 g is required.
The equation is [tex]N=3.5e^{-0.0107t}[/tex]
The time required is 117.08 days.
[tex]t_{1/2}[/tex] = Half-life = 65 days
[tex]N_0[/tex] = Initial amount = 3.5 g
t = Time taken
N = Remaining amount
The equation is
[tex]N=N_0e^{-\dfrac{\ln 2}{t_{1/2}t}}\\\Rightarrow N=3.5e^{-\dfrac{\ln2}{65}t}\\\Rightarrow N=3.5e^{-0.0107t}[/tex]
For [tex]N=1\ \text{g}[/tex]
[tex]1=3.5e^{-0.0107t}\\\Rightarrow \ln\dfrac{1}{3.5}=-0.0107t\\\Rightarrow t=\dfrac{\ln\frac{1}{3.5}}{-0.0107}\\\Rightarrow t=117.08\ \text{days}[/tex]
The time required is 117.08 days.
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