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Chapter 19: Problem 23

Lead-210 has a half-life of \(20.4\) years. This isotope decays by beta particle emission. A counter registers \(1.3 \times 10^{4}\) disintegrations in five minutes. How many grams of \(\mathrm{Pb}-210\) are there?

Short Answer

Expert verified
Answer: The mass of Lead-210 present in the sample is approximately 1.39x10^-12 grams.

Step by step solution

01

Calculate the decay constant

The decay constant \(\lambda\) can be found using the formula relating half-life \(T_{1/2}\) and decay constant(lambda): \(\lambda = \frac{ln 2}{T_{1/2}}\) The half-life of Lead-210 is given as \(20.4\) years. Use this value to calculate the decay constant: \(\lambda = \frac{ln 2}{20.4} = 0.033915\) (per year).
02

Find the number of Lead-210 nuclei

The number of disintegrations in five minutes is given as \(1.3 \times 10^4\). Using this information together with the decay constant, we can determine the number of Lead-210 nuclei, \(N\): \(Disintegrations = \lambda \times N \times time\) The given disintegrations are recorded in five minutes, but the decay constant was calculated in years. Therefore, we need to convert the time to years: \(5 \text{ minutes} = \frac{5}{60\times24\times365} = 9.506x10^{-6} \text{ years}\) Now, replacing the known values in the equation and solving for \(N\): \(1.3 \times 10^4 = 0.033915 \times N \times 9.506x10^{-6}\) \(N = \frac{1.3 \times 10^4}{0.033915 \times 9.506x10^{-6}} = 4.015x10^{9}\) nuclei
03

Convert the number of nuclei to grams

With the number of nuclei calculated, now it is necessary to convert it into grams. Using Avogadro's number, the molar mass of Lead-210, and the calculated number of nuclei, we can find the grams of Lead-210: \( grams = \frac{Number\_of\_nuclei \times Molar\_Mass\_of\_Pb\_210}{Avogadro's\_Number}\) \( grams = \frac{4.015x10^{9} \text{ nuclei} \times 210 \text{ g/mol}}{6.022x10^{23} \text{ nuclei/mol}}\) \( grams = 1.39x10^{-12}\) g The mass of Lead-210 present in the sample is approximately \(1.39x10^{-12}\) grams.

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Most popular questions from this chapter

A sample of a wooden artifact gives \(5.0\) disintegrations \(/ \mathrm{min} / \mathrm{g}\) carbon. The half-life of carbon-14 is 5730 years, and the activity of C-14 in wood just cut down from a tree is \(15.3\) disintegrations \(/ \mathrm{min} / \mathrm{g}\) carbon. How old is the wooden artifact?

Bromine-82 has a half-life of 36 hours. A sample containing Br-82 was found to have an activity of \(1.2 \times 10^{5}\) disintegrations \(/ \mathrm{min}\). How many grams of Br-82 were present in the sample? Assume that there were no other radioactive nuclides in the sample.

Balance the following equations by filling in the blanks. (a) \({ }_{92}^{235} \mathrm{U}+{ }_{0} n \longrightarrow{ }_{54}^{137}=2{ }_{0}^{1} n+\) (b) \({ }_{90}^{232} \mathrm{Th}+{ }_{6}^{12}\) \(\longrightarrow\) \(1 . n+{ }_{96}^{240} \mathrm{Cm}\) (c) \({ }_{2}^{4} \mathrm{He}+{ }_{42}^{96} \mathrm{Mo} \longrightarrow{ }_{43}^{100}\) (d) \(+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{84}^{210}+{ }_{0}^{1} n\)

Radium-226 decays by alpha emission to radon-222. Suppose that \(25.0 \%\) of the energy given off by one gram of radium is converted to electrical energy. What is the minimum mass of lithium that would be needed for the voltaic cell \(\mathrm{Li}\left|\mathrm{Li}^{+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\), at standard conditions, to produce the same amount of electrical work \(\left(\Delta G^{\circ}\right) ?\)

Write balanced nuclear equations for the bombardment of (a) Fe-54 with an alpha particle to produce another nucleus and two protons. (b) Mo-96 with deuterium \(\left({ }_{1}^{2} \mathrm{H}\right)\) to produce a neutron and another nucleus. (c) Ar-40 with an unknown particle to produce \(\mathrm{K}-43\) and a proton. (d) a nucleus with a neutron to produce a proton and \(\mathrm{P}-31 .\)
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