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Ch 39: Particles Behaving as Waves
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 39: Particles Behaving as WavesProblem 17
Chapter 39, Problem 17

A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in 'head-on' to a particular lead nucleus and stops 6.50×10−146.50\(\times\)10^{-14} m away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 8282 protons, remains at rest. The mass of the alpha particle is 6.64×10−276.64\(\times\)10^{-27} kg.
(a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in MeV.
(b) What initial kinetic energy (in joules and in MeV) did the alpha particle have?
(c) What was the initial speed of the alpha particle?

Verified step by step guidance
1
Step 1: Understand the problem and identify the key concepts. The problem involves calculating the electrostatic potential energy between an alpha particle and a lead nucleus, determining the initial kinetic energy of the alpha particle, and finding its initial speed. Key concepts include Coulomb's law, energy conservation, and kinematics.
Step 2: Use Coulomb's law to calculate the electrostatic potential energy. The formula for electrostatic potential energy is U = (k * q1 * q2) / r, where k is Coulomb's constant (8.99 × 10^9 N·m²/C²), q1 and q2 are the charges of the alpha particle and lead nucleus respectively, and r is the distance between them. The charge of the alpha particle is 2e (where e = 1.60 × 10^-19 C), and the charge of the lead nucleus is 82e.
Step 3: Convert the electrostatic potential energy from joules to MeV. To convert joules to MeV, use the conversion factor 1 eV = 1.60 × 10^-19 J, and 1 MeV = 10^6 eV. Divide the calculated energy in joules by 1.60 × 10^-13 to express it in MeV.
Step 4: Apply the principle of energy conservation to determine the initial kinetic energy of the alpha particle. Since the alpha particle stops at the given distance, all its initial kinetic energy is converted into electrostatic potential energy. Therefore, the initial kinetic energy is equal to the calculated electrostatic potential energy.
Step 5: Use the formula for kinetic energy, KE = (1/2) * m * v², to find the initial speed of the alpha particle. Rearrange the formula to solve for v: v = sqrt((2 * KE) / m), where m is the mass of the alpha particle (6.64 × 10^-27 kg) and KE is the initial kinetic energy calculated in step 4. Ensure the speed is expressed in meters per second.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electrostatic Potential Energy

Electrostatic potential energy is the energy stored due to the position of charged particles relative to each other. It can be calculated using the formula U = k * (q1 * q2) / r, where U is the potential energy, k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them. In this scenario, the alpha particle and the lead nucleus create a potential energy due to their charges as the alpha particle approaches the nucleus.
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Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion, expressed as KE = 0.5 * m * v^2, where m is the mass and v is the velocity of the object. For the alpha particle, its initial kinetic energy can be determined from its speed before it interacts with the lead nucleus. This energy is crucial for understanding how the alpha particle behaves as it approaches the nucleus.
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Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this context, the initial kinetic energy of the alpha particle is converted into electrostatic potential energy as it approaches the lead nucleus. By applying this principle, one can relate the initial kinetic energy to the potential energy at the point of closest approach.
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Related Practice
Textbook Question

Through what potential difference must electrons be accelerated if they are to have:

(a) the same wavelength as an x ray of wavelength 0.2200.220 nm; and

(b) the same energy as the x ray in part (a)?

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Textbook Question

Calculate the de Broglie wavelength of a 5.005.00-g bullet that is moving at 340340 m/s. Will the bullet exhibit wavelike properties?

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Textbook Question

A 4.784.78-MeV alpha particle from a 226226Ra decay makes a head-on collision with a uranium nucleus. A uranium nucleus has 9292 protons.

(a) What is the distance of closest approach of the alpha particle to the center of the nucleus? Assume that the uranium nucleus remains at rest and that the distance of closest approach is much greater than the radius of the uranium nucleus.

(b) What is the force on the alpha particle at the instant when it is at the distance of closest approach?

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Textbook Question

A hydrogen atom is in a state with energy −1.51-1.51 eV. In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus?

Textbook Question

A triply ionized beryllium ion, Be3+ (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom except that the nuclear charge is four times as great. What is the ground-level energy of Be3+? How does this compare to the ground-level energy of the hydrogen atom?

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Textbook Question

A triply ionized beryllium ion, Be3+ (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom except that the nuclear charge is four times as great. For the hydrogen atom, the wavelength of the photon emitted in the n=2n = 2 to n=1n = 1 transition is 122122 nm (see Example 39.639.6). What is the wavelength of the photon emitted when a Be3+ ion undergoes this transition?