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The mass of a hydrogen atom is \(1.00814 \mathrm{amu}\), that of a neutron is \(1.00898\) amu, and that of a helium atom (two hydrogen atoms and two neutrons) is \(4.00388\) amu. Find the binding energy as a fraction of the total energy of a helium atom.

Short Answer

Expert verified
Answer: The binding energy of a helium atom is approximately 7.59% of its total energy.

Step by step solution

01

Identify the given information and the required output

We are given the mass of hydrogen atom (\(m_H\)), neutron (\(m_n\)) and helium atom (\(m_{He}\)). We need to find the binding energy as a fraction of the total energy of a helium atom.
02

Find the mass difference between the helium atom's constituents and its actual mass

A helium atom is composed of 2 hydrogen atoms and 2 neutrons, so the combined mass of its constituents is: Combined mass = \(2m_H + 2m_n\) We can plug in the given values for hydrogen atom and neutron masses: Combined mass = \(2 \times 1.00814 \mathrm{amu} + 2 \times 1.00898 \mathrm{amu}\) Combined mass = \(4.03424 \mathrm{amu}\) Now, we can find the mass difference between the combined mass of constituents and helium atom's mass: Mass difference = Combined mass - \(m_{He}\) Mass difference = \(4.03424 \mathrm{amu} - 4.00388 \mathrm{amu}\) Mass difference = \(0.03036 \mathrm{amu}\)
03

Convert mass difference to energy using Einstein's formula

To convert mass difference to energy, we use Einstein's formula: \(E = mc^2\) Here, \(m\) is the mass difference and \(c\) is the speed of light (\(3 \times 10^8 \mathrm{m/s}\)). First, we need to convert mass difference from amu to kg. 1 amu = \(1.66054 \times 10^{-27} \mathrm{kg}\). Mass difference = \(0.03036 \mathrm{amu} \times 1.66054 \times 10^{-27} \mathrm{kg/amu}\) = \(5.04392 \times 10^{-29}\mathrm{kg}\) Now, we can find the energy: \(E = (5.04392 \times 10^{-29}\mathrm{kg}) \times (3 \times 10^8 \mathrm{m/s})^2\) \(E = 4.53886 \times 10^{-12} \mathrm{J}\)
04

Calculate binding energy as a fraction of the total energy of the helium atom.

Finally, we need to find the binding energy as a fraction of the total energy of the helium atom. The total energy of the helium atom is equal to its mass-energy: Total energy = \(m_{He}c^2\) We need to convert \(m_{He}\) to kg, as well: \(m_{He} = 4.00388 \mathrm{amu} \times 1.66054 \times 10^{-27} \mathrm{kg/amu}\) = \(6.64557 \times 10^{-27}\mathrm{kg}\) Now, we can find the total energy: Total energy = \((6.64557 \times 10^{-27}\mathrm{kg}) \times (3 \times 10^8 \mathrm{m/s})^2\) Total energy = \(5.98187 \times 10^{-10}\mathrm{J}\) Now, we find the fraction of binding energy: Fraction = \(\frac{Binding\ energy}{Total\ energy}\) = \(\frac{4.53886 \times 10^{-12} \mathrm{J}}{5.98187 \times 10^{-10} \mathrm{J}}\) Fraction = \(0.0759\) The binding energy of the helium atom is approximately 7.59% of its total energy.

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Key Concepts

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

Special Relativity
At the core of modern physics lies the theory of special relativity, introduced by Albert Einstein in 1905. This groundbreaking theory fundamentally changed our understanding of space and time, asserting that they are interconnected as a four-dimensional continuum known as spacetime. Special relativity brings forth two key postulates: the laws of physics are the same for all observers in uniform motion relative to one another, and the speed of light in a vacuum is constant for all observers, regardless of their relative motion.

These principles imply that measurements of time intervals and lengths can change depending on the relative speed of the observers—effects famously known as time dilation and length contraction. In the context of our exercise, the concept of special relativity gives us a stepping stone to comprehend the significance of Einstein's mass-energy equivalence, which is a critical component for calculating the binding energy of atomic nuclei.
Atomic Mass Unit
The atomic mass unit (amu) is a standard unit of mass that quantifies the mass of atoms and molecules. It's defined as one-twelfth the mass of a carbon-12 atom, which is approximately equal to the mass of a single nucleon (either a proton or neutron). The current definition, as per the International System of Units (SI), states that 1 amu equals 1.66053906660 x 10-27 kilograms.

This precise measurement allows scientists and students to compare the mass of different atoms and molecules on an equal footing. When we look at the exercise, the difference in mass is compared using amu. This comparison is crucial for calculating the binding energy—the energy required to separate a nucleus into its component nucleons. Understanding amu helps students grasp the small scale at which nuclear reactions and changes occur, a concept that is not always intuitive.
Einstein's Mass-Energy Equivalence
One of Einstein's most famous contributions to science is the mass-energy equivalence formula:
\(E = mc^2\).
This deceptively simple equation describes how mass (\(m\)) and energy (\(E\)) are two forms of the same thing and can be converted into each other. Here, \(c\) represents the speed of light in a vacuum, which is about 3 x 108 meters per second. This relationship implies that a small amount of mass can be converted into a vast amount of energy, which is the principle behind both atomic bombs and nuclear power.

In the solution to our problem, we've used this formula to convert the mass defect attributable to the binding energy of a helium nucleus into its energetic equivalent. It allows us to understand how much energy is released when a helium atom is formed from protons and neutrons. The binding energy, thus calculated, provides insight into the stability of an atomic nucleus. The larger the binding energy, the more stable the nucleus. This is vital for students to comprehend the natural processes occurring both on a cosmic scale, such as in stars, and at the atomic scale in nuclear reactors.

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Most popular questions from this chapter

Uniform parallel radiation is observed in two arbitrary inertial frames \(\mathbf{S}\) and \(\mathbf{S}^{\prime}\) in which it has frequencies \(v\) and \(v^{\prime}\) respectively. If \(p, g, \sigma\) denote, respectively, the radiation pressure, momentum density, and energy density of the radiation in \(S\), and primed symbols denote corresponding quantities in \(S^{\prime}\), prove \(p^{\prime} / p=g^{\prime} / g=\sigma^{\prime} / \sigma\) \(=v^{\prime 2} / v^{2} .[\) Hint: Exercise III \((17) .]\)

A particle with four-momentum \(P\) is observed by an observer who moves with four-velocity U \(_{0}\). Prove that the energy of the particle relative to that observer is \(\mathbf{U}_{0} \cdot \mathbf{P}\).

If a photon with four-momentum \(P\) is observed by two observers having four- velocities \(\mathrm{U}_{0}\) and \(\mathrm{U}_{1}\), prove that the observed frequencies are in the ratio \(\mathbf{U}_{0} \cdot \mathbf{P} / \mathbf{U}_{1} \cdot \mathbf{P}\). Hence rederive equation (17.3).

Prove that, in relativistic as in Newtonian mechanics, the time rate of change of the angular momentum \(h=\mathbf{r} \times p\) of a particle about an origin \(\mathrm{O}\) is equal to the couple \(\mathrm{r} \times \mathrm{f}\) of the applied force about \(\mathrm{O}\). If 1 \(^{\mu v}\) is the particle's four- angular momentum, and if we define the fourcouple \(G^{\mu v}=x^{\mu} F^{v}-x^{v} F^{\mu}\), where \(x^{\mu}\) and \(F^{\mu}\) are the four-vectors corresponding to \(\mathbf{r}\) and \(\mathbf{f}\), prove that \((\mathrm{d} / \mathrm{d} \tau) L^{\mu \nu}=G^{\mu \nu}\) and that the spacc-space part of this equation corresponds to the above threevector result.

Consider a head-on elastic collision of a "bullet' of rest mass \(M\) with a stationary 'target' of rest mass \(m\). Prove that the post-collision \(\gamma\)-factor of the bullet cannot exceed \(\left(m^{2}+M^{2}\right) / 2 m M\). This means that for large bullet energies (with \(\gamma\)-factors much larger than this critical value), almost the entire energy of the bullet is transferred to the target. [Hint: if \(\mathbf{P}, \mathbf{P}^{\prime}\) are the pre-and post-collision four-momenta of the bullet, and \(\mathbf{Q}, \mathbf{Q}^{\prime}\) those of the target, show, by going to the \(\mathrm{CM}\) frame, that \(\left(\mathbf{P}^{\prime}-\mathbf{Q}\right)^{2} \geqslant 0\); in fact, in the CM frame \(\mathbf{P}^{\prime}-\mathbf{Q}\) has no spatial components.] The situation is radically different in Newtonian mechanics, where the pre- and post-collision velocities of the bullet are related by \(u / u^{\prime}=(M+m) /(M-m)\). Prove this.

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