Answer:
(a) Amount remaining after 15 days = 27 mg
(b) Amount remaining after t days = [216(0.5)^t/5] mg
Explanation:
Nt = No(0.5)^t/t1/2
No (initial amount) = 216 mg, t = 15 days, t1/2 (half-life) = 5 days
N15 (amount remaining after 15 days) = 216(0.5)^15/5= 216(0.5)^3 = 216 × 0.125 = 27 mg
(b) Nt (amount remaining after t days) = 216(0.5)^t/5
Answer:
(a). 27 mg, (b). N = 216 (1/2)^ (t/5 days).
Explanation:
From the question, we are given that the half-life of bismuth-210, 210Bi = 5 days, a sample mass = 216 mg, the amount remaining after 15 days= ???(unknown).
Using the equation (1) below we can solve for the amount remaining after 15 days.
N= N° (1/2)^(t/th).-------------------------(1).
Where N = is the amount remaining, N° = is the initial amount, t= is time and th= half life.
Therefore,amount remaining, N = 216mg (1/2)^(15 days/5 days).
=====> Amount remaining, N =216 mg (1/2)^3.
=====> Amount remaining, N= 216 mg × 0.125. = 27 mg.
OR
We can solve it by using; 1/2 M°. Where M° is the initial mass.
Therefore, after 5 days, 1/2 × 216 = 108 mg remains.
After 10 days, 1/2 × 108 mg = 54 mg remains.
After 15 days; 1/2 × 54 mg = 27 mg remains.
(b). We are to find the amount remaining after time, t.
The amount remaining after time,t ==> N= N° (1/2)^(t/th)
===> N = 216 (1/2)^ (t/5 days).
With this we can calculate the amount of the sample at particular time if the time is given.
