Examples of alpha particle in the following topics:
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- In alpha decay an atomic nucleus emits an alpha particle and transforms into an atom with smaller mass (by four) and atomic number (by two).
- Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle that consists of two protons and two neutrons, as shown in .
- Because an alpha particle is the same as a helium-4 nucleus, which has mass number 4 and atomic number 2, this can also be written as:
- Alpha decay is the most common cluster decay because of the combined extremely high binding energy and relatively small mass of the helium-4 product nucleus (the alpha particle).
- Alpha particles have a typical kinetic energy of 5 MeV (approximately 0.13 percent of their total energy, i.e., 110 TJ/kg) and a speed of 15,000 km/s.
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- In 1911, Rutherford designed an experiment to further explore atomic structure using the alpha particles emitted by a radioactive element .
- Following his direction, Geiger and Marsden shot alpha particles with large kinetic energies toward a thin foil of gold.
- Under the prevailing plum pudding model, the alpha particles should all have been deflected by, at most, a few degrees.
- Although many of the alpha particles did pass through as expected, many others were deflected at small angles while others were reflected back to the alpha source.
- Top: Expected results -- alpha particles pass through the plum pudding model of the atom undisturbed.
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- Though originally viewed as a particle that cannot be cut into smaller particles , modern scientific usage denotes the atom as composed of various subatomic particles.
- Earlier, Rutherford learned to create hydrogen nuclei as a type of radiation produced as a yield of the impact of alpha particles on hydrogen gas; these nuclei were recognized by their unique penetration signature in air and their appearance in scintillation detectors.
- These experiments began when Rutherford noticed that when alpha particles were shot into air (mostly nitrogen), his scintillation detectors displayed the signatures of typical hydrogen nuclei as a product.
- One hydrogen nucleus was knocked off by the impact of the alpha particle, producing oxygen-17 in the process.
- This was the first reported nuclear reaction, $14\text{N} + \alpha \rightarrow 17\text{O} + \text{p}$.
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- An atom with an unstable nucleus, called a radionuclide, is characterized by excess energy available either for a newly created radiation particle within the nucleus or via internal conversion.
- Radioactive decay results in the emission of gamma rays and/or subatomic particles such as alpha or beta particles, as shown in .
- Alpha decay is one type of radioactive decay.
- An atomic nucleus emits an alpha particle and thereby transforms ("decays") into an atom with a mass number smaller by four and an atomic number smaller by two.
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- $\displaystyle J^\mu (x^\alpha) \equiv c \int \frac{d^3 {\bf p}}{E_{\bf p}} p^\mu f(x^\alpha, {\bf p}).$
- $\displaystyle J^0(x^\alpha) = \int \frac{d^3 {\bf p}}{E_{\bf p}} c p^0 f(x^\alpha, {\bf p}) = \int d^3 {\bf p} f(x^\alpha, {\bf p}) = n(x^\alpha) \\ {\bf J}(x^\alpha) = \int \frac{d^3 {\bf p}}{E_{\bf p}} c {\bf p} f(x^\alpha, {\bf p}) = \frac{1}{c} \int d^3 {\bf p} {\bf v} f(x^\alpha, {\bf p}) = \frac{\langle {\bf v} \rangle}{c} n(x^\alpha) $
- If we assume that the scattering (${\cal C}$) conserves energy, momentum and particles we have
- where $\rho=mn$ where $m$ is the rest mass of the individual particles.
- Because the fourth equation is consistent with the Lioville equation (seventh equation) and more generally with the Boltzmann equation (sixth equation) and $J^\mu_{;\mu}=0$ if particles are conserved, the Lioville and Boltzmann equations cannot hold if $\nabla_{\bf p} \cdot F \neq 0$ and particles are conserved.
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- The phase-space density of particles gives the number of particles in an infinitesimal region of phase space,
- $\displaystyle d N = f(x^\alpha,{\bf p}) d^3 {\bf x} d^3 {\bf p}$
- where ${\bf F}$ is a force that accelerates the particles.
- $\displaystyle \frac{n(x^\alpha)}{{\bar E}_\mbox{har}(x^\alpha)} \equiv \int \frac{d^3 {\bf p}}{E_{\bf p}} f(x^\alpha, {\bf p})$
- that transforms as a scalar where $n(x^\alpha)$ is the number density.
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- When a substance is heated, its constituent particles begin moving more, thus maintaining a greater average separation with their neighboring particles.
- The answer can be found in the shape of the typical particle-particle potential in matter.
- Particles in solids and liquids constantly feel the presence of other neighboring particles.
- Fig 2 illustrates how this inter-particle potential usually takes an asymmetric form rather than a symmetric form, as a function of particle-particle distance.
- In the diagram, (b) shows that as the substance is heated, the equilibrium (or average) particle-particle distance increases.
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- We are going to take the results from the previous section to study particles that are harmonically bound, so their motion satisfies the following equation,
- Let's solve this by assuming that $x(t) = A e^{\alpha t}$.
- $\displaystyle \alpha = \pm i \omega_0 \sqrt{1 - \omega_0^2 \tau^2} - \frac{1}{2} \omega_0^2 \tau.$
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- Here, the velocity of particle is changing - though the motion is "uniform".
- For the length of the arc subtending angle " at the origin and "r" is the radius of the circle containing the position of the particle, we have $s=r\theta $.
- Similarly, we also get $a = \alpha r$ where $a$ stands for linear acceleration, while $\alpha$ refers to angular acceleration (In a more general case, the relationship between angular and linear quantities are given as $\bf{v = \omega \times r}, ~~ \bf{a = \alpha \times r + \omega \times v}$. )
- With the relationship of the linear and angular speed/acceleration, we can derive the following four rotational kinematic equations for constant $a$ and $\alpha$:
- Each particle constituting the body executes a uniform circular motion about the fixed axis.
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- How long does it take the particle to pass through this angle?
- $\gamma m \frac{2 v}{\gamma \Delta t} = \frac{q}{c} v \sin \alpha B \\ \Delta t = \frac{2 m c}{q B \sin\alpha} = \frac{2}{\gamma \omega_B \sin\alpha}$
- $\Delta t^A = \frac{2}{\gamma \omega_B \sin\alpha} \left ( 1 - \frac{v}{c} \right )$