A radioactive isotope undergoes decay by emission of a positron.After 2.00 h,6.720% of the initial amount of the isotope remains undecayed.What is the half-life of this isotope?

Question

Quizplus · Accepted Answer

Since the isotope undergoes decay by emission of a positron, we know that it is a type of beta decay, specifically beta plus decay. This means that the atomic number of the isotope decreases by 1, while the mass number remains constant.

Using the given information, we can set up an equation to solve for the half-life:

0.06720 = 0.5^(t/h)

where t is the time elapsed (2.00 h) and h is the half-life we want to find.

Taking the logarithm of both sides and solving for h, we get:

h = t / log(1/2) * log(0.06720 / 1)

h = 30.8 min

Therefore, the half-life of the isotope is 30.8 min.

A Radioactive Isotope Undergoes Decay by Emission of a Positron