160 mg of radioactive sodium was made and it was found that there was only 20 mg left 45 h later.?

Question

160 mg of radioactive sodium was made and it was found that there was only 20 mg left 45 h later.

a) What is the half life of sodium?

b) Find a function that models the amount A left after t hours

c) If the labratory requires 100 mg of sodium 12 h from now, how much sodium should the research assistant make now? (Ignore the 20 mg she currently has.)

d) How much of the original 20 mg would be left in 12 h?

Answers:
b) A = 160(1/2)t/15

c) 174 mg

d) 11.5 mg

Please show your work. Thanks

Answer 1

a) 20mg/160mg = 1/8.
Since 1/8 = (1/2)^3, it's three half-lives later.
If 45 hours = 3 half-lives, then one half-life = 15 hours.

If the amount were not an even power of 1/2, you couldn't do it in your head and you'd have to solve:

remaining = initial * (1/2)^(t / half_life)
r = n * (1/2)^(t / h)
(r/n) = (1/2)^(t/h)
-log2(r/n) = t/h
h = -t / log2(r/n)
... and since you are given t, r, and n, it's easy to solve for h.

b) Not much work to show. Once you know the half-life and the initial quantity, the same equation as above is used, but with some of the variables replaced by known numbers:

A = (initial_quantity) * 1/2 ^ (t / half_life)
A = 160 * (1/2)^(t/15)

c) Using the same equation, you need to solve:

A = (initial_quantity) * 1/2 ^ (t / half_life)
100mg = x * (1/2)^(12/15)
100mg = x * 0.5743
100mg / 0.5743 = x
174.11mg = x

d) Using the same equation, you need to solve:

A = (initial_quantity) * 1/2 ^ (t / half_life)
x = 20mg * (1/2) ^ (12/15)
x = 20mg * 0.5743
x = 11.486mg

Answer 2

a) We have 1/8 = (1/2)^3 the original amount, so three half-lives have elapsed. So the half-life is 45/3 = 15 hours.

b) A(t) = A(0).(1/2)^(t/15)
= 160 (1/2)^(t/15)

c) A(12) = 100 = A(0) (1/2)^(12/15)
So A(0) = 100 / (1/2)^(4/5)
= 174.1 mg to 1 d.p.

d) A(12) = 20 (1/2)^(12/15) = 11.49 mg to 2 d.p.

Answer 3

hello! i just want to help you but i do not have a scientific calculator here; anyway here's the solution: you should get first the function:
general equation:
P2=P1e^(-kt)
20=160e^(-kt)
20=160e^(-45k)
solving for k:
k= -ln(1/8)/45
particular solution:

P2=160e^{[ln(1/8)/45]t}

solving for its half life which is 160/2=80mg:

80=160e^{[ln(1/8)/45]t}

t=(ln 0.5)/{[ln(1/8)/45]t}

same solution for the next questions.