Question

1. The \(^{{\rm{210}}}\) Po source used in a physics laboratory is labelled as having an activity of \(1.0\,{\rm{\mu Ci}}\) on the date it was prepared. A student measures the radioactivity of this source with a Geiger counter and observes 1500 counts per minute. She notices that the source was prepared 120 days before her lab. What fraction of the decays is she observing with her apparatus?
2. Identify some of the reasons that only a fraction of the αs emitted are observed by the detector.

Short answer

1. The fraction of the decays is\(1.23 \times {10^{ - 3}}\).
2. As it is not possible to cover the entire area of the surface from which decay occurs. So,the counter will show the small fraction of the actual decay only.

Step-by-step solution

Concept Introduction

For a given number of nuclei, the shorter the half-life, the more decays per unit time. As a result, activity R should be proportional to N, the number of radioactive nuclei, and inversely proportional to t1/2, their half-life. In reality, your instincts are spot on. It can be demonstrated that a source's activity is\(R = \frac{{0.693N}}{{{t_{1/2}}}}\).

Find fraction of the decays

1. The following is the relationship between beginning activity and activity after time t:

\(R = {R_0}{e^{ - \lambda t}}\)

Where, \({R_0}\) is the initial activity, \(R\) is the final activity \(\lambda = \frac{{0.693}}{{{t_{1/2}}}}\) and \(t\) is the time.

As we all know, \(1\,{\rm{Bq}} = 2.7 \times {10^{ - 11}}\,{\rm{Ci}}\)

So,

\({R_0} = 1\,{\rm{\mu Ci}} = 1 \times {10^{ - 6}}\,{\rm{Ci}} = 3.70 \times {10^4}\,{\rm{Bq}}\)

Put the value of into the equation now.

\(\lambda = \frac{{0.693}}{{{t_{1/2}}}},{t_{1/2}} = 138.4\;\,{\rm{d}}\) and \(t = 120\,{\rm{d}}\)and find the value of R

\(\begin{aligned}R{\rm{ }} = \left( {3.7 \times {{10}^4}\,{\rm{Bq}}} \right){e^{ - \frac{{0.693(120\,{\rm{d}})}}{{138.4\,{\rm{d}}}}}}\\ = 2.03 \times {10^4}\,{\rm{Bq}}\end{aligned}\)

The Geiger counter is now counting at 1500 counts per minute. As a result, the observed activity is as follows:

\(\begin{aligned}{R}{\rm{ }} = 1500\,{\rm{/min}}\\ = 25\;\,{{\rm{s}}^{{\rm{ - 1}}}}\\ = 25\;\,{\rm{Bq}}\end{aligned}\)

As a result, the observed decay fraction is given by:

\(\begin{aligned}\frac{{{R^t}}}{R}{\rm{ }} = \frac{{25\;\,{\rm{Bq}}}}{{2.03 \times {{10}^4}\,{\rm{Bq}}}}\\ = 1.23 \times {10^{ - 3}}\end{aligned}\)

Therefore, the fraction of the decays is\(1.23 \times {10^{ - 3}}\).

Write reasons for fraction of the \({\bf{\alpha }}\)s emitted are observed by the detector.

Because it is impossible to cover the entire surface from which decay occurs.

As a result, the counter will only display a small fraction of the actual decay.