Problem 65 - University Physics with Modern Physics

Question

A sample of hydrogen atoms is irradiated with light with wavelength 85.5 \(\mathrm{nm}\) , 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?

Step-by-Step Solution

Answer

(a) 0.95 eV; (b) Some atoms were in an excited state (10.2 eV above the ground).

Step 1: Calculate the energy of the incident photons

First, convert the given wavelength of the light, 85.5 nm, into meters: \[ 85.5 \text{ nm} = 85.5 \times 10^{-9} \text{ m}. \] Now, use the formula \(E = \frac{hc}{\lambda}\) to find the energy \(E\) of the photon. Here, \(h = 6.626 \times 10^{-34} \text{ J s}\) is Planck's constant and \(c = 3.0 \times 10^{8} \text{ m/s}\) is the speed of light. \[E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{85.5 \times 10^{-9}} = 2.33 \times 10^{-18} \text{ J}.\] Convert this energy into electron volts (1 eV = \(1.602 \times 10^{-19} \text{ J}\)): \[E = \frac{2.33 \times 10^{-18}}{1.602 \times 10^{-19}} = 14.55 \text{ eV}.\]

Step 2: Determine the ionization energy from the ground state

The ionization energy required to remove an electron from the ground state of hydrogen is 13.6 eV. This energy corresponds to the energy difference between the ground state and the ionization level of the hydrogen atom.

Step 3: Calculate the maximum kinetic energy of the photoelectrons

Using the photoelectric effect equation, the maximum kinetic energy \(K_{\text{max}}\) of the photoelectron is given by: \[K_{\text{max}} = E - \text{Ionization energy}.\] Substitute the values we have: \[K_{\text{max}} = 14.55 \text{ eV} - 13.6 \text{ eV} = 0.95 \text{ eV}.\]

Step 4: Analyze the anomaly in electron energies

Certain electrons have energies 10.2 eV higher than the calculated maximum kinetic energy. This suggests that some hydrogen atoms were initially in an excited state. The difference in energy implies these atoms could have been in the first excited state, an energy level 10.2 eV above the ground state.

Key Concepts

Hydrogen Atom

The hydrogen atom is the simplest atom in the universe, consisting of just one proton and one electron. It's an essential building block in chemistry and physics because it helps us understand more complex atomic structures. In the ground state, the electron is in the closest possible orbit around the nucleus. This is the lowest energy level, also known as the n=1 state.
Hydrogen plays a critical role in the study of atomic spectra. When energy is added to a hydrogen atom, the electron can move to a higher energy level or orbital. This is crucial in the context of the photoelectric effect, where the energy from photons leads to electrons being ejected from atoms.
Since hydrogen has only one electron, its ionization energy is relatively low. This attribute makes it a vital atom for experiments, such as those exploring the photoelectric effect, because the calculations and implications are more straightforward.

Ionization Energy

Ionization energy is the amount of energy needed to remove an electron from an atom. For hydrogen, the ionization energy is 13.6 eV. When discussing the photoelectric effect, ionization energy becomes critical because it defines the threshold energy to eject the electron from the atom's ground state.
This concept explains why, in the exercise given, the photoelectron's maximum kinetic energy proved to be less than the incident photon's energy. The photon must have at least this energy to liberate the electron. If the energy is higher, only the surplus manifests as kinetic energy in the freed electron.
Understanding ionization energy aids in predicting whether a photon of a certain energy can cause ionization. If the photon's energy is below the ionization energy, no electron will be ejected. Higher energy photons will result in more energetic electrons.

Photon Energy

Photon energy refers to the energy carried by a single photon, and it is directly proportional to its frequency and inversely proportional to its wavelength. The formula used to calculate this energy is \[ E = \frac{hc}{\lambda} \]where:

• E is the photon energy,
• h is Planck's constant \(6.626 \times 10^{-34} \text{ J s}\),
• c is the speed of light \(3.0 \times 10^{8} \text{ m/s}\),
• \(\lambda\) is the wavelength of the photon.

In our context, photons with sufficient energy can cause the ejection of electrons through the photoelectric effect. For instance, in the original exercise, photons with an energy equivalent to 14.55 eV strike hydrogen atoms. This energy is enough to overcome the ionization energy and still leave some extra energy that translates into the kinetic energy of the ejected electron.
Photon energy is a foundational concept in quantum physics, helping to explain the interactions between light and matter on a particle level.

Excited State

An excited state refers to the condition where an electron in an atom has absorbed energy and moved to a higher energy level. In hydrogen, this can happen when it's exposed to light or heat, causing the electron to jump from its ground state (n=1) to higher levels (n=2, 3, etc.).
When the electron transitions back to a lower energy state, it releases energy in the form of light or another photon. This emission can be detected as part of the atomic emission spectrum. In the photoelectric effect, if an electron already occupies an excited state, less energy is required to remove it compared to one in the ground state. This situation explains variations in the observed kinetic energy of photoelectrons.
In the exercise, electrons with energies 10.2 eV higher than expected suggest that some hydrogen atoms were in the first excited state. This aligns with the known difference in energy between the ground state and the first excited state being 10.2 eV, facilitating their higher-than-anticipated kinetic energy upon ejection.