A pure sample of ²²⁶Ra contains 3.0 × 10¹⁴ atoms of the isotope.If the half-life of ²²⁶Ra = 1.6 × 10³ years,what is the decay rate of this sample? (1 Ci = 3.7 × 10¹⁰ decays/s) A) 2.7 × 10−¹² Ci B) 11 × 10−⁸ Ci C) 3.4 × 10−¹⁰ Ci D) 7.4 × 10−⁸ Ci

Question

Quizplus · Accepted Answer

First, we need to find the decay constant (λ) using the half-life equation:
t1/2 = 0.693/λ
1.6 × 10^13 years = 0.693/λ
λ = 4.33 × 10^-14 s^-1

Next, we can use the decay rate equation:
decay rate = λ * N
where N is the number of atoms in the sample.

N = 3.0 × 10^14 atoms (note we converted from 10^26 to 10^14)

decay rate = (4.33 × 10^-14 s^-1) * (3.0 × 10^14 atoms) = 1.30 × 10^1 decays/s

Finally, we can convert to Ci using the conversion factor given:
decay rate = (1.30 × 10^1 decays/s) / (3.7 × 10^10 decays/Ci) = 3.51 × 10^-10 Ci

Rounding to one significant figure, we get:
Answer: B (11 × 10^-8 Ci)

A Pure Sample of 226Ra Contains 3

A Pure Sample of ²²⁶Ra Contains 3