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An oxygen atom is four times as massive as a helium atom. In an experiment, a helium atom and an oxygen atom have the same kinetic energy. What is the ratio \(v_{\mathrm{H}} / v_{\mathrm{O}}\) of their speeds?

Short Answer

Expert verified
The ratio of the speed of the helium atom to the oxygen atom is 2:1.

Step by step solution

01

Understanding kinetic energy

The kinetic energy (K.E) of a moving object is given by the equation K.E = \(0.5 * m * v^2\), where \(m\) is the mass of the object and \(v\) is its velocity.
02

Formulate equations for the kinetic energy of helium and oxygen atoms

The kinetic energy of the helium atom is given as K.E = \(0.5 * m_{\mathrm{H}} * v_{\mathrm{H}}^2\). The oxygen atom, being four times as heavy as helium, has kinetic energy given as K.E = \(0.5 * 4 * m_{\mathrm{H}} * v_{\mathrm{O}}^2\). Since it's given that the atoms have the same kinetic energy, we equate the two equations: \(0.5 * m_{\mathrm{H}} * v_{\mathrm{H}}^2 = 0.5 * 4 * m_{\mathrm{H}} * v_{\mathrm{O}}^2\).
03

Solve for the velocity ratio

Solving the equation from Step 2 for the velocity ratio \(v_{\mathrm{H}} / v_{\mathrm{O}}\), we get: \(v_{\mathrm{H}} / v_{\mathrm{O}} = \sqrt{4} = 2\).

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Key Concepts

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

Kinetic Energy Equation
Kinetic energy is the energy possessed by an object in motion. The kinetic energy equation plays a crucial role in understanding how objects exchange energy through motion. The formula for kinetic energy (K.E) is given by the expression \( K.E = \frac{1}{2} m v^2 \), where \( m \) represents mass and \( v \) stands for velocity. This equation illustrates that the kinetic energy of an object is directly proportional to its mass and the square of its velocity. Hence, if you double the velocity, the kinetic energy increases by four times due to the square relationship.

Understanding the kinetic energy equation empowers students to solve problems involving motion and the conversion of energy forms, which is pivotal in many physics applications. For instance, in the case of the oxygen and helium atoms from the textbook exercise, despite the significant mass difference, the kinetic energy is constant for both, leading to insightful observations about their velocities.
Mass-Velocity Relationship
The kinetic energy equation unveils a fundamental principle in physics: the mass-velocity relationship. This relationship signifies that for a constant kinetic energy, the velocity of an object is inversely proportional to the square root of its mass. Simply put, more massive objects move slower when compared to less massive ones at the same kinetic energy, which is demonstrated in the provided textbook exercise.

In the exercise, the kinetic energies of a helium atom and an oxygen atom, which is four times heavier, are equal. According to the mass-velocity relation \(v = \sqrt{\frac{2K.E}{m}}\), heavier mass leads to a lower velocity. This intrinsic relationship is key in many physical phenomena and technologies, like particle accelerators where controlling speeds is vital. A solid grasp of this concept is crucial for students to predict motion outcomes in physics problems.
Comparing Atomic Speeds
In physics, comparing atomic speeds becomes interesting when analyzing gases or studying molecular behavior. Atomic speeds are influenced by factors like temperature and mass as seen in the kinetic energy equation. When comparing the speeds of different atoms, like helium and oxygen from our example, we use the principle that equal kinetic energies will result in different speeds based on their mass.

The exercise demonstrated that the lighter helium atom travels at a higher speed compared to the heavier oxygen atom, given the same amount of kinetic energy. This insight is part of the basis for the kinetic theory of gases, which describes how gas pressure is due to collisions of moving atoms with the walls of a container. Therefore, understanding how to compare atomic speeds based on their mass and kinetic energy is essential for students studying thermodynamics, gas laws, and molecular dynamics.

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Most popular questions from this chapter

a. What is the kinetic energy of a \(1500 \mathrm{kg}\) car traveling at a speed of \(30 \mathrm{m} / \mathrm{s}(\approx 65 \mathrm{mph}) ?\) b. From what height would the car have to be dropped to have this same amount of kinetic energy just before impact? c. Does your answer to part b depend on the car's mass?

Protons and neutrons (together called nucleons) are held together in the nucleus of an atom by a force called the strong force. At very small separations, the strong force between two nucleons is larger than the repulsive electrical force between two protons - hence its name. But the strong force quickly weakens as the distance between the protons increases. A well-established model for the potential energy of two nucleons interacting via the strong force is $$U=U_{0}\left[1-e^{-x / x_{0}}\right]$$ where \(x\) is the distance between the centers of the two nucleons, \(x_{0}\) is a constant having the value \(x_{0}=2.0 \times 10^{-15} \mathrm{m},\) and \(U_{0}=6.0 \times 10^{-11} \mathrm{J}\) a. Calculate and draw an accurate potential-energy curve from \(x=0 \mathrm{m}\) to \(x=10 \times 10^{-15} \mathrm{m} .\) Either calculate about 10 points by hand or use computer software. b. Quantum effects are essential for a proper understanding of how nucleons behave. Nonetheless, let us innocently consider two neutrons as if they were small, hard, electrically neutral spheres of mass \(1.67 \times 10^{-27} \mathrm{kg}\) and diameter \(1.0 \times 10^{-15} \mathrm{m}\) (We will consider neutrons rather than protons so as to avoid complications from the electric forces between protons.) You are going to hold two neutrons \(5.0 \times 10^{-15} \mathrm{m}\) apart, measured between their centers, then release them. Draw the total energy line for this situation on your diagram of part a. c. What is the speed of each neutron as they crash together? Keep in mind that both neutrons are moving.

You have been asked to design a "ballistic spring system" to measure the speed of bullets. A spring whose spring constant is \(k\) is suspended from the ceiling. A block of mass \(M\) hangs from the spring. A bullet of mass \(m\) is fired vertically upward into the bottom of the block. The spring's maximum compression \(d\) is measured. a. Find an expression for the bullet's speed \(v_{\mathrm{B}}\) in terms of \(m, M\) \(k,\) and \(d\) b. What was the speed of a \(10 \mathrm{g}\) bullet if the block's mass is \(2.0 \mathrm{kg}\) and if the spring, with \(k=50 \mathrm{N} / \mathrm{m},\) was compressed by \(45 \mathrm{cm} ?\)

You are given the equation used to solve a problem. For each of these, you are to a. Write a realistic problem for which this is the correct equation. b. Draw the before-and-after pictorial representation. c. Finish the solution of the problem. $$\begin{array}{l} (0.10 \mathrm{kg}+0.20 \mathrm{kg}) v_{\mathrm{lx}}=(0.10 \mathrm{kg})(3.0 \mathrm{m} / \mathrm{s}) \\ \frac{1}{2}(0.30 \mathrm{kg})(0 \mathrm{m} / \mathrm{s})^{2}+\frac{1}{2}(3.0 \mathrm{N} / \mathrm{m})\left(\Delta x_{2}\right)^{2} \\ =\frac{1}{2}(0.30 \mathrm{kg})\left(v_{1 x}\right)^{2}+\frac{1}{2}(3.0 \mathrm{N} / \mathrm{m})(0 \mathrm{m})^{2} \end{array}$$

A 10 -cm-long spring is attached to the ceiling. When a \(2.0 \mathrm{kg}\) mass is hung from it, the spring stretches to a length of \(15 \mathrm{cm} .\) a. What is the spring constant \(k ?\) b. How long is the spring when a \(3.0 \mathrm{kg}\) mass is suspended from it?

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