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Chapter 28: Problem 56

An unknown element has a spectrum for absorption from its ground level with lines at \(2.0,5.0,\) and 9.0 eV. Its ionization energy is 10.0 eV. (a) Draw an energy-level diagram for this element. (b) If a 9.0 eV photon is absorbed, what energies can the subsequently emitted photons have?

Short Answer

Expert verified
Possible emitted photon energies are 4.0 eV, 7.0 eV, and 9.0 eV.

Step by step solution

01

Analyze the Given Information

We are provided with an unknown element where its absorption spectrum shows lines at 2.0 eV, 5.0 eV, and 9.0 eV. The ionization energy is 10.0 eV. We are required to draw an energy-level diagram and determine the energies of emitted photons after absorption of a 9.0 eV photon.
02

Draw the Energy-Level Diagram

Start by labeling the ground state energy level at 0 eV. The element has energy levels at 2.0 eV, 5.0 eV, and 9.0 eV based on the absorption lines. The ionization level is at 10.0 eV. Arrange these levels vertically with the ground state at the bottom and the ionization level at the top, marking each intermediate level (2.0 eV, 5.0 eV, and 9.0 eV).
03

Absorption of a 9.0 eV Photon

When a 9.0 eV photon is absorbed by the element starting from the ground state (0 eV), the electron transitions to the 9.0 eV energy level since it matches the energy of the photon absorbed.
04

Determine Possible Emission Energies

After reaching the 9.0 eV level, the electron can fall back to any lower energy level (including the ground state). Possible transitions are: from 9.0 eV to 5.0 eV (emitting 4.0 eV), from 9.0 eV to 2.0 eV (emitting 7.0 eV), and from 9.0 eV to 0 eV (emitting 9.0 eV). Thus, the possible energies for emitted photons are 4.0 eV, 7.0 eV, and 9.0 eV.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Absorption Spectrum
An absorption spectrum provides vital clues about the energy levels in an atom. When light passes through a substance, certain wavelengths are absorbed by the electrons of the atoms, causing them to jump to higher energy levels. These absorbed wavelengths manifest as dark lines in the spectrum. For our unknown element, the absorption spectrum reveals energy levels where electrons can move from the ground level. These are given as 2.0 eV, 5.0 eV, and 9.0 eV.
  • The absorption of a photon by an electron triggers an energy transition upward to a specific energy level.
  • Only photons with just the right energy can be absorbed and cause these transitions.
The process reflects the quantized nature of energy levels in atoms, meaning not every energy amount can cause a transition, only precise ones.
Ionization Energy
Ionization energy is a crucial parameter in understanding an atom's energy-level diagram. It is the energy required to remove an electron from an atom completely, which corresponds to a transition from the atom's highest energy level to an infinite distance (the point of ionization). Our unknown element requires 10.0 eV for ionization.
  • When the ionization energy is reached, the electron escapes the atomic pull and leaves the atom.
  • Higher ionization energy indicates a stronger force holding the electrons within the atom.
Knowing the ionization energy helps in determining the highest energy level of the atom before an electron is lost and assists in plotting the maximum level on an energy-level diagram.
Photon Emission
Photon emission occurs when an electron in an excited state loses energy and returns to a lower energy level. This process releases energy in the form of a photon鈥攍ight with a specific energy corresponding to the difference in energy levels. After a 9.0 eV photon is absorbed by the unknown element, several emission possibilities arise as the electron reverts to lower states.
  • From 9.0 eV to 5.0 eV, a photon of 4.0 eV is emitted.
  • Going from 9.0 eV to 2.0 eV results in the emission of a 7.0 eV photon.
  • Returning to the ground state (0 eV) releases a 9.0 eV photon.
Identifying these emission energies helps in verifying the energy-level diagram and understanding the element's atomic structure.
Energy Transition
Energy transitions are the essence of atomic behavior in electric and magnetic fields and within the absorption and emission spectra. They involve electrons moving between different quantized energy levels, either absorbing or releasing energy. In our scenario, when a 9.0 eV photon is absorbed, the electron transitions from the ground state (0 eV) to a higher level (9.0 eV). This movement signifies an upward transition.
  • Upward transitions occur upon absorption, as seen with our 9.0 eV photon.
  • Downward transitions occur during emission when the electron drops from a higher to a lower level.
Understanding these transitions is vital for interpreting spectral lines and predicting the behavior of elements when interacting with light.

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Most popular questions from this chapter

An incident \(x\) -ray photon is scattered from a free electron that is initially at rest. The photon is scattered straight back at an angle of \(180^{\circ}\) from its initial direction. The wavelength of the scattered photon is 0.0830 \(\mathrm{nm.}\) (a) What is the wavelength of the incident photon? (b) What is the magnitude of the momentum of the electron after the collision? (c) What is the kinetic energy of the electron after the collision?

\(\bullet\) The photoelectric work function of potassium is 2.3 \(\mathrm{eV}\) . If light having a wavelength of 250 \(\mathrm{nm}\) falls on potassium, find (a) the stopping potential in volts; (b) the kinetic energy, in electron volts, of the most energetic electrons ejected; (c) the speeds of these electrons.

\(\cdot\) What is the de Broglie wavelength of a red blood cell with a mass of \(1.00 \times 10^{-11} \mathrm{g}\) that is moving with a speed of 0.400 \(\mathrm{cm} / \mathrm{s} ?\) Do we need to be concerned with the wave nature of the blood cells when we describe the flow of blood in the body?

\(\bullet\) A certain atom has an energy level 3.50 eV above the ground state. When excited to this state, it remains \(4.0 \mu s,\) on the average, before emitting a photon and returning to the ground state. (a) What is the energy of the photon? What is its wavelength? (b) What is the smallest possible uncertainty in energy of the photon?

\(\bullet\) Exposing photographic film. The light-sensitive com- pound on most photographic films is silver bromide (AgBr). A film is "exposed" when the light energy absorbed dissociates this molecule into its atoms. (The actual process is more complex, but the quantitative result does not differ greatly.) The energy of dissociation of AgBr is \(1.00 \times 10^{5} \mathrm{J} / \mathrm{mol}\) . For a photon that is just able to dissociate a molecule of silver bromide, find (a) the photon's energy in electron volts, (b) the wavelength of the photon, and (c) the frequency of the photon. (d) Light from a firefly can expose photographic film, but the radiation from an FM station broadcasting \(50,000 \mathrm{W}\) at 100 \(\mathrm{MHz}\) cannot. Explain why this is so, basing your answer on the energy of the photons involved.
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