3) Potassium-40 decays to argon-40 with a half-life of 1.27 x 109 yr. The age of

Question

3) Potassium-40 decays to argon-40 with a half-life of 1.27 x 109 yr. The age of a mineral sample that has a mass ratio of 40Ar to 40K of 0.330 is yr. Answer: 5.23 x 108 4) If we start with 1.000 g of strontium-90,0.805 g will remain after 9.00 yr. This means that the half- life of strontium-90 is yr. Answer: 28.8 5) A freshly prepared sample of curium-243 undergoes 3312 disintegrations per second. After 8.00 yr the activity of the sample declines to 2591 disintegrations per second. The half-life of curium-243 is years. Answer: 22.6

Explanation / Answer

Solution:- (3) let's say initial mass of Potassium-40 is 1 g. Let's say, X mass of Potassium-40 is decomposed to Argon-40.

So, remaining mass of Potassium-40 would be = 1 - X

and mass of Argon-40 formed would be = X

It's given that, mass rartio of Argon-40 to Potassium-40 is 0.330

It means, mass of Argon/mass of potassium = 0.330

From here, mass of Argon = mass of potassium* 0.330

So, X = (1-X)*0.330

X = 0.330 - 0.330X

X + 0.330X = 0.330

1.330X = 0.330

X = 0.330/1.330 = 0.248

radioactive decay is first order reaction for which..

ln[A] = - kt + ln[A]0

where [A]0 is initial amount and [A] is final amount.

Initial amount of Potassium-40 is 1.0 g and it's final amount is 1.0 - 0.248 = 0.752 g

Half like is given as 1.27 x 109 years.

decay constant, k = 0.693/half life

k = 0.693/1.27 x 109 year = 5.46 x 10-10 year-1

plugging in the values in the first order equation written above..

ln(0.752) = -(5.46 x 10-10 year-1)t + ln(1.0)

-0.285 = -(5.46 x 10-10 year-1)t + 0

0.285 = (5.46 x 10-10 year-1)t

t = 0.285/5.46 x 10-10 year-1

t = 5.22 x 108 year

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.