A radioactive sample of 141Cs has a half-life of 32.5 days. In a sample that contains 8.00*10^7 nuclei of 141Cs, how many beta particles per second will be emitted?

A 19.8
B 1.7*10^6
C 71 076
D 1185
E 2.47*10^-7

The 141 supposed to be a superscript text, but firefox won't let me type it as such. It represents an isotope of cesium.

To determine the number of beta particles emitted per second, we need to use the concept of radioactive decay and the given information.

First, let's understand what half-life means. The half-life of a radioactive isotope is the time required for half of the sample to undergo radioactive decay. In this case, the half-life of 141Cs is 32.5 days.

Given that there are 8.00*10^7 nuclei of 141Cs, we can determine the number of decays that will occur in one second.

To do that, we need to calculate the decay constant (λ), which is the probability of decay per unit time. The decay constant (λ) is related to the half-life (t1/2) by the formula:

λ = ln(2) / t1/2

Substituting the given half-life, we have:

λ = ln(2) / 32.5 days

Next, we can calculate the number of decays per second using the formula:

Decays per second = λ * N

where N is the initial number of nuclei in the sample.

Decays per second = (ln(2) / 32.5 days) * 8.00*10^7

Now let's calculate this value:

Decays per second = (ln(2) / 32.5 days) * 8.00*10^7 = 1.7*10^6

Therefore, the answer is choice B: 1.7*10^6 beta particles per second will be emitted.