How much energy would be released if six hydrogen atoms and six neutrons were combined to form \(^{12}_{6} \mathrm{C} ?\)

Short Answer

Expert verified
To calculate the energy released when six hydrogen atoms and six neutrons combine to form a carbon-12 nucleus, we first find the total mass of the reactants: \(M_{initial} = (6 \times m_p) + (6 \times m_n)\). Then, we calculate the mass defect: \(\Delta m = M_{initial} - m_{C}\). Finally, we convert the mass defect into energy using Einstein's equation, E=mc^2: \(E = \Delta m \times c^2\).

Step by step solution

01

Identify given information

We are given the following information: 1. Six hydrogen atoms (each hydrogen atom has 1 proton) 2. Six neutrons 3. The final product is a carbon-12 nucleus
02

Calculate the mass of the initial particles

To find out how much energy is released, we first need to find the total mass of the reactants (hydrogen atoms and neutrons). For hydrogen atoms: Mass of 1 proton = mass of 1 hydrogen atom = \(m_p = 1.007276466812 \, amu\) For neutrons: Mass of 1 neutron = \(m_n = 1.008664915 \, amu\) Now, calculate the total mass (M_initial) of six hydrogen atoms and six neutrons: \(M_{initial} = (6 \times m_p) + (6 \times m_n)\)
03

Calculate the mass of the final product (carbon-12)

To determine the mass of the final product, carbon-12, we can use its atomic mass unit (amu) value: Mass of 1 carbon-12 nucleus = \(m_{C} = 12 \, amu\)
04

Calculate the mass defect

The mass defect is the difference between the total mass of the reactants and the mass of the final product. Calculate the mass defect (Δm): \(\Delta m = M_{initial} - m_{C}\)
05

Convert mass defect to energy

To calculate the energy released, we need to convert the mass defect into energy using Einstein's equation, E=mc^2, where E is energy, m is mass, and c is the speed of light: Speed of light (c) = \(3 \times 10^8 \, m/s\) First, convert the mass defect from amu to kilograms by multiplying it by the conversion factor \(1 \, amu = 1.66054 \times 10^{-27} \, kg\). Next, calculate the energy (E) released: \(E = \Delta m \times c^2\) The result is the amount of energy released when six hydrogen atoms and six neutrons combine to form a carbon-12 nucleus.

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.

Mass-Energy Equivalence
Imagine the famous equation from Albert Einstein, E=mc^2, which allows us to connect the realms of mass and energy, a concept known as mass-energy equivalence. Essentially, it tells us that mass can be converted into energy and vice versa. That 'c' in the equation stands for the speed of light, which is a rather large number (\(3 \times 10^8 \text{ m/s}\)), making 'c^2' extremely large. Therefore, even a small amount of mass (\(m\)) can be converted into a huge amount of energy (\(E\)). This isn't just a theoretical concept — it's the core principle behind nuclear power and weapons, and it is also why the sun and other stars shine, through the process of nuclear fusion. The energy calculated in our textbook exercise is derived from this fundamental principle.

In practical terms, for the exercise we discussed, this means that when several small particles like hydrogen atoms come together to form a heavier nucleus, like that of carbon, they lose some mass. That lost mass is not gone — instead, it is transformed into energy, which follows from the mass-energy equivalence concept.
Mass Defect
Mass defect is a term we use to describe the phenomenon where the mass of a nucleus is less than the sum of its individual nucleons (protons and neutrons). This 'missing' mass is key to understanding nuclear fusion and fission. The steps of the textbook problem guide us through calculating the mass defect by subtracting the mass of the final nucleus from the total mass of the individual protons and neutrons before they combined. This difference represents the binding energy, the energy needed to hold the nucleus together, or conversely, the energy released when the nucleus is formed from these protons and neutrons.

Using the example from our exercise, the combined initial mass of six hydrogen atoms and six neutrons is greater than the mass of the resulting carbon-12 nucleus. The mass defect quantifies that difference and, when plugged into Einstein's equation, tells us exactly how much energy we can expect to be released during the process.
Atomic Mass Unit
The atomic mass unit (amu) is a standard unit of mass that quantifies mass on an atomic or molecular scale. It's defined as one twelfth of the mass of an unbound neutral atom of carbon-12 at rest and in its ground state. Because atoms and subatomic particles are so minuscule, standard units of mass like grams or kilograms aren't practical — they would involve inconveniently small decimal numbers. This is where amus come in handy, helping scientists and students work with more manageable figures.

In our example, we used the fact that the mass of a carbon-12 nucleus is defined as 12 amu. By comparing this with the initial mass of the hydrogen atoms and neutrons in amus, we were able to calculate the mass defect in the same unit, making it simpler to then convert that mass into energy (using Einstein's equation again), a crucial step for solving the problem at hand.
Nuclear Fusion
Nuclear fusion is a powerful process that occurs when two light atomic nuclei combine to form a heavier nucleus, releasing energy in the process. It's the exact opposite of nuclear fission, where a heavy nucleus splits into two lighter nuclei. Fusion powers the sun and stars, as hydrogen atoms fuse to form helium and other elements in stellar cores under extremely high pressure and temperature conditions.

In the educational exercise we're considering, the fusion of six hydrogen atoms and six neutrons to form a carbon-12 nucleus is a simplified model of what happens in stars, albeit under different conditions. Understanding fusion is not only fundamental to astrophysics but also has practical implications. Researchers are working tirelessly to harness nuclear fusion as a clean energy source here on Earth. This technology has the potential to provide a near-limitless supply of energy with relatively benign waste products compared to current nuclear fission reactors.

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

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?

The \(^{210}\) Po source used in a physics laboratory is labeled as having an activity of \(1.0 \mu \mathrm{Ci}\) on the date it was prepared. A student measures the radioactivity of this source with a Geiger counter and observes 1500 counts per minute. She notices that the source was prepared 120 days before her lab. What fraction of the decays is she observing with her apparatus?

Verify that the total number of nucleons, and total charge are conserved for each of the following fusion reactions in the proton-proton chain. (i) \(^{1} \mathrm{H}+^{1} \mathrm{H} \rightarrow^{2} \mathrm{H}+e^{+}+v_{\mathrm{e}},\) (ii) \(^{1} \mathrm{H}+^{2} \mathrm{H} \rightarrow^{3} \mathrm{He}+\gamma,\) and (iii) \(^{3} \mathrm{He}+^{3} \mathrm{He} \rightarrow^{4} \mathrm{He}+^{1} \mathrm{H}+^{1} \mathrm{H}\). (List the value of each of the conserved quantities before and after each of the reactions.)

To obtain the most precise value of the binding energy per nucleon, it is important to take into account forces between nucleons at the surface of the nucleus. Will surface effects increase or decrease estimates of BEN?

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?

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