The fusion reaction \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+\mathrm{n}\) releases about \(17.5 \mathrm{MeV}\). The available energy is \(14.1 \mathrm{MeV}\) given to the neutron. The world's energy consumption in 2013 has been estimated as \(4 \times 10^{20} \mathrm{~J}\). About one out of every 6420 hydrogen atoms in the earth's oceans is deuterium. The number of kilograms of water that would be enough to supply the world's annual energy needs is most nearly (A) \(10^8 \mathrm{~kg}\) (B) \(10^9 \mathrm{~kg}\) (C) \(10^{10} \mathrm{~kg}\) (D) \(10^{11} \mathrm{~kg}\) (E) \(10^{12} \mathrm{~kg}\)

Short Answer

Expert verified
Answer: Approximately \(10^{11} \mathrm{~kg}\) of water are needed to supply the world's annual energy needs using the given fusion reaction.

Step by step solution

01

Find the energy released per kg of deuterium in the fusion reaction

The fusion reaction releases 17.5 MeV of energy, and it is given that 14.1 MeV is transferred to the neutron. The remaining energy, which is 17.5 - 14.1 = 3.4 MeV, is given to the helium nucleus. Using the conversion factor 1 MeV = 1.60218 x 10^{-13} J, we find the energy released per reaction in joules: 3.4 MeV * 1.60218 x 10^{-13} J/MeV = 5.4478 × 10^{-13} J
02

Determine the amount of deuterium in a kg of water

It is given that one out of every 6420 hydrogen atoms in the earth's oceans is deuterium. An atom of hydrogen weighs approximately 1 amu (atomic mass unit) and deuterium weighs approximately 2 amu. The molecular weight of water (H2O) is 18 amu (1 + 1 + 16 = 18). The weight of deuterium in a kg of water can be calculated as follows: ((1 kg of water * 1000) / 18) * (1 / 6420) * 2
03

Calculate the number of fusion reactions per kg of water

Now that the weight of deuterium in a kg of water is known, we can find the number of fusion reactions per kg of water by dividing the moles of deuterium by Avogadro's number (approximately 6.022 x 10^23/mol): ( (1 kg of water * 1000) / 18) * (1 / 6420) * 2 * Avogadro's number
04

Calculate the energy released per kg of water

Now, we can multiply the number of fusion reactions carried out per kg of water by the energy released per reaction to find the energy released per kg of water: Energy released per kg of water = ( (1 kg of water * 1000) / 18) * (1 / 6420) * 2 * Avogadro's number * 5.4478 × 10^{-13} J
05

Find the number of kg of water needed to supply the world's annual energy consumption

We are given that the world's energy consumption in 2013 was 4 x 10^20 J. By dividing this by the energy released per kg of water, we can find the number of kg of water required to supply this energy: Number of kg of water = (4 x 10^20 J) / ( (1 kg of water * 1000) / 18) * (1 / 6420) * 2 * Avogadro's number * 5.4478 × 10^{-13} J The result is approximately \(10^{11} \mathrm{~kg}\). Hence the correct answer is (D).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Nuclear Fusion Reaction
Nuclear fusion is a process where two light atomic nuclei combine to form a heavier nucleus, releasing energy in the process. This is the same type of reaction that powers the sun and other stars, where hydrogen atoms fuse to form helium. In the textbook exercise, a fusion reaction combines deuterium (hydrogen with one neutron) and tritium (hydrogen with two neutrons) to create a helium nucleus and a free neutron.

Each fusion reaction releases a considerable amount of energy, characteristically measured in millions of electron volts (MeV). In our day-to-day experience, energy is more commonly referred to in joules (J). The conversion from MeV to joules is crucial to understand the large-scale energy yield of fusion reactions. Converting these immense energies into more tangible units allows one to appreciate the potential of nuclear fusion as an energy source for our planet.
Grasping Energy Conversion Calculations
Energy conversion calculations are essential to compare different forms and scales of energy. As mentioned, physicists often use units like MeV to quantify the energy released in nuclear reactions. However, to relate this to practical energy requirements, such as the world's annual energy consumption, we convert MeV to joules.

To convert the energy released per fusion reaction to joules, we use the relationship 1 MeV = 1.60218 x 10^{-13} J. This step transforms the abstract concept of nuclear energy into a form that can be used to calculate how much water would be needed to fulfil global energy needs through nuclear fusion. It highlights the incredible efficiency of nuclear fusion, where a small amount of matter can release vast amounts of energy.
Applying Avogadro's Number
Avogadro's number, approximately 6.022 x 10^23, is a fundamental constant in chemistry that represents the number of particles (usually atoms or molecules) in one mole of a substance. It serves as a bridge between the macroscopic and atomic scales. When we calculate the number of fusion reactions that can occur per kilogram of water, Avogadro's number is used to determine the number of deuterium atoms available to fuse.

In the calculation from the exercise, deuterium atoms within water are counted because only they are involved in the specific fusion reaction considered. By dividing the moles of deuterium in a kilogram of water by Avogadro's number, we can estimate the number of fusion reactions, and thus the energy, we can harness from a given amount of water. This not only solidifies the linkage between atomic-scale phenomena and real-world applications but also illustrates the potential power tucked away in the water we encounter every day.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A hypothetical atom has energy levels at \(-12 \mathrm{eV},-8 \mathrm{eV},-3 \mathrm{eV},-1 \mathrm{eV}\). a) Draw the energy levels of the atom. Label the levels with the principal quantum number. b) An electron with velocity \(v=1.326 \times 10^6 \mathrm{~m} / \mathrm{s}\) is incident on the atom. What is the de Broglie wavelength of the electron? Ignore any relativistic effects. c) Can the electron excite an electron in any of the energy levels to a higher state? If so, which are the two levels involved? d) An electron decays from the \(-3 \mathrm{eV}\) state to the \(-8 \mathrm{eV}\) state and emits a photon. What is its wavelength? What part of the electromagnetic spectrum is it in?

A laser used in retinal surgery emits photons of wavelength \(632.8 \mathrm{~nm}\) in pulses of \(0.5 \mathrm{~W}\) that last \(2 \times 10^{-2} \mathrm{~s}\). The number of photons in each pulse is nearest to (A) \(10^{10}\) (B) \(10^{15}\) (C) \(10^{20}\) (D) \(10^{25}\) (E) \(10^{30}\)

A photon of energy \(1.75 \mathrm{eV}\) has a wavelength of most nearly (A) \(710 \mathrm{~nm}\) (B) \(700 \mathrm{~nm}\) (C) \(650 \mathrm{~nm}\) (D) \(600 \mathrm{~nm}\) (E) \(550 \mathrm{~nm}\)

A boron atom bombarded by a slow neutron can undergo the following reaction: $$ \mathrm{n}+{ }^{10} \mathrm{~B} \rightarrow{ }^4 \mathrm{He}+{ }^7 \mathrm{Li}+\gamma . $$ The reaction products receive \(2.31 \mathrm{MeV}\) of kinetic energy. This kinetic energy should be computed as (A) equal to the kinetic energy of the neutron. (B) the total energy of the neutron. (C) the energy deficit of the reaction. (D) the energy deficit of the reaction minus the photon energy. (E) the energy deficit of the reaction minus the kinetic energy of the neutron.

Meteorites created in the early solar system contaned aluminum-26, which is a radioactive isotope of aluminum with a half-life of \(7.2 \times 10^5 \mathrm{yrs}\). Aluminum-26 decays first into an excited state of magnesium-26 via the reaction \({ }_{13}^{26} \mathrm{Al} \rightarrow{ }_{12}^{26} \mathrm{Mg}^*+\mathrm{e}^{+}\), where the \(e^{+}\)has energy \(2.99 \mathrm{MeV}\). (The \(e^{+}\)is a positron; see previous problem. The asterisk (*) indicates "excited.") The \({ }_{12}^{26} \mathrm{Mg}^*\) then decays into the stable isotope magnesium- 26 via the reaction \({ }_{12}^{26} \mathrm{Mg}^* \rightarrow{ }_{12}^{26} \mathrm{Mg}+\gamma\). The \(\gamma\) has energy \(1.8 \mathrm{MeV}\). a) If you were asked to calculate the de Broglie wavelength of the positron, would it be permissible to use Newtonian physics? Justify your answer. b) What is the wavelength of the photon emitted when the excited magnesium-26 decays into its ground state? What is its momentum? c) \({ }_{12}^{26} \mathrm{Mg}\) has an atomic mass of \(25.9826 \mathrm{u}\). What is the speed of the recoiling nucleus when the photon is emitted? d) What is the nucleus' kinetic energy in electron volts? e) Precise measurements indicate that for a certain meteorite \(A\) the present ratio \({ }^{26} \mathrm{Mg} /{ }^{27} \mathrm{Al}=5 \times 10^{-5}\), where \({ }^{27} \mathrm{Al}\) is the common, stable isotope of aluminum. For a meteorite \(B\) the ratio is \({ }^{26} \mathrm{Mg} /{ }^{27} \mathrm{Al}=1.55 \times 10^{-7}\). Assuming that the different ratios are due to the difference in the meteorites' times of creation, how much older is meteorite \(B\) than \(A\) ?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free