Part C Americium-241 is used in some smoke detectors. It is an a emitter with a

Question

Part C Americium-241 is used in some smoke detectors. It is an a emitter with a halflife of 432 years. How long will it take in years for 38.0% of an Am-241 sample to decay? Express your answer with the appropriate units t Value Units Submit Hints My Answers Give Up Review Part Part D A fossil was analyzed and determined to have a carbon-14 level that is 40% that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil? Express your answer with the appropriate units Value Units

Explanation / Answer

Solution:

Half life of Americium= 432 years

Nuclear reactions are considered as first order kinetics.

So we use integrated form of fist law kinetics.

Ln ([A]t/ [A]0 ) = -kt

[A]t and [A]0 are final and initial amount

k is decay constant and t is time

Lets assume initial amount of [A]0 100

After time t its 38.0% amount is reduced

So final amount = 100 – 38 = 62

We find k by using t half

We know the relation ,

t ½ = 0.693 / k

or k = 0.693 / t ½

=0.693 / 432 yrs

= 0.001604 per yrs

We plug all values in above equation to get value of time

Ln ([A]t/ [A]0 ) = -kt

Ln ( 62/100) = - 0.001604 per yrs * t

-0.478 = - 0.001604 per yrs * t

0.478 = 0.001604 per yrs * t

t = 297.99 years.

Question 2 )

Solution:

If we assume 100 as initial amount then 40 will be final amount. By using same equation we get time to decay upto 40 from 100 and that time is the age.

Ln ([A]t/ [A]0 ) = -kt

We calculate decay constant

k = 0.693 / 5730 yrs = 0.000121 per yrs

Ln ( 40 / 100 ) = - 0.000121 per yers * t

-0.916 =   - 0.000121 per yers * t

t = 7576 years

The age of organism is 7576 yrs

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---