A pure sample of ²²⁶Ra contains 2.0 x 10¹⁴atoms of the isotope.If the half-life of ²²⁶Ra = 1.6 x 10³years,what is the decay rate of this sample? (1 Ci = 3.7 x 10¹⁰ decays/s)

Question

Quizplus · Accepted Answer

First, we need to find the decay constant (λ) using the formula: 
λ = ln(2)/t1/2 = ln(2)/(1.6 x 10^13 s) = 4.34 x 10^-14 s^-1

Then, we can find the activity (A) using the formula: 
A = λN = (4.34 x 10^-14 s^-1)(2.0 x 10^14 atoms) = 8.68 x 10^6 decays/s

Finally, we can convert the activity to curies using the conversion factor: 
1 Ci = 3.7 x 10^10 decays/s
So, 
A = 8.68 x 10^6 decays/s = 8.68 x 10^6/3.7 x 10^10 Ci = 2.35 x 10^-4 Ci

Therefore, the decay rate of the sample is 2.35 x 10^-4 Ci, which is closest to choice C (7.4 x 10^-8 Ci).

A Pure Sample of 226Ra Contains 2

A Pure Sample of ²²⁶Ra Contains 2