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Chapter 29: Problem 33

A certain radioactive nuclide has a half-life of \(200.0 \mathrm{s}\) A sample containing just this one radioactive nuclide has an initial activity of \(80,000.0 \mathrm{s}^{-1} .\) (a) What is the activity 600.0 s later? (b) How many nuclei were there initially? (c) What is the probability per second that any one of the nuclei decays?

Short Answer

Expert verified
Answer: (a) 10,000 s⁻¹, (b) 2.3 × 10⁷, and (c) 0.00347 s⁻¹.

Step by step solution

01

Identify the given parameters

The problem states that the radioactive nuclide has a half-life of \(200.0\mathrm{s}\) and an initial activity of \(80,000.0\mathrm{s^{-1}}\). We will be using these values to solve for (a), (b), and (c).
02

Calculate the decay constant

To solve this problem, we first need to find the decay constant (λ). The decay constant is related to the half-life of the radioactive substance by the following formula: $$ λ = \frac{\ln{2}}{t_{1/2}} $$ Where \(t_{1/2}\) is the half-life, and λ is the decay constant. In this case, the half-life is \(200.0\mathrm{s}\). Plugging these values into the equation, we get: $$ λ = \frac{\ln{2}}{200.0\mathrm{s}} = 0.00347\, \mathrm{s}^{-1} $$
03

Calculate the activity after 600 s

Now we will calculate the activity (A) after \(600.0\mathrm{s}\). The activity of a radioactive nuclide at any time can be found using the following equation: $$ A(t) = A_0 e^{-λt} $$ Where \(A_0\) is the initial activity, \(A(t)\) is the activity at time \(t\), and \(λ\) is the decay constant. Using the given initial activity (\(80,000.0\mathrm{s^{-1}}\)) and decay constant obtained in step 2, we can find the activity after \(600.0 \mathrm{s}\): $$ A(600) = 80,000.0\, \mathrm{s}^{-1} \times e^{-0.00347 \times 600} $$ $$ A(600) = 10,000 \, \mathrm{s}^{-1} $$ So, the activity after 600 s is \(10,000\mathrm{s^{-1}}\). (Answer for (a))
04

Calculate the number of initial nuclei

The next step is to find the number of initial nuclei (N0). We can do this using the formula: $$ A_0 = λN_0 $$ We already have the values for the initial activity (\(A_0 = 80,000.0\, \mathrm{s}^{-1}\)) and decay constant (\(λ = 0.00347\, \mathrm{s}^{-1}\)) from previous steps, so we can solve for \(N_0\): $$ N_0 = \frac{A_0}{λ} = \frac{80,000.0\, \mathrm{s}^{-1}}{0.00347\, \mathrm{s}^{-1}} = 2.3 \times 10^7 $$ The number of initial nuclei is \(2.3 \times 10^7\). (Answer for (b))
05

Calculate the probability per second of any nucleus decaying

Finally, we need to find the probability per second that any one nucleus decays. This probability is given by the decay constant, λ, which we calculated in step 2. Therefore, the probability per second of any nucleus decaying is \(0.00347\, \mathrm{s}^{-1}\). (Answer for (c)) In conclusion, (a) the activity after 600 s is \(10,000\, \mathrm{s}^{-1}\), (b) the number of initial nuclei is \(2.3\times10^7 \), and (c) the probability per second of any nucleus decaying is \(0.00347\, \mathrm{s}^{-1}\).

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Most popular questions from this chapter

The nucleus in a \({ }_{7}^{12} \mathrm{~N}\) atom captures one of the atom's electrons, changing the nucleus to \({ }_{6}^{12} \mathrm{C}\) and emitting a neutrino. What is the total energy of the emitted neutrino? [Hint:You can use the classical expression for the kinetic energy of the ${ }_{6}^{12} \mathrm{C}$ atom and the extremely relativistic expression for the kinetic energy of the neutrino.]
The Physics at Home in Section 29.4 suggests tossing coins as a model of radioactive decay. An improved version is to toss a large number of dice instead of coins: each die that comes up a "one" represents a nucleus that has decayed. Suppose that \(N\) dice are tossed. (a) What is the average number of dice you expect to decay on one toss? (b) What is the average number of dice you expect to remain undecayed after three tosses? (c) What is the average number of dice you expect to remain undecayed after four tosses? (d) What is the half-life in numbers of tosses?
Calculate the activity of \(1.0 \mathrm{g}\) of radium- 226 in Ci.
The last step in the carbon cycle that takes place inside stars is \(\mathrm{p}+^{15} \mathrm{N} \rightarrow^{12} \mathrm{C}+(?) .\) This step releases \(5.00 \mathrm{MeV}\) of energy. (a) Show that the reaction product "(?)" must be an \(\alpha\) particle. (b) Calculate the atomic mass of helium-4 from the information given. (c) In order for this reaction to occur, the proton must come into contact with the nitrogen nucleus. Calculate the distance \(d\) between their centers when they just "touch." (d) If the proton and nitrogen nucleus are initially far apart, what is the minimum value of their total kinetic energy necessary to bring the two into contact?
A water sample is found to have \(0.016 \%\) deuterium content (that is, $0.016 \%\( of the hydrogen nuclei in the water are \)^{2} \mathrm{H}$ ). If the fusion reaction \(\left(^{2} \mathrm{H}+^{2} \mathrm{H}\right)\) yields 3.65 MeV of energy on average, how much energy could you get from \(1.00 \mathrm{L}\) of the water? (There are two reactions with approximately equal probabilities; one yields \(4.03 \mathrm{MeV}\) and the other \(3.27 \mathrm{MeV} .\) ) Assume that you are able to extract and fuse \(87.0 \%\) of the deuterium in the water. Give your answer in kilowatt hours.
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