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Chapter 21: Problem 38

How much time is required for a \(5.00-g\) sample of \({ }^{233}\) Pa to decay to \(0.625 \mathrm{~g}\) if the half-life for the beta decay of ${ }^{233} \mathrm{~Pa}$ is 27.4 days?

Short Answer

Expert verified
The time required for the \(5.00\,\mathrm{g}\) sample of \({ }^{233}\mathrm{~Pa}\) to decay to \(0.625\,\mathrm{g}\) is approximately 82.2 days.

Step by step solution

01

Understand the decay process and half-life formula

In radioactive decay, the half-life is the time taken for half of a sample to decay. In other words, if a sample starts with some initial mass, M_i, after one half-life, its mass will reduce to half, and after another half-life, the remaining half will again reduce to half. The formula relating half-life (t_half), initial mass (M_i), final mass (M_f) and decay time (t) is : \[t = \frac{(\ln(\frac{M_i}{M_f})) \times t_{half}}{\ln{2}}\]
02

Identify the given values in the problem

We are given: 1. The initial mass of the sample, M_i = 5.00 g 2. The final mass of the sample, M_f = 0.625 g 3. The half-life of the beta decay process, t_half = 27.4 days
03

Substitute the given values into the formula

Now, we can substitute the given values into the formula: \[t = \frac{(\ln(\frac{5.00g}{0.625g})) \times 27.4\,days}{\ln{2}}\]
04

Compute the time required for the decay process

Calculate the time using the formula: \(t = \frac{(\ln(8)) \times 27.4\,days}{\ln{2}}\) \(t =\frac{(\ln(2^3)) \times 27.4\,days}{\ln{2}}\) We can use the logarithmic property: \(\ln(a^b)=b\ln(a)\) \(t =\frac{(3\ln{2}) \times 27.4\,days}{\ln{2}}\) Now, divide by \(\ln{2}\) on both sides: \(t = 3 \times 27.4\,days\) Finally, multiply: \(t = 82.2\,days\) The time required for the \(5.00\,\mathrm{g}\) sample of \({ }^{233}\mathrm{~Pa}\) to decay to \(0.625\,\mathrm{g}\) is approximately 82.2 days.

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Most popular questions from this chapter

Despite the similarities in the chemical reactivity of elements in the lanthanide series, their abundances in Earth's crust vary by two orders of magnitude. This graph shows the relative abundance as a function of atomic number. Which of the following statements best explains the sawtooth variation across the series? (a) The elements with an odd atomic number lie above the belt of stability. (b) The elements with an odd atomic number lie below the belt of stability. (c) The elements with an even atomic number have a magic number of protons. (d) Pairs of protons have a special stability.

Indicate the number of protons and neutrons in the following nuclei: $(\mathbf{a}){ }^{214} \mathrm{Bi},(\mathbf{b}){ }^{210} \mathrm{~Pb},(\mathbf{c})\( uranium- \)235 .$.

What particle is produced during the following decay processes: (a) actinium-215 decays to francium-211; (b) boron- 13 decays to carbon-13; (c) holmium-151 decays to terbium- \(147 ;\) (d) carbon-11 decays to boron-11?

A radioactive decay series that begins with 90 Th ends with formation of the stable nuclide \({ }^{208} \mathrm{~Pb} .\) How many alpha-particle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays?

In 2002 , a team of scientists from Russia and the United States reported the creation of the first atom of element 118 , which is named oganesson, and whose symbol is Og. The synthesis involved the collision of californium- 249 atoms with accelerated ions of an atom which we will denote X. In the synthesis, an oganesson-294 is formed together with three neutrons. $$ { }_{98}^{249} \mathrm{Cf}+\mathrm{X} \longrightarrow{ }_{118}^{294} \mathrm{Og}+3{ }_{0}^{1} \mathrm{n} $$ (a) What are the identities of isotopes X? (b) Isotope \(X\) is unusual in that it is very long-lived (its half-life is on the order of \(10^{19} \mathrm{yr}\) ) in spite of having an unfavorable neutron-to-proton ratio (Figure 21.1). Can you propose a reason for its unusual stability? (c) Oganesson-294 decays into livermorium-290 by alpha decay. Write a balanced equation for this.
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