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Consider a transition of the electron in the hydrogen atom from \(n=8\) to \(n=3\). (a) Is \(\Delta E\) for this process positive or negative? (b) Determine the wavelength of light that is associated with this transition. Will the light be absorbed or emitted? (c) In which portion of the electromagnetic spectrum is the light in part (b)?

Short Answer

Expert verified
(a) 螖E is negative. (b) The wavelength of light is 957.3 nm and is emitted. (c) The light is in the infrared portion of the electromagnetic spectrum.

Step by step solution

01

Calculate the energy of the initial and final levels

We use the formula for the energy of an electron in the nth energy level of a hydrogen atom: \[E_n = - \frac{13.6 eV}{n^2}\] For the initial level (n=8): \[E_{initial} = - \frac{13.6 eV}{8^2} = -\frac{13.6 eV}{64} = -0.2125 eV\] For the final level (n=3): \[E_{final} = - \frac{13.6 eV}{3^2} = -\frac{13.6 eV}{9} = -1.511 eV\]
02

Calculate the change in energy, 螖E

The change in energy during the transition is given as: \[\Delta E = E_{final} - E_{initial}\] \[\Delta E = -1.511 eV - (-0.2125 eV) = -1.511 eV + 0.2125 eV = -1.2985 eV\] Since 螖E is negative, that means energy is released during the transition.
03

Calculate the wavelength associated with the change in energy

To calculate the wavelength of light associated with the transition, we use the energy-wavelength formula: \[\Delta E = \frac{hc}{\lambda}\] Where h is Planck's constant (h = 4.1357 脳 10鈦宦光伒 eV路s), c is the speed of light (c = 2.998 脳 10鈦 m/s), and 位 is the wavelength. Solving for the wavelength: \[\lambda = \frac{hc}{\Delta E}\] Now we plug in the numbers: \[\lambda = \frac{(4.1357 脳 10^{-15} eV路s)(2.998 脳 10^8 m/s)}{-1.2985 eV} = 9.573 脳 10^{-8} m = 957.3 nm\]
04

Determine if the light is absorbed or emitted

Since 螖E is negative and energy was released during the transition, the light associated with this transition must be emitted.
05

Identify the portion of the electromagnetic spectrum

The wavelength of the light emitted (957.3 nm) falls in the infrared (IR) portion of the electromagnetic spectrum, which typically has wavelengths between 700 nm and 1 mm. Final answers: (a) The change in energy, 螖E, is negative. (b) The wavelength of light associated with this transition is 957.3 nm, and the light is emitted. (c) The light is in the infrared (IR) portion of the electromagnetic spectrum.

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Key Concepts

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

Energy Level Calculation
When we talk about the energy levels of a hydrogen atom, we're delving into the quantum world where electrons exist in specific orbits around the nucleus. These orbits or "energy levels" are denoted by the principal quantum number, denoted as \( n \). Each energy level \( E_n \) is quantified by the relation:\[ E_n = - \frac{13.6 eV}{n^2} \]This formula tells us the energy associated with an electron at any given level \( n \). The negative sign signifies that the electrons are bound to the nucleus, meaning it would require 13.6 eV to remove an electron from the ground state.To understand energy changes during electron transitions, like from \( n=8 \) to \( n=3 \), we calculate the energy at each level. This is done in steps:
  • Find the energy at the initial level: \( E_{initial} = \frac{-13.6 eV}{8^2} \)
  • Find the energy at the final level: \( E_{final} = \frac{-13.6 eV}{3^2} \)
  • Calculate the change in energy \( \Delta E = E_{final} - E_{initial} \)
In our example, \( \Delta E = -1.2985 \ eV \), indicating the electron moves to a lower energy level and thus releases energy, making the process exothermic. Insights like these are foundational in quantum mechanics and help us understand atomic spectra.
Wavelength Determination
Understanding wavelength is crucial when studying light transitions because it defines the color and type of radiation emitted or absorbed. The energy change in an electron's transition within a hydrogen atom directly influences the wavelength of light involved. This concept is quantified by the energy-wavelength relationship:\[ \Delta E = \frac{hc}{\lambda} \]Where:
  • \( h \) is Planck's constant \( = 4.1357 \times 10^{-15} \,eV\cdot s \)
  • \( c \) is the speed of light \( = 2.998 \times 10^{8} \ \text{m/s} \)
  • \( \lambda \) is the wavelength of the light.
To determine the wavelength, rearrange the formula to find \( \lambda \):\[ \lambda = \frac{hc}{\Delta E} \]Substituting the values, we get a wavelength of about 957.3 nm for our given transition from \( n=8 \) to \( n=3 \). This value indicates that the hydrogen atom emits infrared light. The relationship between energy changes and wavelength is pivotal in understanding atomic emission and absorption spectra in various applications from astronomy to medical imaging.
Electromagnetic Spectrum
The electromagnetic spectrum is an extensive range of wavelengths and frequencies of electromagnetic radiation. It stretches from the very short wavelengths, like gamma rays, to very long wavelengths, like radio waves. In the context of hydrogen atom electron transitions, understanding where a specific wavelength fits within this spectrum provides valuable insights into the nature of the emitted or absorbed light. In our example, the wavelength calculated was approximately 957.3 nm. This positions it in the infrared (IR) section of the spectrum, which includes wavelengths from about 700 nm to 1 mm. Infrared is not visible to the human eye, but it's experienced as heat. This characteristic makes infrared radiation useful in various technologies such as night-vision equipment and remote controls. Understanding where different wavelengths fall within the electromagnetic spectrum is important for fields like spectroscopy, which relies on precise wavelength measurements to identify substances. When energy transitions in hydrogen atoms lead to specific wavelengths, scientists can make deductions about energy levels and atomic structure, adding to the intricate knowledge of quantum physics and the nature of light.

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Most popular questions from this chapter

In the television series Star Trek, the transporter beam is a device used to "beam down" people from the Starship Enterprise to another location, such as the surface of a planet. The writers of the show put a "Heisenberg compensator" into the transporter beam mechanism. Explain why such a compensator (which is entirely fictional) would be necessary to get around Heisenberg's uncertainty principle.

For orbitals that are symmetric but not spherical, the contour representations (as in Figures 6.23 and 6.24 ) suggest where nodal planes exist (that is, where the electron density is zero). For example, the \(p_{x}\) orbital has a node wherever \(x=0\). This equation is satisfied by all points on the \(y z\) plane, so this plane is called a nodal plane of the \(p_{x}\) orbital. (a) Determine the nodal plane of the \(p_{z}\) orbital. (b) What are the two nodal planes of the \(d_{x y}\) orbital? (c) What are the two nodal planes of the \(d_{x^{2}-y^{2}}\) orbital?

Give the values for \(n, l,\) and \(m_{l}\) for \((\mathbf{a})\) each orbital in the \(3 p\) subshell, (b) each orbital in the \(4 f\) subshell.

(a) What is the frequency of radiation that has a wavelength of \(10 \mu \mathrm{m}\), about the size of a bacterium? \((\mathbf{b})\) What is the wavelength of radiation that has a frequency of \(5.50 \times 10^{14} \mathrm{~s}^{-1} ?\) (c) Would the radiations in part (a) or part \((b)\) be visible to the human eye? (d) What distance does electromagnetic radiation travel in \(50.0 \mu \mathrm{s} ?\)

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of \(325 \mathrm{nm}\) (a) What is the energy of a photon of this wavelength? (b) What is the energy of a mole of these photons? (c) How many photons are in a \(1.00 \mathrm{~mJ}\) burst of this radiation? \((\mathbf{d})\) These \(\mathrm{UV}\) photons can break chemical bonds in your skin to cause sunburn-a form of radiation damage. If the \(325-\mathrm{nm}\) radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in \(\mathrm{kJ} / \mathrm{mol}\).

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