Question

A beam of electrons with a speed of \(676.01 \mathrm{~km} / \mathrm{s}\) is incident on a system of excited hydrogen atoms. If an electron hits a hydrogen atom in the \(n=3\) state, what is the highest level \(n\), to which this hydrogen atom can be excited in this collision? (In the collision of the electron with the hydrogen atom, you may neglect the recoil energy of the hydrogen atom, because it has a mass much greater than that of the electron.)

Short answer

Answer: To determine the highest energy level (n') the hydrogen atom can be excited, follow these steps: 1. Calculate the kinetic energy of the incident electron using the formula: \(K_e = \frac{1}{2} m_e v^2\) 2. Find the maximum energy change of the hydrogen atom, assuming all the kinetic energy of the electron is transferred to it. 3. Use the energy levels of the hydrogen atom to find the initial energy and the energy of the new state. 4. Solve for the largest possible integer value of n' which satisfies the energy change equation.

Step-by-step solution

Calculate the kinetic energy of the incident electron

The kinetic energy of the incident electron can be calculated using the formula: \(K_e = \frac{1}{2} m_e v^2\), where \(m_e\) is the mass of the electron and \(v\) is its velocity. The velocity of the electron is given in km/s, we need to convert it to m/s: \(v = 676.01 \mathrm{~km/s} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} = 676010 \mathrm{~m/s}\). Now, we can calculate the kinetic energy of the incident electron: \(K_e = \frac{1}{2} m_e \times (676010 \mathrm{~m/s})^2\)

Calculate the maximum energy change of hydrogen atom

In order to find the highest energy level the hydrogen atom can be excited, we need to find the maximum energy change that can happen to the hydrogen atom. As the mass of the hydrogen atom is much greater than that of the electron, we can assume that all the kinetic energy of the electron is transferred to the hydrogen atom in the collision. Therefore, the maximum energy change of the hydrogen atom is equal to the kinetic energy of the incident electron: \(\Delta E = K_e\)

Use Hydrogen atom energy levels

The energy levels of the hydrogen atom are given by the formula: \(E_n = -\frac{13.6 \mathrm{eV}}{n^2}\), where \(n\) is the principal quantum number. In this problem, we are given that the hydrogen atom is initially in the \(n=3\) state. So, the initial energy of the hydrogen atom is: \(E_{initial} = -\frac{13.6 \mathrm{eV}}{(3)^2} = -1.51 \mathrm{eV}\) Now, let's assume that the hydrogen atom is excited to a new level \(n'\) due to the energy transfer. The energy of this new state is: \(E_{final} = -\frac{13.6 \mathrm{eV}}{(n')^2}\) The energy change experienced by the hydrogen atom can be expressed as: \(\Delta E = E_{final} - E_{initial}\)

Find the highest level n'

Now we need to find the largest value for n' that satisfies the Eq.(3) considering the maximum energy that can be transferred to the hydrogen atom. Plugging \(\Delta E\) in Eq.(3): \(K_e = -\frac{13.6 \mathrm{eV}}{(n')^2} - (-1.51 \mathrm{eV})\) Solving for the largest possible integer value of \(n'\) will give us the answer.

Key concepts

Kinetic Energy of Electrons

Understanding the kinetic energy of electrons is crucial when exploring atomic collisions and excitations. Kinetic energy is the energy that an object possesses due to its motion. For electrons, this energy can be expressed mathematically using the formula:
\[ K_e = \frac{1}{2} m_e v^2 \]
where \( K_e \) is the kinetic energy, \( m_e \) is the mass of the electron, and \( v \) is the velocity of the electron. In atomic physics, the kinetic energy of an electron can often lead to the excitation of atoms when an electron collides with an atom. The faster the electron moves, the greater its kinetic energy will be. For instance, if an electron with a given velocity hits a hydrogen atom, the kinetic energy of the electron can potentially be transferred to the atom, leading to the excitation of the atom to a higher energy state, which brings us to understanding the principal quantum number and hydrogen atom energy levels.

Principal Quantum Number

The principal quantum number \( n \) plays a pivotal role in defining the energy levels of electrons in an atom. It is a positive integer (starting from 1, 2, 3, ...) that corresponds to the electron shells around the nucleus. The principal quantum number indicates the relative distance of the electron from the nucleus; a higher value of \( n \) indicates that the electron is in a higher energy level, further away from the nucleus.
This quantum number is not just a mere ranking of shells but rather is deeply connected to the energy of the electron in its respective shell. Each jump from a lower to a higher energy level indicates that an electron has gained energy. Conversely, when an electron drops from a higher to a lower energy level, it emits energy, often in the form of light. Therefore, the principal quantum number gives us valuable insights into the excitation potential of an atom and the likely transitions that may occur when energy is applied, such as during an electron collision.

Hydrogen Atom Energy Levels

Delving into the energy levels of a hydrogen atom, we're looking at discrete regions where an electron can exist around the atom's nucleus. Contrary to a classical approach where an electron can have any random value of energy, quantum mechanics restricts electrons to specific energy levels.
For a hydrogen atom, these energy levels are described by the formula:
\[ E_n = -\frac{13.6 \text{ eV}}{n^2} \]
Here, \( E_n \) is the energy associated with the \( n^{th} \) level and \( n \) is the principal quantum number. The negative sign indicates that these energy levels are bound states; the electron is trapped within the potential well of the atom. It is essential to note that the lower the value of \( n \), the lower (more negative) the energy level. When electrons transition between these levels, they can absorb or emit energy quantized to the difference between the initial and final states' energy.