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I need some help with the following question:

A dangerous radioactive substance has a half-life of 90 years. It will be deemed safe when its activity is down to 0.05 of its initial value. How long will it be before it is deemed safe?

I don't understand the half life bit so am unsure how to set up the equation thing. Any help would be appreciated :smile:

Reply 1

Kvothe the Arcane

Original post by Super199

Half life is the time taken for a quantity to fall to half of its value.
So if we call the initial value

I

then we know that the time taken for it to reach

\dfrac{I}{2}

90

years. Similarly, the time taken to reach

\dfrac{I}{4}

180

years.
So we can say that

I \times \left( \dfrac{1}{2} \right)^n=90n

. Because each time you halve the initial value, you increase the time by 90 years. If it's asking you to work out the time taken for it to get the 0.05 of its original value, you need to first work out what value n takes if

\left( \dfrac{1}{2} \right)^n=0.05

(edited 9 years ago)

Reply 2

ghostwalker

Original post by keromedic

So we can say that

I \times \left( \dfrac{1}{2} \right)^n=90n

Appreciate what you're trying to say there, but :eek:

Reply 3

Kvothe the Arcane

Original post by ghostwalker

Appreciate what you're trying to say there, but :eek:

What sorry? :smile:

Reply 4

ghostwalker

Original post by keromedic

What sorry? :smile:

On the left we have the level of activity. On the right we have time in years.
To reach a certain level of activity takes a certain length of time, but the two are not numerically equal.

The left hand side is decreasing as n increases, and the right is increasing as n increases, so they can't by true for all n.

Reply 5

Kvothe the Arcane

Sorry OP

Original post by ghostwalker

Oh right, I see what you mean.

Thanks. :facepalm:

Reply 6

ghostwalker

Original post by keromedic

Oh right, I see what you mean.

Sorry for not making it clear in my first post. It feels like a sly sort of an error - if there is such a thing; slips past without being noticed. The rest of the post was fine.

(edited 9 years ago)

Reply 7

Kvothe the Arcane

Original post by ghostwalker

...

So I'm guessing it's more like the below?

I=I_0 \times \left( \dfrac{1}{2} \right)^n

T=90n

Original post by ghostwalker

Sorry for not making it clear in my first post. It feels like a sly sort of an error - if there is such a thing; slips past without being noticed. The rest of the post was fine.

Oh no, it's fine. I'm just glad you said something or the OP would have received the wrong information.

(edited 9 years ago)

Reply 8

ghostwalker

Original post by keromedic

So I'm guessing it's more like the below?

I=I_0 \times \left( \dfrac{1}{2} \right)^n

T=90n

Yep. And in the question we have

I = 0.05I_0

@OP: Using that to solve the first equation gives you the number of multiples of the half-life, and then sub into the second to get the number of years.

Reply 9

Super199

Original post by ghostwalker

Yep. And in the question we have

I = 0.05I_0

@OP: Using that to solve the first equation gives you the number of multiples of the half-life, and then sub into the second to get the number of years.