The decay constant of a radioactive nuclide is $4.6 \times 10 ^ { - 3 } \mathrm {~s} ^ { - 1 }$. What is the half-life of this isotope, in minutes?

Question

Quizplus · Accepted Answer

The half-life of a radioactive isotope is given by the equation:$$t_{1/2} = \frac{ln(2)}{\lambda}$$where $t_{1/2}$ is the half-life, $\lambda$ is the decay constant, and $ln$ is the natural logarithm.Substituting the given values into the equation, we get:$$t_{1/2} = \frac{ln(2)}{4.6 \times 10 ^ { - 3 } \mathrm {~s} ^ { - 1 }} = 150 \mathrm {~s} = 2.5 \mathrm {~min}$$Therefore, the half-life of the isotope is 2.5 minutes.

The Decay Constant of a Radioactive Nuclide Is 4.6×10−3 s−14.6 \times 10 ^ { - 3 } \mathrm {~s} ^ { - 1 }4.6×10−3 s−1

The Decay Constant of a Radioactive Nuclide Is $4.6 \times 10 ^ { - 3 } \mathrm {~s} ^ { - 1 }$