Decay word problem

[acunningham19]

A radioactive substance decays in such a way that after 20 years there is just 30% of the original amount left. What is the half-life of the substance? round answer to the 4th decimal place.

 

[srmichael]

A radioactive substance decays in such a way that after 20 years there is just 30% of the original amount left. What is the half-life of the substance? round answer to the 4th decimal place.

What have you tried so far? Do you at least know the decay formula to work with?

 

[acunningham19]

What have you tried so far? Do you at least know the decay formula to work with?

What we have is P=F(t)=ab^t

 

[HallsofIvy]

What we have is P=F(t)=ab^t

Good! So "after 20 years there is just 30% of the original amount left" tells you that F(20)= .7a= ab^20 giving b^20= .7. You can solve for b by taking a logarithm of both sides. The "half life" will be a number, T, such that ab^T= .5a or b^T= .5.

 

[acunningham19]

Good! So "after 20 years there is just 30% of the original amount left" tells you that F(20)= .7a= ab^20 giving b^20= .7. You can solve for b by taking a logarithm of both sides. The "half life" will be a number, T, such that ab^T= .5a or b^T= .5.

I am still not sure how to put it in my calculator.... I am confused

 

[stapel]

I am still not sure how to put it in my calculator....

You're still not sure how to put what in your calculator?

 

[srmichael]

I am still not sure how to put it in my calculator.... I am confused

There was a slight error in Halls post.

So we know that:

\(\displaystyle F(20)=0.3a=ab^{20}\)

Thus, after dividing both sides by "a" you get:

\(\displaystyle 0.3=b^{20}\)

Now take the 20th root of both sides and that is your b:

\(\displaystyle 0.3^{1/20}=b\)

You can then use this to find the half-life time like Halls explained.

Let us know what you get.