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Because the neutron has no charge, its mass must be found in some way other than by using a mass spectrometer. When a neutron and a proton meet (assume both to be almost stationary), they combine and form a deuteron, emitting a gamma ray whose energy is \(2.2233 \mathrm{MeV}\). The masses of the proton and the deuteron are 1.007276467 u and \(2.013553212 \mathrm{u},\) respectively. Find the mass of the neutron from these data.

Short Answer

Expert verified
The mass of the neutron is approximately 1.00866 u.

Step by step solution

01

Write the Reaction Equation

The nuclear reaction occurring here is a neutron () and a proton () combining to form a deuteron () while emitting a gamma ray (). This can be written as: \[ n + p \rightarrow D + \gamma \] where \( n \) is the neutron, \( p \) is the proton, \( D \) is the deuteron, and \( \gamma \) is the gamma ray.
02

Understand Energy-Mass Relationship

According to Einstein's mass-energy equivalence principle, any loss in mass during a nuclear reaction is converted into energy. This can be described by the equation: \[ E = \Delta mc^2 \] where \( E \) is the energy emitted (2.2233 MeV), \( \Delta m \) is the change in mass, and \( c \) is the speed of light.
03

Calculate Mass Defect

First, calculate the mass defect \( \Delta m \), which is the difference in mass on the left and right sides of the reaction equation. Using the given masses:\[ \Delta m = (m_n + m_p) - m_D \] where \( m_n \) is the mass of the neutron, \( m_p = 1.007276467 \text{ u} \) is the mass of the proton, and \( m_D = 2.013553212 \text{ u} \) is the mass of the deuteron.
04

Relate Mass Defect to Energy using MeV

Convert energy to mass units to find \( \Delta m \). The energy of the emitted gamma ray is given as 2.2233 MeV. 1 u is equivalent to \( 931.5 \text{ MeV} \), thus the change in mass \( \Delta m \) can be expressed as: \[ \Delta m \times 931.5 = 2.2233 \] Therefore, \( \Delta m = \frac{2.2233}{931.5} \approx 0.00239 \text{ u} \).
05

Solve for the Mass of the Neutron

Using the mass defect, we can calculate the mass of the neutron \( m_n \) as follows:\[ m_n = \Delta m + m_D - m_p \]Substitute the known values: \[ m_n = 0.00239 \text{ u} + 2.013553212 \text{ u} - 1.007276467 \text{ u} \]Thus, \( m_n \approx 1.00866 \text{ u} \).

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Key Concepts

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

Neutron Mass
The mass of a neutron is fundamental in nuclear physics. Neutrons are neutral particles found within atomic nuclei. They slightly outweigh protons, yet their masses are similar. Determining a neutron's mass can be challenging due to its lack of electric charge. While protons and charged particles are easily analyzed with a mass spectrometer, neutrons are not. Instead, scientists use indirect methods like nuclear reactions to deduce neutron mass. In such nuclear reactions, the known masses of other particles help calculate the neutron’s mass. The assumed almost stationary state of particles facilitates simpler calculations, allowing researchers to achieve accurate mass estimations for neutrons.
Mass-Energy Equivalence
Mass-energy equivalence, introduced by Albert Einstein, reveals an intimate relationship between mass and energy. Represented by the famous equation \( E = mc^2 \), this principle explains how mass converts to energy and vice versa. In nuclear reactions, a small change in mass, or mass defect, results in significant energy release. This is because energy output is amplified by the square of the speed of light, \( c \). In the context of our exercise, the mass lost during the formation of a deuteron is converted to the energy of the emitted gamma ray. Understanding this principle is crucial for comprehending energy dynamics in nuclear reactions and other high-energy processes.
Nuclear Reaction
Nuclear reactions involve the transformation of one or more atomic nuclei into different nuclei or particles. Unlike chemical reactions, which primarily involve electron exchanges, nuclear reactions significantly alter atomic composition and release or absorb large energy quantities. In the exercise, a nuclear reaction showcases a neutron and a proton creating a deuteron and a gamma ray. During this process, the combined mass of the neutron and proton reduces slightly to form a deuteron, illustrating a mass defect. This defect is then converted into energy, manifesting as the emitted gamma ray. By employing nuclear reactions, scientists can unravel insights into nucleic properties and particle interactions.
Gamma Ray Emission
Gamma rays are high-energy electromagnetic waves released from atomic nuclei in certain reactions. They harbor greater energy than visible light, placing them at the short-wavelength end of the electromagnetic spectrum. Often produced in nuclear reactions, gamma rays originate from energy transitions within a nucleus, such as excited nuclear states reverting to stability. In the provided exercise, the fusion of a neutron and proton releasing a gamma ray illustrates such a transition. The gamma ray's energy, quantified as 2.2233 MeV in this scenario, corresponds to the mass-energy equivalence principle. It represents the excess energy after mass defect conversion. Understanding gamma ray emission is vital, as it plays significant roles in astrophysics, medical imaging, and radiation therapy.

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Most popular questions from this chapter

Under certain rare circumstances, a nucleus can decay by emitting a particle more massive than an alpha particle. Consider the decays $${ }^{223} \mathrm{Ra} \rightarrow{ }^{209} \mathrm{~Pb}+{ }^{14} \mathrm{C} \quad \text { and } \quad{ }^{223} \mathrm{Ra} \rightarrow{ }^{219} \mathrm{Rn}+{ }^{4} \mathrm{He}$$ Calculate the \(Q\) value for the (a) first and (b) second decay and determine that both are energetically possible. (c) The Coulomb barrier height for alpha-particle emission is \(30.0 \mathrm{MeV}\). What is the barrier height for \({ }^{14} \mathrm{C}\) emission? (Be careful about the nuclear radii.) The needed atomic masses are $$ \begin{aligned} &\begin{array}{llll} { }^{223} \mathrm{Ra} & 223.01850 \mathrm{u} & { }^{14} \mathrm{C} & 14.00324 \mathrm{u} \end{array}\\\ &{ }^{209} \mathrm{~Pb} \quad 208.98107 \mathrm{u} \quad{ }^{4} \mathrm{He} \quad 4.00260 \mathrm{u}\\\ &{ }^{219} \mathrm{Rn} \quad 219.00948 \mathrm{u} \end{aligned} $$

The nuclide \({ }^{198} \mathrm{Au},\) with a half-life of \(2.70 \mathrm{~d},\) is used in cancer therapy. What mass of this nuclide is required to produce an activity of \(250 \mathrm{Ci} ?\)

If the unit for atomic mass were defined so that the mass of \({ }^{1} \mathrm{H}\) were exactly \(1.000000 \mathrm{u},\) what would be the mass of (a) \({ }^{12} \mathrm{C}\) (actual mass \(12.000000 \mathrm{u}\) ) and (b) \({ }^{238} \mathrm{U}\) (actual mass \(238.050785 \mathrm{u}\) )?

Radioactive element \(A A\) can decay to either element \(B B\) or element \(C C\). The decay depends on chance, but the ratio of the resulting number of \(B B\) atoms to the resulting number of \(C C\) atoms is always \(2 / 1 .\) The decay has a half-life of 8.00 days. We start with a sample of pure \(A A .\) How long must we wait until the number of \(C C\) atoms is 1.50 times the number of \(A A\) atoms?

The radionuclide \({ }^{64} \mathrm{Cu}\) has a half-life of \(12.7 \mathrm{~h}\). If a sample contains \(5.50 \mathrm{~g}\) of initially pure \({ }^{64} \mathrm{Cu}\) at \(t=0,\) how much of it will decay between \(t=14.0 \mathrm{~h}\) and \(t=16.0 \mathrm{~h} ?\)

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