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Chapter 19: Problem 34

Phosphorus-32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is \(14.3\) days. What mass of phosphorus-32 is left from an original sample of \(175 \mathrm{mg}\) \(\mathrm{Na}_{3}{ }^{32} \mathrm{PO}_{4}\) after \(35.0\) days? Assume the atomic mass of \({ }^{32} \mathrm{P}\) is \(32.0 \mathrm{u}\).

Short Answer

Expert verified
The mass of phosphorus-32 remaining after 35 days = (\(175 \,mg \times \frac{32}{(3\times23) + (32 + 16 \times4)}\)) x \((\frac{1}{2})^{\frac{35}{14.3}}\) ≈ 21.15 mg.

Step by step solution

01

Find the total number of half-lives elapsed during the 35 days

We are given that half-life of the phosphorus-32 is 14.3 days. We need to determine how many half-lives have passed in 35 days. To do this, we can divide the total time elapsed (35 days) by the half-life (14.3 days): Number of half-lives = \(\frac{35}{14.3}\)
02

Calculate the fraction of phosphorus-32 remaining

Since the decay is exponential in form, we can use the formula: remaining fraction = \((\frac{1}{2})^{number \, of \,half-lives}\) plug-in the number of half-lives to get the remaining fraction, \((\frac{1}{2})^{\frac{35}{14.3}}\)
03

Determine the mass of the original phosphorus-32

We are given that the original mass of \(\mathrm{Na}_{3}{ }^{32}\mathrm{PO}_{4}\) is \(175 \, mg\) One Phosphorus atom is present in one molecule of \(\mathrm{Na}_{3}{ }^{32}\mathrm{PO}_{4}\) Hence the mass of the Phosphorus-32 in the original sample can be calculated using the formula: Mass of phosphorus-32 = \(175 \,mg \times \frac{32}{(3\times23) + 32 + 16 \times4}\) Mass of phosphorus-32 = \(175 \,mg \times \frac{32}{(3\times23) + (32 + 16 \times4)}\)
04

Determine the mass of phosphorus-32 remaining

Now that we have the mass of the original phosphorus-32 and the remaining fraction, we can find the mass of phosphorus-32 left after 35 days. Mass of phosphorus-32 remaining = (Mass of original phosphorus-32) x (Remaining fraction) Mass of phosphorus-32 remaining = (\(175 \,mg \times \frac{32}{(3\times23) + (32 + 16 \times4)}\)) x \((\frac{1}{2})^{\frac{35}{14.3}}\) After solving the above expression, we get the mass of phosphorus-32 remaining in the sample after 35 days.

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Most popular questions from this chapter

In each of the following radioactive decay processes, supply the missing particle. a. \({ }^{60} \mathrm{Co} \rightarrow{ }^{60} \mathrm{Ni}+\) ? b. \({ }^{97} \mathrm{Tc}+? \rightarrow{ }^{97} \mathrm{Mo}\) c. \({ }^{99} \mathrm{Tc} \rightarrow{ }^{99} \mathrm{Ru}+\) ? d. \({ }^{239} \mathrm{Pu} \rightarrow{ }^{235} \mathrm{U}+\) ?

Which type of radioactive decay has the net effect of changing a neutron into a proton? Which type of decay has the net effect of turning a proton into a neutron?

In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, \(\mathrm{Sg}\), in honor of Glenn \(\mathrm{T}\). Seaborg, discoverer of the transuranium elements. a. \({ }^{263} \mathrm{Sg}\) was produced by the bombardment of \({ }^{249} \mathrm{Cf}\) with a beam of \({ }^{18} \mathrm{O}\) nuclei. Complete and balance an equation for this reaction. b. \({ }^{263} \mathrm{Sg}\) decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \({ }^{263} \mathrm{Sg}\) ?

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The radioactive isotope \({ }^{247} \mathrm{Bk}\) decays by a series of \(\alpha\) -particle and \(\beta\) -particle productions, taking \({ }^{247} \mathrm{Bk}\) through many transformations to end up as \({ }^{207} \mathrm{~Pb}\). In the complete decay series, how many \(\alpha\) particles and \(\beta\) particles are produced?
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