Answer:
[tex]y = 50(0.85)^{t}[/tex]
Step-by-step explanation:
Let the equation that models the decay of the substance is given by
[tex]y = y_{0} \times e^{-kt}[/tex] ........... (1)
where y is the amount of the substance remaining after t hours and [tex]y_{0}[/tex] represents the initial amount of the substance and k is the decay rate constant.
Now, given that the half-life period of the substance is 4 hours.
So, from equation (1) we can write
[tex]0.5 = e^{- 4k}[/tex]
Now, taking ln on both sides we get
ln 0.5 = -4k
⇒ k = 0.17328
Therefore, the equation (1) becomes
[tex]y =50\times e^{-0.17328t}[/tex] {Since the initial amount of the substance was 50 gm}
⇒ [tex]y = 50(0.85)^{t}[/tex] (Approximate) (Answer)
