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Chapter 21: Problem 26

A freshly isolated sample of \(90 \mathrm{Y}\) was found to have an activity of \(9.8 \times 10^{5}\) disintegrations per minute at 1: 00 P.M. on December \(3,2000 .\) At 2: 15 P.M. on December \(17,2000,\) its activity was redetermined and found to be \(2.6 \times 10^{4}\) disintegrations per minute. Calculate the half-life of \(90 \mathrm{Y}\).

Short Answer

Expert verified
The half-life of \(90 \mathrm{Y}\) is around \(64\) minutes.

Step by step solution

01

Understand Definitions & Constants

Here we are dealing with the concept of half-life of a radioactive element, which is the time required for the activity of a given quantity of a radioactive substance to decay to half of its initial value. We need to find this half-life for \(90 \mathrm{Y}\). Take note of the decay constant \(\lambda\), which is related to half-life \(T_{1/2}\) by the formula \(T_{1/2} = \frac{0.693}{\lambda}\).
02

Define the knowns

We know the initial activity \(N_0 = 9.8 \times 10^{5}\) decays per minute, the final activity \(N = 2.6 \times 10^{4}\) decays per minute, and the time interval which is 14 days and 15 mins equal to \(20085\) minutes. And we are to solve for \(T_{1/2}\), the half-life.
03

Apply the Radioactive Decay Law

Since the decay laws states \(N = N_0 e^{-\lambda t}\), after rearranging it to solve for decay constant we have \(\lambda = - \frac{1}{t} \ln \frac{N}{N_0}\), substituting the known values and calculating will yield \(\lambda\).
04

Find the half-life

Having found \(\lambda\), we can substitute it in the relation \(T_{1/2} = \frac{0.693}{\lambda}\) to find \(T_{1/2}\), the half-life of the \(90 \mathrm{Y}\).

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Most popular questions from this chapter

Tritium contains one proton and two neutrons. There is no proton-proton repulsion present in the nucleus. Why, then, is tritium radioactive?

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