Watering HOle

Question

Suppose that a local water hole for cattle has a half-life of 14 days If the water hole initially contained 11,000 gallons of water, how much water would it contain after 20 days?

 

and

 

After 50 days?

 

Also..... if you have time, i'm sooo stumped on this question.

 

Consider that the half-life of Polonium-210 is 140 days. What approximate daily percentage of decay does Polonium-210 demonstrate? Hint: use the Rule of 70 backwards by dividing by the half-life period instead of the rate.)

 

and

 

 
What percentage of the original amount of Polonium-210 would be left after 475 days of decay?

Answer 1

The formula for half life is:

 

P_t = P₀(1/2)^t/d

 

Where t is the number of days, d is how long it takes to halve the amount of stuff, P₀ is your initial amount of stuff and P_t is how much is left at that time you're referring to.

 

For the first problem,

 

d = 14 days

P₀ = 11,000 gallons

 

P₂₀ = 11,000 * (1/2)^20/14

P₂₀ = 4,086 gallons

 

P₅₀ = 11,000 * (1/2)^50/14

P₅₀ = 925 gallons

 

For the second problem,

 

We know that d = 140 days

 

If our initial amount of polonium-210 is 1 kg (I'm using 1kg in this case because the problem is asking us for a percentage. We can find said percentage by dividing the amount after one day by the initial amount, subtracting from 1, and converting to a percentage), the amount of polonium-210 we see after a day is equal to:

 

P₁ = 1 kg * (1/2)^{1 / 140}

P₁ = 0.995 kg

 

P₁/P₀ = 0.995 / 1 = 0.995

 

% lost = 100% - 99.5%

% lost = 0.5%

 

Now we simply plug in 475 days instead of 1 day to find our total amount of polonium left.

 

P₄₇₅ = 1 kg * (1/2)^(475/140)

P₄₇₅ = 0.095 kg

 

Dividing this by the 1 kg gives us our ratio, and we convert to a percentage

 

% of polonium 210 left after 475 days = 9.5%