Respuesta :
Represent the decay of the substance by the equation
[tex]m(t)=m_{0}e^{-kt}[/tex]
where
m₀ = initial mass
k = decay constant
t = time, years
Because the half life is 18 years, therefore
[tex]e^{-18k}= \frac{1}{2} \\ -18k =ln(0.5)\\ k=- \frac{ln(0.5)}{18} =0.0385[/tex]
When m(t) = 63% of m₀, obtain
[tex]e^{-0.0385t}=0.63\\-0.0385t=ln(0.63)\\t=- \frac{ln(0.63)}{0.0385}= 11.998[/tex]
Answer:
The substance decays to 63% of its original amount in 12 years.
[tex]m(t)=m_{0}e^{-kt}[/tex]
where
m₀ = initial mass
k = decay constant
t = time, years
Because the half life is 18 years, therefore
[tex]e^{-18k}= \frac{1}{2} \\ -18k =ln(0.5)\\ k=- \frac{ln(0.5)}{18} =0.0385[/tex]
When m(t) = 63% of m₀, obtain
[tex]e^{-0.0385t}=0.63\\-0.0385t=ln(0.63)\\t=- \frac{ln(0.63)}{0.0385}= 11.998[/tex]
Answer:
The substance decays to 63% of its original amount in 12 years.
Answer:
It will take 12 years to decay 63% of the substance.
Step-by-step explanation:
Given information:
The half life of substance = 18 years
Now , we know that
[tex]m(t)=m_0e^{-kt}[/tex]
where,
[tex]m_0=[/tex] initial mass
[tex]k=[/tex] decay constant
[tex]t =[/tex] time
now put the values in above equation:
As:
[tex]0.5=e^{-18k}\\\\k=-\frac{ln(0.5)}{18}\\\\k=0.0385\\[/tex]
Now for 63% decay , we can write as:
[tex]e^{-0.0385 \times t}=ln(0.63)\\\\\\t=-(ln(0.63))/0.0385\\t=12[/tex]
Hence, it will take 12 years to decay 63% of the substance.
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