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Electron Collision An electron undergoes a one-dimensional elastic collision with an initially stationary hydrogen atom. What percentage of the electron's initial kinetic energy is transferred to kinetic energy of the hydrogen atom? (The mass of the hydrogen atom is 1840 times the mass of the electron.)

Short Answer

Expert verified
Approximately 0.109% of the electron's initial kinetic energy is transferred to the kinetic energy of the hydrogen atom.

Step by step solution

01

Define variables and given data

Let the mass of the electron be denoted as \( m_e \) and the mass of the hydrogen atom as \( m_H \). We know that \( m_H = 1840 \, m_e \). The initial velocity of the electron is \( v_{e,i} \), and the hydrogen atom is initially stationary (\( v_{H,i} = 0 \)). The initial kinetic energy of the electron is \( K_{e,i} = \frac{1}{2} m_e v_{e,i}^2 \).
02

Write down conservation of momentum

In an elastic collision, the total momentum before and after the collision remains constant. Therefore, we write:\[ m_e v_{e,i} = m_e v_{e,f} + m_H v_{H,f} \]where \( v_{e,f} \) and \( v_{H,f} \) are the final velocities of the electron and hydrogen atom, respectively.
03

Write down conservation of kinetic energy

Total kinetic energy is also conserved in elastic collisions. Therefore, we have:\[ \frac{1}{2} m_e v_{e,i}^2 = \frac{1}{2} m_e v_{e,f}^2 + \frac{1}{2} m_H v_{H,f}^2 \]
04

Solve momentum and energy equations

First, isolate \( v_{H,f} \) from the momentum equation:\[ m_e v_{e,i} = m_e v_{e,f} + 1840 m_e v_{H,f} \]\[ v_{e,i} = v_{e,f} + 1840 v_{H,f} \]Substitute \( v_{e,f} = v_{e,i} - 1840 v_{H,f} \) into the energy equation:\[ \frac{1}{2} m_e v_{e,i}^2 = \frac{1}{2} m_e (v_{e,i} - 1840 v_{H_f})^2 + \frac{1}{2} (1840 m_e) v_{H_f}^2 \]Simplify and solve for \( v_{H,f} \):\[ v_{H,f} = \frac{2 v_{e,i}}{1841} \]
05

Calculate the kinetic energy of hydrogen atom

Now compute the final kinetic energy of the hydrogen atom:\[ K_{H,f} = \frac{1}{2} m_H v_{H,f}^2 = \frac{1}{2} (1840 m_e) \left( \frac{2 v_{e,i}}{1841} \right)^2 \]Simplify the equation:\[ K_{H,f} = \frac{1}{2} (1840 m_e) \cdot \frac{4 v_{e,i}^2}{1841^2} = \frac{3680 m_e v_{e,i}^2}{2 \cdot 1841^2} = \frac{3680 m_e v_{e_i}^2}{2 \cdot 3383361} \]
06

Determine the percentage of initial kinetic energy

Compute the percentage of the initial kinetic energy transferred to the hydrogen atom:\[ \text{Percentage} = \frac{K_{H,f}}{K_{e,i}} \times 100 = \frac{\frac{3680 m_e v_{e_i}^2}{2 \cdot 3383361}}{\frac{1}{2} m_e v_{e_i}^2} \times 100 \]Simplify:\[ \text{Percentage} = \frac{3680}{3383361} \times 100 \approx 0.109 \% \]

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Key Concepts

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

Conservation of Momentum
Momentum is a fundamental concept in physics that tells us about the motion of objects. In an elastic collision, the total momentum before and after the collision remains constant. This is known as the conservation of momentum principle.

For our exercise, we define the mass of the electron as\(m_e\) and the mass of the hydrogen atom as\(m_H = 1840 \, m_e\). The electron initially has a velocity of \(v_{e,i}\), while the hydrogen atom is at rest \(v_{H,i} = 0\). The law of conservation of momentum can be expressed as:

oh\(m_e v_{e,i} = m_e v_{e,f} + m_H v_{H,f}\)ohwhere \(v_{e,f}\) and \(v_{H,f}\) are the final velocities of the electron and hydrogen atom, respectively. This equation ensures that the total momentum of the system does not change due to the collision.
Conservation of Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In an elastic collision, not only is momentum conserved, but the total kinetic energy is also conserved. This means the sum of kinetic energies before and after the collision remains the same.

Initially, the electron has kinetic energy \(K_{e,i} = \frac{1}{2} m_e v_{e_i}^2\). After the collision, the kinetic energies are given by:

oh\(\frac{1}{2} m_e v_{e_i}^2 = \frac{1}{2} m_e v_{e_f}^2 + \frac{1}{2} m_H v_{H_f}^2\)ohThis equation means that the kinetic energy lost by the electron is gained by the hydrogen atom, ensuring no net loss of kinetic energy in the system. Understanding this concept is crucial to solving the problem and finding how much energy is transferred to the hydrogen atom.
Electron Collision
An electron collision involves an electron colliding with another particle, like a hydrogen atom in our problem. When an electron hits a hydrogen atom head-on in an elastic collision, it's crucial to apply conservation laws to determine their velocities and energy transfers.

Given that the mass of the hydrogen atom is much larger (1840 times) than the mass of the electron \(m_H = 1840 \, m_e\), this significant mass difference greatly affects the collision outcome. By analyzing this scenario, we see that the electron transfers only a small fraction of its kinetic energy to the hydrogen atom because of their mass disparity. The calculation reveals this percentage.
Kinetic Energy Transfer
One of the key aspects of solving collision problems is understanding how kinetic energy is transferred between objects. In our specific case of an electron colliding with a hydrogen atom, the percentage of kinetic energy transferred is crucial.

After using conservation laws to find the final velocities, we can calculate the kinetic energy of the hydrogen atom post-collision. The result shows the fraction of the electron's initial kinetic energy transferred to the hydrogen atom, which turns out to be approximately 0.109%.

This minimal transfer is due to the considerable mass difference, emphasizing that lighter objects (like electrons) will generally transfer less energy to heavier objects (like hydrogen atoms) in elastic collisions.
Mass Ratio
The mass ratio between colliding bodies significantly impacts the collision results. In our problem, the mass ratio is extremely large, with the hydrogen atom being 1840 times more massive than the electron. This mass ratio shapes how momentum and kinetic energy are distributed post-collision.

Given that \(m_H = 1840 \, m_e\), it simplifies calculations and highlights that the hydrogen atom's final velocity \((v_{H_f})\) will be much smaller compared to the electron's initial velocity \((v_{e_i})\).

This ratio is directly proportional to the kinetic energy transfer. Therefore, understanding and using mass ratios can give insights into how energy and momentum redistribute in collision events.

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Most popular questions from this chapter

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