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Chapter 40: Problem 34

The half-life of a sample of \(10^{11}\) atoms that decay by alpha emission is \(10 \mathrm{~min} .\) How many alpha particles are emitted between the time interval 100 min and 200 min?

Short Answer

Expert verified
Answer: Approximately \(8.95 \times 10^9\) alpha particles are emitted between the time interval 100 and 200 minutes.

Step by step solution

01

Write down the radioactive decay formula

The radioactive decay formula is the following: \(N(t) = N_0 e^{-\lambda t}\) Here, \(N(t)\) is the number of remaining atoms at time t, \(N_0\) is the initial number of atoms, \(\lambda\) is the decay constant, and \(t\) is the time.
02

Find the decay constant using the half-life

The decay constant \(\lambda\) can be found using the following half-life formula: \(\lambda = \frac{\ln(2)}{T_{1/2}}\) Given that the half-life (\(T_{1/2}\)) is 10 minutes, we can calculate the decay constant: \(\lambda = \frac{\ln(2)}{10} = 0.0693\)
03

Find the number of remaining atoms at t = 100 minutes

Now we will use the radioactive decay formula with the initial number of atoms (\(N_0 = 10^{11}\) atoms), the decay constant (\(\lambda = 0.0693\)), and the time interval t = 100 minutes: \(N(100) = 10^{11}e^{-(0.0693)(100)} = 10^{11}e^{-6.93} \approx 9.93 \times 10^9\)
04

Find the number of remaining atoms at t = 200 minutes

Similarly, we will find the number of remaining atoms at t = 200 minutes using the radioactive decay formula: \(N(200) = 10^{11}e^{-(0.0693)(200)} = 10^{11}e^{-13.86} \approx 9.86 \times 10^8\)
05

Find the number of alpha particles emitted between t = 100 and t = 200 minutes

To find the number of alpha particles emitted during the time interval, we will find the difference between the remaining atoms at the beginning and end of the interval: Number of alpha particles emitted = \(N(100) - N(200) = 9.93 \times 10^9 - 9.86 \times 10^8 = (9.93 - 0.986) \times 10^9 \approx 8.95 \times 10^9\) Therefore, approximately \(8.95 \times 10^9\) alpha particles are emitted between the time interval 100 and 200 minutes.

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Most popular questions from this chapter

Consider the Bethe-Weizsäcker formula for the case of odd \(A\) nuclei. Show that the formula can be written as a quadratic in \(Z\) -and thus, that for any given \(A\), the binding energies of the isotopes having that \(A\) take a quadratic form, \(B=a+b Z+c Z^{2} .\) Find the most deeply bound isotope (the most stable one) having \(A=117\) using your result.

Pu decays with a half-life of 24,100 yr via a 5.25 \(\mathrm{MeV}\) alpha particle. If you have a \(1.00 \mathrm{~kg}\) spherical sample of \({ }^{239} \mathrm{Pu},\) find the initial activity in \(\mathrm{Bq} .\)

A neutron star is essentially a gigantic nucleus with mass 1.35 times that of the Sun, or mass number of order \(10^{57} .\) It consists of approximately \(99 \%\) neutrons, the rest being protons and an equal number of electrons. Explain the physics that determines these features.

Calculate the binding energy for the following two uranium isotopes: a) \({ }_{92}^{238} \mathrm{U},\) which consists of 92 protons, 92 electrons, and 146 neutrons, with a total mass of \(238.0507826 \mathrm{u}\). b) \({ }^{235} \mathrm{U},\) which consists of 92 protons, 92 electrons, and 143 neutrons, with a total mass of \(235.0439299 \mathrm{u} .\) The atomic mass unit \(\mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg} .\) Which isotope is more stable (or less unstable)?

Consider the following fusion reaction, which allows stars to produce progressively heavier elements: \({ }_{2}^{3} \mathrm{He}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{4}^{7} \mathrm{Be}+\gamma\). The mass of \({ }_{2}^{3} \mathrm{He}\) is \(3.016029 \mathrm{u}\), the mass of \({ }_{2}^{4}\) He is \(4.002603 \mathrm{u}\), and the mass of \({ }_{4}^{7} \mathrm{Be}\) is \(7.0169298 \mathrm{u}\). The atomic mass unit is \(u=1.66 \cdot 10^{-27} \mathrm{~kg} .\) Assuming the Be atom is at rest after the reaction and neglecting any potential energy between the atoms and kinetic energy of the He nuclei, calculate the minimum possible energy and maximum possible wavelength of the photon \(\gamma\) that is released in this reaction.
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