A radioactive isotope undergoes decay by emission of a positron.After 2.00 h,6.650% of the initial amount of the isotope remains undecayed.What is the half-life of this isotope?

Question

Quizplus · Accepted Answer

Since the isotope emits a positron, it undergoes beta plus decay. This means that the atomic number of the daughter nucleus will be one less than that of the parent nucleus. The percent of the initial amount remaining after a certain time t is given by the following equation:

% remaining = (1/2)^(t/T1/2) * 100

where T1/2 is the half-life of the isotope.

We can use the given information to solve for the half-life:

6.650% = (1/2)^(2/T1/2) * 100
0.0665 = (1/2)^(2/T1/2)
log(0.0665) = (2/T1/2) * log(1/2)
log(0.0665) = -2/T1/2
T1/2 = -2/log(0.0665)
T1/2 = 30.7 min

Therefore, the half-life of the isotope is 30.7 min, which corresponds to choice B.

A Radioactive Isotope Undergoes Decay by Emission of a Positron