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Chapter 10: Problem 37

A radioactive sample initially contains \(2.40 \times 10^{-2}\) mol of a radioactive material whose half-life is 6.00 h. How many moles of the radioactive material remain after \(6.00 \mathrm{h}\) ? After 12.0 h? After 36.0 h?

Short Answer

Expert verified
After 6 hours, approximately \(1.20 \times 10^{-2}\) mol of radioactive material remain. After 12 hours, approximately \(6.00 \times 10^{-3}\) mol remain. After 36 hours, approximately \(3.75 \times 10^{-4}\) mol remain.

Step by step solution

01

Understand the radioactive decay formula

The radioactive decay equation is given by: \[N = N_0 \times 2^{-\frac{t}{T}}\] Where \(N\) is the remaining number of moles after time \(t\), \(N_0\) is the initial amount of moles, \(T\) is the half-life, and \(t\) is the elapsed time.
02

Identify given values

The given values are: Initial moles of radioactive material (\(N_0\)) = \(2.40 \times 10^{-2}\) mol Half-life (\(T\)) = 6.00 hours We need to calculate the remaining moles after time periods of 6 hours, 12 hours, and 36 hours.
03

Calculate the remaining moles after 6 hours

Using the radioactive decay equation and the given values, we can calculate the remaining moles after 6 hours. \(N = N_0 \times 2^{-\frac{t}{T}}\) \(N = 2.40 \times 10^{-2} \times 2^{\frac{-6}{6}}\) \(N = 2.40 \times 10^{-2} \times 2^{-1}\) \(N = 2.40 \times 10^{-2} \times 0.5\) After 6 hours, the remaining moles are: \(N \approx 1.20 \times 10^{-2}\) mol
04

Calculate the remaining moles after 12 hours

Similarly, we can calculate the remaining moles after 12 hours. \(N = N_0 \times 2^{-\frac{t}{T}}\) \(N = 2.40 \times 10^{-2} \times 2^{\frac{-12}{6}}\) \(N = 2.40 \times 10^{-2} \times 2^{-2}\) \(N = 2.40 \times 10^{-2} \times 0.25\) After 12 hours, the remaining moles are: \(N \approx 6.00 \times 10^{-3}\) mol
05

Calculate the remaining moles after 36 hours

Finally, we can calculate the remaining moles after 36 hours. \(N = N_0 \times 2^{-\frac{t}{T}}\) \(N = 2.40 \times 10^{-2} \times 2^{\frac{-36}{6}}\) \(N = 2.40 \times 10^{-2} \times 2^{-6}\)
\(N = 2.40 \times 10^{-2} \times \frac{1}{64}\) After 36 hours, the remaining moles are: \(N \approx 3.75 \times 10^{-4}\) mol Thus, the remaining moles of radioactive material after 6 hours are approximately \(1.20 \times 10^{-2}\) mol, after 12 hours are approximately \(6.00 \times 10^{-3}\) mol, and after 36 hours are approximately \(3.75 \times 10^{-4}\) mol.

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Most popular questions from this chapter

According to your lab partner, a 2.00-cm-thick sodium-iodide crystal absorbs all but \(10 \%\) of rays from a radioactive source and a 4.00-cm piece of the same material absorbs all but \(5 \%\) ? Is this result reasonable?

A rare decay mode has been observed in which \(^{222} \mathrm{Ra}\) emits a \(^{14} \mathrm{C}\) nucleus. (a) The decay equation is \(^{222} \mathrm{Ra} \rightarrow^{A} \mathrm{X}+^{14} \mathrm{C} .\) Identify the nuclide \(^{A} \mathrm{X} .\) (b) Find the energy emitted in the decay. The mass of \(222 \mathrm{Ra}\) is 222.015353 u.

One half the \(\gamma\) rays from \(^{99 \mathrm{m}} \mathrm{Tc}\) are absorbed by a 0.170-mm-thick lead shielding. Half of the \(\gamma\) rays that pass through the first layer of lead are absorbed in a second layer of equal thickness. What thickness of lead will absorb all but one in 1000 of these \(\gamma\) rays?

(a) Calculate the energy released in the neutron-induced fission reaction \(n+^{235} \mathrm{U} \rightarrow^{92} \mathrm{Kr}+^{142} \mathrm{Ba}+2 n,\) given \(m\left(^{92} \mathrm{Kr}\right)=91.926269 \mathrm{u}\) and \(m\left(^{142} \mathrm{Ba}\right)=141.916361 \mathrm{u} .\) (b) Confirm that the total number of nucleons and total charge are conserved in this reaction.

(a) Calculate BEN for \(^{235} \mathrm{U}\), the rarer of the two most common uranium isotopes; (b) Calculate BEN for \(^{238} \mathrm{U}\) (Most of uranium is \(^{238} \mathrm{U}\).)
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