Ch 37: The Foundations of Modern Physics

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 37: The Foundations of Modern Physics

Problem 44a

Chapter 37, Problem 44a

The oxygen nucleus ¹⁶O has a radius of 3.0 fm. With what speed must a proton be fired toward an oxygen nucleus to have a turning point 1.0 fm from the surface? Assume the nucleus remains at rest.

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Determine the total distance between the proton and the center of the oxygen nucleus at the turning point. Since the radius of the oxygen nucleus is 3.0 fm and the turning point is 1.0 fm from the surface, the total distance is $ r = 3.0 \text{ fm} + 1.0 \text{ fm} = 4.0 \text{ fm} $.

Use the concept of energy conservation. At the turning point, the proton's kinetic energy is completely converted into electric potential energy due to the repulsion between the proton and the oxygen nucleus. The total energy equation is $ K_i + U_i = K_f + U_f $, where $ K $ is kinetic energy and $ U $ is potential energy. At the turning point, $ K_f = 0 $.

Write the expression for the electric potential energy between the proton and the oxygen nucleus: $ U = \frac{k_e q_1 q_2}{r} $, where $ k_e $ is Coulomb's constant $ (8.99 \times 10^9 \text{ N·m}^2/\text{C}^2) $, $ q_1 $ is the charge of the proton $ (+e = 1.6 \times 10^{-19} \text{ C}) $, $ q_2 $ is the charge of the oxygen nucleus $ (+8e) $, and $ r $ is the distance between the charges.

Set the initial kinetic energy of the proton equal to the electric potential energy at the turning point: $ \frac{1}{2} m v^2 = \frac{k_e q_1 q_2}{r} $, where $ m $ is the mass of the proton $ (1.67 \times 10^{-27} \text{ kg}) $ and $ v $ is the initial speed of the proton. Rearrange this equation to solve for $ v $: $ v = \sqrt{\frac{2 k_e q_1 q_2}{m r}} $.

Substitute the known values into the equation for $ v $: $ k_e = 8.99 \times 10^9 \text{ N·m}^2/\text{C}^2 $, $ q_1 = 1.6 \times 10^{-19} \text{ C} $, $ q_2 = 8 \times 1.6 \times 10^{-19} \text{ C} $, $ m = 1.67 \times 10^{-27} \text{ kg} $, and $ r = 4.0 \times 10^{-15} \text{ m} $. Simplify the expression to find the speed $ v $.

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Key Concepts

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

Nuclear Radius

The nuclear radius is a measure of the size of an atomic nucleus, typically on the order of femtometers (fm). For the oxygen nucleus ¹⁶O, the radius is given as 3.0 fm, which indicates the distance from the center of the nucleus to its surface. Understanding the nuclear radius is crucial for calculating distances in nuclear interactions, such as the turning point of a proton approaching the nucleus.

Turning Point in Potential Energy

The turning point in a particle's motion occurs when it can no longer continue moving toward a potential barrier, such as a nucleus, due to the repulsive forces at play. In this context, the turning point is defined as being 1.0 fm from the surface of the oxygen nucleus, which means the proton must have enough kinetic energy to reach this point before being repelled. This concept is essential for determining the required speed of the proton.

Conservation of Energy

The principle of conservation of energy states that the total energy in a closed system remains constant. In this scenario, the kinetic energy of the incoming proton will convert into potential energy as it approaches the nucleus. By applying this principle, one can calculate the necessary speed of the proton to ensure it reaches the specified turning point before being repelled by the nucleus.

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