Question

Consider an electron in the ground state of the hydrogen atom, separated from the proton by a distance of \(0.0529 \mathrm{nm} .\) a) Viewing the electron as a satellite orbiting the proton in the electric potential, calculate the speed of the electron in its orbit. b) Calculate an effective escape speed for the electron. c) Calculate the energy of an electron having this speed, and from it determine the energy that must be given to the electron to ionize the hydrogen atom.

Short answer

a) The speed of an electron in the ground state of a hydrogen atom is approximately: $$ v \approx 2.18 \times 10^6 \, \text{m/s} $$ b) The effective escape speed of the electron from the hydrogen atom is approximately: $$ v_{esc} \approx 3.09 \times 10^6 \, \text{m/s} $$ c) The energy required to ionize the hydrogen atom is approximately \(13.6 \, \text{eV}\).

Step-by-step solution

a) Calculate the Speed of the Electron

We will use the centripetal force formula and Coulomb's law to calculate the speed of an electron in the ground state of a hydrogen atom. The centripetal force required to keep the electron in orbit around the proton is equal to the Coulomb force of attraction between them: $$ m_e v^2/r = k_e \frac{e^2}{r^2} $$ where \(m_e\) is the mass of the electron, \(v\) is its speed, \(r\) is the radial distance between the electron and proton, \(k_e\) is the Coulomb's constant, and \(e\) is the elementary charge. From this equation, we can solve for the speed of the electron: $$ v = \sqrt{\frac{k_e e^2}{m_e r}} $$ Now, we can plug in the values given and constants to calculate the electron's speed: $$ v = \sqrt{\frac{(8.99 \times 10^9 \, \text{N m}^2 \text{/C}^2)(1.602 \times 10^{-19} \, \text{C})^2}{(9.109 \times 10^{-31} \, \text{kg})(0.0529 \times 10^{-9} \, \text{m})}}\text{.} $$ After evaluating this expression, we find that the speed of the electron is approximately: $$ v \approx 2.18 \times 10^6 \, \text{m/s} $$

b) Calculate Effective Escape Speed

To calculate the effective escape speed of the electron, we will use the gravitational analogy of the escape speed and the electric potential energy in the hydrogen atom. The escape speed formula is: $$ v_{esc} = \sqrt{\frac{2 G M}{r}} $$ where \(G\) is the gravitational constant, \(M\) is the mass that the electron is escaping from, and \(r\) is the distance from the center of the mass. In the case of the electron orbiting the proton, we will replace the gravitational force with the electric force and the mass of the electron with the proton's mass: $$ v_{esc} = \sqrt{\frac{2 k_e e^2}{m_p r}} $$ Now, we can substitute the given values and constants to calculate the escape velocity: $$ v_{esc} = \sqrt{\frac{(2)(8.99 \times 10^9 \, \text{N m}^2 \text{/C}^2)(1.602 \times 10^{-19} \, \text{C})^2}{(1.67 \times 10^{-27} \, \text{kg})(0.0529 \times 10^{-9} \, \text{m})}}\text{.} $$ After evaluating this expression, we find that the effective escape speed of the electron is approximately: $$ v_{esc} \approx 3.09 \times 10^6 \, \text{m/s} $$

c) Calculate Ionization Energy

Now that we have calculated the escape speed of the electron, we can find the kinetic energy associated with this speed. The kinetic energy formula is: $$ K.E = \frac{1}{2} m_e v_{esc}^2 $$ Substitute the values and constants into the formula: $$ K.E = \frac{1}{2}(9.109 \times 10^{-31} \, \text{kg})(3.09 \times 10^6 \, \text{m/s})^2\text{.} $$ After evaluating this expression, we find that the kinetic energy of the electron is approximately: $$ K.E \approx 1.37 \times 10^{-18} \, \text{J} $$ The energy required to ionize the hydrogen atom can be determined by the difference between the electron's kinetic energy in the ground state and that associated with the escape speed. The energy of an electron in the ground state of a hydrogen atom is: $$ E_{ground} = -13.6 \, \text{eV} = -13.6 (1.602 \times 10^{-19} \, \text{J/eV}) = -2.18 \times 10^{-18} \, \text{J} $$ Then, the energy required to ionize the hydrogen atom is: $$ \Delta E = K.E - E_{ground} = 1.37 \times 10^{-18} \, \text{J} - (-2.18 \times 10^{-18} \, \text{J}) = 3.55 \times 10^{-18} \, \text{J} $$ Converting the energy to electron volts, we get: $$ \Delta E = 3.55 \times 10^{-18} \, \text{J} (1.602 \times 10^{19} \, \text{eV/J}) = 13.6 \, \text{eV} $$ So, the energy required to ionize the hydrogen atom is approximately \(13.6 \, \text{eV}\).

Key concepts

Electron Orbit Speed

When we examine an electron hurdling around the nucleus of an atom, conceptualizing it as a tiny satellite in orbit can be quite illuminating. The electron orbit speed is the velocity at which the electron travels around the nucleus. In a hydrogen atom, the simplest atomic system with just one electron and one proton, this speed is crucial for the stability of the atom. It is the balance between the electrostatic attraction pulling the electron towards the nucleus and the centrifugal force pushing it away that dictates this constant speed. What's fascinating is that in the ground state of hydrogen, the electron's orbit speed is calculated using a combination of equations from physics describing centripetal force and the electrostatic force outlined by Coulomb's law.

Electric Potential Energy

Delving into the realm of electric potential energy, similar to gravitational potential energy in celestial mechanics, it paints a picture of the energy landscape within an atom. In an electric field, an electron possesses electric potential energy by virtue of its position relative to a positive charge, such as a proton. This form of energy plays a central role when examining the forces at play within an atom. It is also essential for understanding concepts such as ionization energy since the work needed to move an electron away from its nucleus (ionizing the atom) relates directly to its electric potential energy. If one were to imagine the potential energy as a valley or well, the deeper the well, the more energy would be required to climb out of it—that's essentially what’s happening at an atomic scale when we speak of ionization.

Coulomb's Law

At heart of predicting the behavior of charged particles is Coulomb's law. This fundamental principle of electromagnetism describes the force of attraction or repulsion between two point charges. The formula encapsulates a simple notion: alike charges repel, opposite charges attract, and this force diminishes with the square of the distance between the charges. Precisely, it states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Understanding this law is instrumental when unraveling the intricacies of an atomic structure and the forces holding electrons in place around the nucleus. It helps us calculate the speed of an electron in orbit by balancing this electrostatic force with the centripetal force needed to keep the electron in its circular path.

Centripetal Force

The concept of centripetal force might conjure images of twirling objects tethered to strings, but at the subatomic level, it takes on a vital role. It's the invisible tether that holds an electron in its circular path around the nucleus and is always directed towards the center of the orbit. In the context of an atom, this isn't a physical force but a result of the electron's linear momentum and the electrostatic force of attraction exerted by the nucleus—a perfect example of how Coulomb's law manifests as the centripetal force in a hydrogen atom. It's pivotal for the electron to have just the right speed to balance the centripetal force with the electrostatic force, keeping it in a stable orbit and preventing it from spiraling into the nucleus.