Cesium- 137 undergoes beta decay and has a half- life of 30 years. How many beta particles are emitted by a 14.0- g sample of cesium- 137 in three minutes?

Question

Quizplus · Accepted Answer

First, we need to use the concept of half-life to find out the fraction of the sample that remains after three minutes.

Since the half-life of cesium-137 is 30 years, we can use the equation:

N = N0 * (1/2)^(t/t1/2)

where N is the remaining fraction of the sample, N0 is the initial fraction of the sample, t is the time elapsed, and t1/2 is the half-life.

Plugging in the values, we get:

N = 1 * (1/2)^(3/(30*365*24*60)) ≈ 0.9867

This means that about 98.67% of the sample remains after three minutes.

Now, we can find the number of beta particles emitted by multiplying the initial number of atoms by the decay constant and the elapsed time:

Nβ = λ * N0 * t

where Nβ is the number of beta particles emitted, λ is the decay constant, and t is the elapsed time.

The decay constant can be found using the half-life:

λ = ln(2)/t1/2 ≈ 0.0231/yr

We need to convert the time to years:

3 minutes ≈ 3/60 ≈ 0.05 hours ≈ 0.05/24 ≈ 0.00208 days ≈ 0.00208/365 ≈ 5.68 x 10^-6 years

Plugging in the values, we get:

Nβ = 0.0231/yr * (14.0 g / 137.327 g/mol) * 6.02 x 10^23 atoms/mol * 5.68 x 10^-6 yr

Nβ ≈ 8.11 x 10^15

Therefore, the number of beta particles emitted by a 14.0-g sample of cesium-137 in three minutes is approximately 8.1 x 10^15. The best choice is A.

Cesium- 137 Undergoes Beta Decay and Has a Half- Life