Question

A sample of hydrogen atoms is irradiated with light with wavelength 85.5 nm, and electrons are observed leaving the gas. (a) If each hydrogen atom were initially in its ground level, what would be the maximum kinetic energy in electron volts of these photoelectrons? (b) A few electrons are detected with energies as much as 10.2 eV greater than the maximum kinetic energy calculated in part (a). How can this be?

Short answer

(a) Calculate the maximum kinetic energy using the photon energy minus 13.6 eV. (b) Electrons can be in excited states, providing them additional kinetic energy.

Step-by-step solution

Calculate the Energy of the Incident Photon

First, we need to calculate the energy of the incident photon using the given wavelength (85.5 nm). We use the formula: \[ E = \frac{hc}{\lambda} \]where \( h = 6.626 \times 10^{-34} \; \text{Joule} \cdot \text{second} \) is Planck's constant, \( c = 3.00 \times 10^{8} \; \text{meters/second} \) is the speed of light, and \( \lambda = 85.5 \times 10^{-9} \; \text{meters} \) is the wavelength. After calculation, convert the energy from joules to electron volts using the conversion factor \( 1 \; \text{eV} = 1.602 \times 10^{-19} \; \text{joules} \).

Determine the Work Function of Hydrogen

For hydrogen atoms in their ground state, the work function, which is the ionization energy or binding energy for the ground state, is 13.6 eV. The energy required to remove an electron from this ground state is thus 13.6 eV.

Calculate the Maximum Kinetic Energy of Ejected Electrons

Using the previously calculated photon energy and the work function of hydrogen, apply the photoelectric equation:\[ K_{\text{max}} = E_{\text{photon}} - \phi \]where \( \phi = 13.6 \; \text{eV} \) is the work function of hydrogen. Substitute the photon energy from Step 1 to find the maximum kinetic energy \( K_{\text{max}} \).

Explain Why Electrons Have Higher Energy than Expected

It's observed that some electrons have kinetic energies 10.2 eV greater than the computed maximum. This can occur if the electrons initially occupied higher energy excited states rather than the ground state. Thus, when irradiated, these electrons require less energy to escape compared to those in the ground state, releasing additional energy as kinetic energy.

Key concepts

Hydrogen Atom

The hydrogen atom is the simplest atom, comprising only one proton and one electron. In its ground state, the electron of a hydrogen atom is in its lowest energy level, closest to the nucleus. This setup makes hydrogen an excellent subject for studying atomic physics and quantum mechanics. When dealing with experiments involving hydrogen, such as the photoelectric effect, it's important to understand how energy levels work. Each level has a specific energy, and transitions of the electron between these levels absorb or release photons.
In these transitions, key measurements like energy are often noted in electron volts (eV). This offers a clear picture of the changes occurring at this atomic scale. Knowing the arrangement of energy levels helps predict how a hydrogen atom might respond to different stimuli, such as irradiation with light.

Photon Energy

Photon energy is a key element in understanding interactions within atomic physics. A photon is a particle of light that carries energy proportional to its frequency. The essential formula for calculating photon energy is given by: \[ E = \frac{hc}{\lambda} \] where:

• \( h = 6.626 \times 10^{-34} \; \text{Joule} \cdot \text{second} \) is Planck's constant,
• \( c = 3.00 \times 10^{8} \; \text{meters/second} \) is the speed of light,
• \( \lambda \) is the wavelength of the light.

This equation shows that energy is inversely proportional to wavelength; shorter wavelengths correspond to higher energy photons. In the context of photoelectric experiments, such as our exercise with hydrogen, knowing the energy of the photon allows you to determine the potential kinetic energy transferred to an ejected electron.
Understanding these interactions is crucial for comprehending phenomena like the ionization process and the emission of photoelectrons.

Ionization Energy

Ionization energy refers to the energy required to remove an electron from an atom in its gaseous state. For the hydrogen atom in its ground state, this energy is precisely calculated as 13.6 eV. It represents the "work function" for hydrogen, providing a benchmark for understanding how much external energy is needed to strip away the electron.
In the case of our photoelectric effect exercise, the ionization energy is used as the work function. This represents the minimum energy threshold needed by incident photons to eject an electron from its ground state. If the photon energy exceeds this threshold, the excess is converted into kinetic energy of the emitted electron. Therefore, knowing the ionization energy helps predict and explain the behavior of electrons under external stimuli such as light.

Kinetic Energy of Electrons

The kinetic energy of electrons, especially in the scope of the photoelectric effect, is pivotal for understanding electron movement after photon interaction. Once a photon with sufficient energy strikes an electron, and the energy exceeds the ionization threshold, the leftover energy transforms into the electron's kinetic energy.
The formula depicting this relationship is: \[ K_{\text{max}} = E_{\text{photon}} - \phi \] where:

• \( K_{\text{max}} \) is the maximum kinetic energy of the electron,
• \( E_{\text{photon}} \) is the energy of the photon,
• \( \phi \) is the work function, or ionization energy.

For the hydrogen atom, this calculation helps assess the speed and movement of electrons post-ionization. If electrons gain more kinetic energy than expected, it suggests they originated from excited states, signifying lower energy thresholds needed to free them. Understanding this concept is essential to mastering photoelectric experiments and atomic energy distribution.