Respuesta :
The mass at time may be expressed by
[tex]m(t) = m_{0} e^{-kt}[/tex]
where
m₀ = the original mass
k = decay constant
t = time, hours
Because the half-life is 2.6 h, therefore
[tex]e^{-2.6k} = 1/2 \\ -2.6k = ln(1/2) \\ k = \frac{ln(1/2)}{-2.6} =0.2666[/tex]
Since m₀ = 1.0 mg, the mass after 10.4 hours is
[tex](1.0 \, mg)e^{-0.2666*10.4} = 0.0625 \, mg[/tex]
Answer: 0.0625 mg
[tex]m(t) = m_{0} e^{-kt}[/tex]
where
m₀ = the original mass
k = decay constant
t = time, hours
Because the half-life is 2.6 h, therefore
[tex]e^{-2.6k} = 1/2 \\ -2.6k = ln(1/2) \\ k = \frac{ln(1/2)}{-2.6} =0.2666[/tex]
Since m₀ = 1.0 mg, the mass after 10.4 hours is
[tex](1.0 \, mg)e^{-0.2666*10.4} = 0.0625 \, mg[/tex]
Answer: 0.0625 mg
