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Go In a Rutherford scattering experiment, assume that an incident alpha particle (radius \(1.80 \mathrm{fm}\) ) is headed directly toward a target gold nucleus (radius \(6.23 \mathrm{fm}) .\) What energy must the alpha particle have to just barely "touch" the gold nucleus?

Short Answer

Expert verified
The alpha particle must have approximately 27.49 MeV of energy to just barely "touch" the gold nucleus.

Step by step solution

01

Determine the Distance of Closest Approach

The distance of closest approach, where the alpha particle just "touches" the gold nucleus, is the sum of the radii of the alpha particle and the gold nucleus. The given radius of the alpha particle is \(1.80 \mathrm{fm}\) and the radius of the gold nucleus is \(6.23 \mathrm{fm}\). So, the closest approach is \(1.80 + 6.23 = 8.03 \mathrm{fm}\).
02

Convert the Distance to Meters

First, convert the combined radius (distance of closest approach) from femtometers to meters. \(1 \mathrm{fm} = 10^{-15} \mathrm{m}\), therefore \(8.03 \mathrm{fm} = 8.03 \times 10^{-15} \mathrm{m}\).
03

Use the Coulomb's Law to Calculate Potential Energy

The potential energy at the distance of closest approach is given by \( U = \frac{k \cdot Z_1 \cdot Z_2 \cdot e^2}{r} \), where \(k = 8.99 \times 10^9 \mathrm{Nm}^2/\mathrm{C}^2\), \(Z_1 = 2\) (for alpha particle, which has 2 protons), \(Z_2 = 79\) (for gold, which has 79 protons), \(e = 1.60 \times 10^{-19} \mathrm{C}\), and \(r = 8.03 \times 10^{-15} \mathrm{m}\).
04

Compute the Required Energy in Joules

Plug the values into the formula: \[ U = \frac{(8.99 \times 10^9 \mathrm{Nm}^2/\mathrm{C}^2) \cdot (2) \cdot (79) \cdot (1.60 \times 10^{-19} \mathrm{C})^2}{8.03 \times 10^{-15} \mathrm{m}} \]. Calculate this value to find the energy in joules.
05

Convert the Energy to MeV

Finally, convert the energy from joules to mega-electronvolts (MeV) using the conversion \(1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J}\) and \(1 \mathrm{MeV} = 10^6 \mathrm{eV}\). This involves dividing the computed energy in joules by \(1.602 \times 10^{-13}\).

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Key Concepts

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

Alpha Particles
Alpha particles are one of the fundamental types of atomic nuclei that consist of two protons and two neutrons. This compact structure gives them a positive charge of +2e where e is the elementary charge, approximately equal to 1.60 x 10^{-19} C.
Due to this composition, alpha particles are identical to the helium nucleus. In nature, alpha particles are commonly emitted by radioactive materials as a part of the alpha decay process. This decay occurs when heavier unstable nuclei release an alpha particle to achieve a more stable state.
The properties of alpha particles make them significant in various nuclear physics experiments, like Rutherford scattering, where they help probe the structure of target nuclei such as gold. Here are some key attributes of alpha particles:
  • They have a relatively high mass compared to other forms of radiation, such as beta particles.
  • They are positively charged and can be deflected by electric and magnetic fields.
  • Because of their charge, they can strongly interact with matter, losing energy rapidly and having limited penetration power.
Understanding these properties is crucial for interpreting experiments that involve interactions with materials.
Coulomb's law
Coulomb's law is a pivotal principle in electromagnetism that describes the electrostatic force between charged particles. It stipulates that the force F between two point charges, q1 and q2, is directly proportional to the product of the magnitudes of the two charges and inversely proportional to the square of the distance r between their centers.
Mathematically, this is expressed as:\[F = \frac{k \, q_1 \, q_2}{r^2}\]Where \(k\) represents Coulomb's constant, approximately equal to 8.99 x 10^9 Nm^2/C^2. This law not only explains the attraction or repulsion between charged bodies but also helps calculate potential energy, which is pivotal in high-energy physics experiments, like those involving alpha particles and gold nuclei interactions.A few important points about Coulomb's law:
  • The force is attractive if the charges are of opposite signs and repulsive if their signs are the same.
  • The greater the magnitudes of charges, the stronger the force; the greater the distance, the weaker the force.
By using Coulomb's law, one can derive the potential energy needed to interpret phenomena such as the one in the Rutherford scattering experiment.
Potential Energy
In nuclear physics, potential energy is a key concept when analyzing interactions between particles, like alpha particles and nuclei. The potential energy provides insights into the forces at play when particles approach close to each other.
For charged particles, potential energy is due to their electric potential, defined by:\[U = \frac{k \, Z_1 \, Z_2 \, e^2}{r}\]Where \(Z_1\) and \(Z_2\) are the atomic number of the interacting nuclei, \(e\) is the elementary charge, and \(r\) is the distance between them. In the context of Rutherford scattering, potential energy explains how the alpha particle interacts with the gold nucleus.Consider these key aspects:
  • The potential energy increases as the distance r decreases, emphasizing the high repulsion when like-charged particles come very close.
  • It reflects the energy required for an alpha particle to closely approach or barely "touch" another nucleus.
Understanding potential energy in such contexts is vital to calculating interaction strengths and energies needed in scattering experiments.
Nuclear Physics
Nuclear physics explores the constituents and behavior of atomic nuclei. This field delves deep into understanding nuclear forces, decay modes, and particle interactions. Experiments like Rutherford scattering are classic studies within this domain.
Nuclear physics encompasses:
  • Radioactivity: the process through which unstable nuclei release energy, often emitting alpha, beta, or gamma radiation.
  • Nuclear reactions: interactions that change the structure or energy state of a nucleus, involving a variety of particles.
  • Models of the nucleus: efforts to understand and mathematically model the complex structures within an atomic nucleus.
In Rutherford scattering, nuclear physics principles are applied to determine how alpha particles interact with heavier nuclei like gold. This experiment was crucial in discovering the nuclear atom model, significantly reshaping our understanding of atomic structure.
Modern nuclear physics builds upon such pioneering experiments to further explore nuclear energy and its applications, lending insights into both fundamental physics and practical energy production.

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Most popular questions from this chapter

The isotope \({ }^{238} \mathrm{U}\) decays to \({ }^{206} \mathrm{~Pb}\) with a half-life of \(4.47 \times 10^{9}\) y. Although the decay occurs in many individual steps, the first step has by far the longest half-life; therefore, one can often consider the decay to go directly to lead. That is, $$ { }^{238} \mathrm{U} \rightarrow{ }^{206} \mathrm{~Pb}+\text { various decay products. } $$ A rock is found to contain \(4.20 \mathrm{mg}\) of \({ }^{238} \mathrm{U}\) and \(2.135 \mathrm{mg}\) of \({ }^{206} \mathrm{~Pb}\). Assume that the rock contained no lead at formation, so all the lead now present arose from the decay of uranium. How many atoms of (a) \({ }^{238} \mathrm{U}\) and \((\mathrm{b})^{206} \mathrm{~Pb}\) does the rock now contain? (c) How many atoms of \({ }^{238} \mathrm{U}\) did the rock contain at formation? (d) What is the age of the rock?

What is the mass excess \(\Delta_{1}\) of \({ }^{1} \mathrm{H}\) (actual mass is \(\left.1.007825 \mathrm{u}\right)\) in (a) atomic mass units and (b) \(\mathrm{MeV} / \mathrm{c}^{2}\) ? What is the mass excess \(\Delta_{\mathrm{n}}\) of a neutron (actual mass is \(1.008665 \mathrm{u}\) ) in (c) atomic mass units and (d) \(\mathrm{MeV} / \mathrm{c}^{2} ?\) What is the mass excess \(\Delta_{120}\) of \({ }^{120} \mathrm{Sn}\) (actual mass is \(119.902197 \mathrm{u})\) in (e) atomic mass units and (f) \(\mathrm{MeV} / \mathrm{c}^{2}\) ?

Plutonium isotope \({ }^{239}\) Pu decays by alpha decay with a halflife of \(24100 \mathrm{y}\). How many milligrams of helium are produced by an initially pure \(12.0 \mathrm{~g}\) sample of \({ }^{239} \mathrm{Pu}\) at the end of \(20000 \mathrm{y}\) ? (Consider only the helium produced directly by the plutonium and not by any by-products of the decay process.)

The electric potential energy of a uniform sphere of charge \(q\) and radius \(r\) is given by $$ U=\frac{3 q^{2}}{20 \pi \varepsilon_{0} r} $$ (a) Does the energy represent a tendency for the sphere to bind together or blow apart? The nuclide \({ }^{239} \mathrm{Pu}\) is spherical with radius \(6.64\) \(\mathrm{fm}\). For this nuclide, what are (b) the electric potential energy \(U\) according to the equation, (c) the electric potential energy per proton, and (d) the electric potential energy per nucleon? The binding energy per nucleon is \(7.56 \mathrm{MeV}\). (e) Why is the nuclide bound so well when the answers to \((\mathrm{c})\) and \((\mathrm{d})\) are large and positive?

\(.\) co The radionuclide \({ }^{11} \mathrm{C}\) decays according to $$ { }^{11} \mathrm{C} \rightarrow{ }^{11} \mathrm{~B}+\mathrm{e}^{+}+\nu, \quad T_{1 / 2}=20.3 \mathrm{~min} $$ The maximum energy of the emitted positrons is \(0.960 \mathrm{MeV}\). (a) Show that the disintegration energy \(Q\) for this process is given by $$ Q=\left(m_{\mathrm{C}}-m_{\mathrm{B}}-2 m_{\mathrm{e}}\right) c^{2} $$ where \(m_{\mathrm{C}}\) and \(m_{\mathrm{B}}\) are the atomic masses of \({ }^{11} \mathrm{C}\) and \({ }^{11} \mathrm{~B}\), respectively, and \(m_{\mathrm{e}}\) is the mass of a positron. (b) Given the mass values \(m_{\mathrm{C}}=11.011434 \mathrm{u}, m_{\mathrm{B}}=11.009305 \mathrm{u}\), and \(m_{\mathrm{e}}=0.0005486 \mathrm{u}\), calculate \(Q\) and compare it with the maximum energy of the emitted positron given above. (Hint: Let \(\mathbf{m}_{\mathrm{C}}\) and \(\mathbf{m}_{\mathrm{B}}\) be the nuclear masses and then add in enough electrons to use the atomic masses.)

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