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When an electron makes a transition from the \(n=4\) to the \(n=2\) hydrogen atom Bohr orbit, the energy difference between these two orbits \(\left(4.1 \times 10^{-19} \mathrm{~J}\right)\) is emitted as a photon of light. The relationship between the energy of a photon and its wavelength is given by \(E=h c / \lambda\), where \(E\) is the energy of the photon in \(J, h\) is Planck's constant \(\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\), and \(c\) is the speed of light \(\left(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)\). Find the wavelength of light emitted by hydrogen atoms when an electron makes this transition.

Short Answer

Expert verified
\(\lambda = 4.84 \times 10^{-7} \text{m}\) or 484 nm

Step by step solution

01

Understand the Photonic Energy-Wavelength Relationship

The energy of the photon emitted during an electron transition in the hydrogen atom is related to the wavelength by the formula: \(E = \frac{h c}{\lambda}\), where \(E\) is the energy of the photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the photon.
02

Rearrange the Formula to Solve for Wavelength

Rearrange the relationship to solve for the wavelength (\(\lambda\)): \(\lambda = \frac{h c}{E}\).
03

Substitute the Known Values

Substitute the values of Planck's constant \(h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s} \), the speed of light \(c = 3.00 \times 10^{8} \text{ m/s} \), and the energy difference \(E = 4.1 \times 10^{-19} \text{ J} \) into the rearranged equation to calculate the wavelength.
04

Calculate the Wavelength

Perform the calculation using the provided values to find the wavelength: \(\lambda = \frac{6.626 \times 10^{-34} \text{ J}\cdot\text{s} \cdot 3.00 \times 10^{8} \text{ m/s}}{4.1 \times 10^{-19} \text{ J}}\).
05

Simplify the Expression and Find the Wavelength

Simplify the expression to get the wavelength in meters, which is the solution.

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Key Concepts

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

Photonic energy-wavelength relationship
Understanding the relationship between photonic energy and wavelength is essential for grasping the physics of light and atomic transitions. Let's simplify this concept using the case of an electron transition, such as from a higher to a lower energy level within an atom. When this happens, a photon of light is emitted, and the energy of this photon is inversely proportional to its wavelength. To put it simply, the shorter the wavelength of the light, the higher the energy of the photon emitted.

The mathematical formula that represents this relationship is: \[\begin{equation}E = \frac{h c}{\text{lambda}}\end{equation}\]Where:
  • E stands for the energy of the photon (measured in Joules, J)
  • h is Planck's constant
  • c is the speed of light (measured in meters per second, m/s)
  • \text{lambda} represents the wavelength of the light (measured in meters, m)
The beauty of this relationship is that by knowing any two of the values, you can solve for the third. This principle is invaluable in the field of spectroscopy and in understanding the properties of light emitted or absorbed by substances.

The electron transition exercise provided is a great example. An electron drops from one energy level to another and emits a photon with energy that can be calculated using Planck's constant and the speed of light. By rearranging the formula, we can solve for the wavelength of the emitted light, giving us insights into the type of light that would be observed, such as ultraviolet, visible, or infrared.
Planck's constant
The concept of Planck's constant is pivotal when delving into the microscopic world of quantum physics. It bridges the gap between the macroscopic and quantum world, providing a unit of measurement for the action in quantum mechanics. The value of Planck's constant is:\[\begin{equation}h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}\end{equation}\]This very small number may not seem important at first glance, but it holds the key to understanding phenomena like the quantization of energy levels in an atom and the duality of light acting as both a wave and a particle.

Planck's constant enables us to calculate the energy of photons given their frequency or, as seen in the photonic energy-wavelength relationship, their wavelength. It is named after Max Planck, who proposed that energy is exchanged in discrete amounts, or 'quanta'. This discovery sparked the development of quantum theory, revolutionizing our understanding of the atomic and subatomic world.

In terms of the exercise, Planck's constant is used to determine the energy of the photon emitted when an electron transitions between energy levels, which in turn can reveal the photon's wavelength. Without this constant, our ability to decipher the quantum signatures of different elements would be significantly hampered.
Speed of light
The speed of light, denoted as c, is a fundamental constant in the universe. It is the speed at which all massless particles and waves, including light, propagate in the vacuum of space and it is an essential component of many equations in physics. The recognized value of the speed of light is:\[\begin{equation}c = 3.00 \times 10^{8} \text{ m/s}\end{equation}\]The speed of light is not just awe-inspiringly fast, but it's also a crucial constant in physics that influences our understanding of space, time, and causality. Beyond that, it's vital for calculations involving light and electromagnetic waves, as we see in the photonic energy-wavelength relationship.

The immutability of the speed of light is central to Albert Einstein's theory of relativity and has profound implications for the concepts of simultaneity and time dilation. In practical terms, the speed of light is used in technologies such as GPS and in the study of distant cosmic objects. It also allows us to calculate how long it takes for light from stars or other celestial bodies to reach us, providing a cosmic 'speedometer' that defines the scale of the universe.

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Most popular questions from this chapter

Which type of electromagnetic radiation has the longest wavelength? (a) visible (b) ultraviolet (c) infrared (d) X-ray

Write full orbital diagrams and indicate the number of unpaired electrons for each element. (a) \(\mathrm{F}\) (b) \(C\) (c) \(\mathrm{Ne}\) (d) \(\mathrm{Be}\)

Write the electron configuration for each ion. What do all of the electron configurations have in common? (a) \(\mathrm{Ca}^{2+}\) (b) \(\mathrm{K}^{+}\) (c) \(\mathrm{S}^{2-}\) (d) \(\mathrm{Br}^{-}\)

List the quantum-mechanical orbitals through \(5 s\), in the correct energy order for multi-electron atoms.

When atoms lose more than one electron, the ionization energy to remove the second electron is always more than the ionization energy to remove the first. Similarly, the ionization energy to remove the third electron is more than the second and so on. However, the increase in ionization energy upon the removal of subsequent electrons is not necessarily uniform. For example, consider the first three ionization energies of magnesium: \(\begin{array}{ll}\text { First ionization energy } & 738 \mathrm{~kJ} / \mathrm{mol} \\ \text { Second ionization energy } & 1450 \mathrm{~kJ} / \mathrm{mol} \\ \text { Third ionization energy } & 7730 \mathrm{~kJ} / \mathrm{mol}\end{array}\) The second ionization energy is roughly twice the first ionization energy, but then the third ionization energy is over five times the second. Use the electron configuration of magnesium to explain why this is so. Would you expect the same behavior in sodium? Why or why not?

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