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An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is 1837 times that of the electron. Assume that all motion, before and after the collision, occurs along the same straight line. What is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision?

Short Answer

Expert verified
The ratio is \( \frac{4}{1837} \).

Step by step solution

01

Identify Known Variables

Let's define the problem variables: the mass of the electron is denoted as \( m_e \) and the mass of the hydrogen atom is \( m_H = 1837 m_e \). Initially, the hydrogen atom is stationary, meaning its initial velocity, \( u_H = 0 \). The initial velocity of the electron is \( u_e \). After the collision, the velocities are \( v_e \) for the electron and \( v_H \) for the hydrogen atom.
02

Apply Conservation of Momentum

The total momentum before the collision is equal to the total momentum after the collision. The equation is:\[ m_e u_e = m_e v_e + m_H v_H. \]Substituting \( m_H = 1837 m_e \) into the equation, we get:\[ m_e u_e = m_e v_e + 1837 m_e v_H. \]
03

Apply Conservation of Kinetic Energy

The kinetic energy is also conserved in an elastic collision. Therefore, the equation for kinetic energy is:\[ \frac{1}{2} m_e u_e^2 = \frac{1}{2} m_e v_e^2 + \frac{1}{2} m_H v_H^2. \]Substituting \( m_H = 1837 m_e \) gives:\[ m_e u_e^2 = m_e v_e^2 + 1837 m_e v_H^2. \]
04

Determine Final Velocities

To solve for the final velocities, solve the two equations from the conservation of momentum and kinetic energy:1. \( m_e u_e = m_e v_e + 1837 m_e v_H \)2. \( u_e^2 = v_e^2 + 1837 v_H^2. \)Using algebraic manipulation or substituting one equation into another help find \( v_e \) and \( v_H \).
05

Calculate Kinetic Energy Ratio

Once the final velocities are determined, calculate the kinetic energy of the hydrogen atom: \[ KE_H = \frac{1}{2} m_H v_H^2 = \frac{1}{2} \cdot 1837 m_e v_H^2 \]and the initial kinetic energy of the electron:\[ KE_{e0} = \frac{1}{2} m_e u_e^2. \]The ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision is:\[ \frac{KE_H}{KE_{e0}}. \]
06

Substitute and Simplify

Substitute the expressions for kinetic energies into the ratio:\[ \frac{1837 m_e v_H^2}{m_e u_e^2}. \]Simplify the equation using relationships between \( u_e, v_H, \) and ratios derived from previous equations. Solving gives the ratio \( \frac{4}{1837} \).

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Key Concepts

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

Conservation of Momentum
Conservation of momentum is a fundamental principle that states the total momentum before a collision is equal to the total momentum after the collision, provided no external forces act on the system. In an elastic collision, like the one between an electron and a hydrogen atom, this principle is crucial.

Momentum is calculated as the product of mass and velocity. For our problem, we have an electron with mass \(m_e\) and initial velocity \(u_e\), and a hydrogen atom with mass \(m_H = 1837m_e\) initially at rest. The conservation equation becomes:
  • Initial momentum: \(m_e u_e\)
  • Final momentum: \(m_e v_e + m_H v_H\)
Setting these equal gives:\[ m_e u_e = m_e v_e + 1837 m_e v_H. \]
This equation helps us understand that any gain in momentum by the hydrogen atom after the collision must be balanced by a loss in momentum of the electron. It's the principle that allows us to solve for the velocities after the collision.
Kinetic Energy
Kinetic energy, the energy an object has due to its motion, is also conserved in elastic collisions. This is another crucial law because it allows us to explore not only the exchange of momentum but also how energy is redistributed between the colliding particles.

Kinetic energy is given by the equation \( KE = \frac{1}{2} mv^2 \). Before the collision, only the electron has kinetic energy since the hydrogen atom is stationary. After the collision, both have kinetic energy:
  • Initial: \( \frac{1}{2} m_e u_e^2 \)
  • Final: \( \frac{1}{2} m_e v_e^2 + \frac{1}{2} m_H v_H^2 \)
Equating these gives:\[ m_e u_e^2 = m_e v_e^2 + 1837 m_e v_H^2. \]
Using this principle, we can solve for velocities of both particles post-collision and ultimately find the ratio of the hydrogen atom's kinetic energy after the collision to the electron's initial kinetic energy. This energy ratio turns out to be \( \frac{4}{1837} \), highlighting how a greater mass (like hydrogen's) carries away less kinetic energy proportionally from a colliding smaller mass (like an electron).
Electron-Hydrogen Collision
When considering an electron-hydrogen collision, we focus on the interaction between a small and a large particle. The electron is significantly lighter, with the hydrogen atom being 1837 times more massive. This gives the hydrogen atom a larger inertia and results in unique outcomes in terms of velocity and energy exchange during collisions.

An elastic collision implies both momentum and kinetic energy are conserved. Unlike inelastic collisions where energy is transformed into other forms, elastic collisions maintain the mechanical energy within the system.

For our scenario, initially, the electron has kinetic energy while the hydrogen is at rest. After the collision, both the electron and the hydrogen atom move but with different velocities. The light electron ends up transferring some of its energy to the much heavier hydrogen atom.

The striking feature of such collisions is how they demonstrate the interplay between mass disparity and kinetic energy transfer. This understanding is not only fundamental to physics but also helps us grasp concepts applicable in atomic-scale phenomena and provides insights for various technologies that hinge on particle interactions, such as particle accelerators and radiation therapy. This case study showcases these principles through a practical and easily understandable example of basic physics at work.

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Most popular questions from this chapter

A ball is attached to one end of a wire, the other end beingfastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, \(1.60 \mathrm{kg}\) and \(2.40 \mathrm{kg},\) and the length of the wire is \(1.20 \mathrm{m} .\) Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

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