Question

Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius \(r = 5.29 \times \space 10^{-11}\) m around a stationary proton. What is the speed of the electron in its orbit?

Short answer

The speed of the electron is approximately \(2.18 \times 10^6\) m/s.

Step-by-step solution

Identify Known Quantities

We have the radius of the orbit, which is \(r = 5.29 \times 10^{-11}\) m. We also know the mass of the electron \(m_e = 9.11 \times 10^{-31}\) kg and the charge of the electron \(e = 1.6 \times 10^{-19}\) C. The electrostatic constant is \(k = 8.99 \times 10^9\) N m²/C².

Use Coulomb's Law for Centripetal Force

In a hydrogen atom model, the force keeping the electron in orbit is the electrostatic force. Using Coulomb's Law, we have:\[ F = \frac{k \, e^2}{r^2} \] This force acts as the centripetal force on the electron:

Solve for Centripetal Force Equation

The centripetal force needed to keep an object moving in a circular path is given by:\[ F_{c} = \frac{m_e \, v^2}{r} \]Set the electrostatic force equal to the centripetal force:\[ \frac{k \, e^2}{r^2} = \frac{m_e \, v^2}{r} \]

Rearrange and Solve for Velocity

Simplify and solve the equation for \(v\):\[v^2 = \frac{k \, e^2}{m_e \, r}\]\[v = \sqrt{\frac{k \, e^2}{m_e \, r}}\]

Substitute Values and Calculate

Substitute known values into the equation:\[ v = \sqrt{\frac{(8.99 \times 10^9) \times (1.6 \times 10^{-19})^2}{(9.11 \times 10^{-31}) \times (5.29 \times 10^{-11})}} \]Calculate to find:\[ v \approx 2.18 \times 10^6 \text{ m/s} \]

Key concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in physics that describes the force between two charged particles. In a hydrogen atom, this law helps us understand the attraction between the positively charged proton and the negatively charged electron. Coulomb's Law is given by the equation:

• \( F = \frac{k \, q_1 \, q_2}{r^2} \)

where:

• \( F \) is the electrostatic force
• \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \)
• \( q_1 \) and \( q_2 \) are the charges of the particles
• \( r \) is the distance between the charges

In the model of a hydrogen atom, the proton and electron are the two charges involved, thus their attraction forms the force that keeps the electron in orbit.

Centripetal Force

Centripetal force is the force required to keep an object moving in a circular path. This force acts towards the center of the rotation. In a hydrogen atom, as the electron orbits the proton, it needs a force to pull it inward and maintain its circular path. This is where centripetal force comes into play.The formula for centripetal force can be expressed as:

• \( F_c = \frac{m \, v^2}{r} \)

where:

• \( F_c \) is the centripetal force
• \( m \) is the mass of the electron
• \( v \) is the velocity of the electron
• \( r \) is the radius of the orbit

In the hydrogen atom model, this centripetal force is provided by the electrostatic attraction force as described by Coulomb's Law. By setting these forces equal, we can derive expressions for orbital characteristics such as the electron's velocity.

Hydrogen Atom

The hydrogen atom is the simplest atom and consists of just one electron orbiting a single proton. Its simplicity makes it a valuable system for studying basic principles of quantum mechanics and atomic physics. According to Bohr's model, the electron moves in circular orbits around the nucleus, which is the proton in this case.Several key features of the hydrogen atom include:

• Its single electron orbits the proton at specific energy levels.
• The electron's orbit is characterized by a particular radius, approximately \(5.29 \times 10^{-11}\) meters, called the Bohr radius.
• The attraction between the proton and electron comes from the electrostatic force described by Coulomb's Law.

Understanding the motion of the electron in a hydrogen atom helps illustrate fundamental physical concepts, such as quantization of energy levels and the balance of forces within atomic structures.