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One of the emission lines of the hydrogen atom has a wavelength of \(93.07 \mathrm{~nm}\). (a) In what region of the electromagnetic spectrum is this emission found? (b) Determine the initial and final values of \(n\) associated with this emission.

Short Answer

Expert verified
The given wavelength of 93.07 nm falls within the Ultraviolet region of the electromagnetic spectrum. Using the Rydberg formula, we find that the corresponding initial and final values of the principal quantum number (n) for this emission in the hydrogen atom are \(n_1 = 1\) and \(n_2 = 5\).

Step by step solution

01

Identify the region of the electromagnetic spectrum

The given wavelength for the hydrogen spectrum is 93.07 nm. We need to find in which region of the electromagnetic spectrum this emission is found. The electromagnetic spectrum can be divided into the following regions (approximate values): - Gamma Rays: less than 0.01 nm - X-rays: 0.01 nm - 10 nm - Ultraviolet: 10 nm - 400 nm - Visible light: 400 nm - 700 nm - Infrared: 700 nm - 1 mm - Microwaves: 1 mm - 1 m - Radio waves: more than 1 m Comparing the given value with these regions, we find that 93.07 nm falls in the Ultraviolet region.
02

Find the initial and final values of n using the Rydberg formula

To determine the initial and final values of n, we use the Rydberg formula for hydrogen: \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) Where: - λ is the wavelength of the emission (in meters) - RH is the Rydberg constant for hydrogen, approximately \(1.097 \times 10^7 m^{-1}\) - n1 is the smaller quantum number - n2 is the larger quantum number (n2 > n1) First, convert the given wavelength to meters: \( 93.07 nm = 93.07 \times 10^{-9} m \) Now, using the Rydberg formula, we can find the possible values of n1 and n2: \( \frac{1}{93.07 \times 10^{-9}} = 1.097 \times 10^7 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) We can start by assuming n1 = 1 and finding n2: \( n_2^2 = \frac{1}{\frac{1}{n_1^2} - \frac{1}{\lambda \cdot R_H}} = \frac{1}{\frac{1}{(1)^2} - \frac{1}{(93.07 \times 10^{-9})(1.097 \times 10^7)}} \approx 4.98 \) Since n2 must be an integer value, we round n2 to the nearest integer, obtaining n2 = 5. Therefore, n1 = 1, and n2 = 5. Hence, the emission originates from an electron transitioning from the energy level with n = 5 to the energy level with n = 1.

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Key Concepts

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

Electromagnetic Spectrum
The electromagnetic spectrum is a continuum of all electromagnetic waves arranged according to frequency or wavelength. An electromagnetic wave consists of electric and magnetic fields oscillating at right angles to each other, moving through space. The spectrum ranges from gamma rays, which have the shortest wavelengths, to radio waves, with the longest wavelengths.

In this spectrum, the ultraviolet region lies between visible light and X-rays. It's characterized by wavelengths from about 10 nm to 400 nm. This is also where we find the hydrogen emission line of 93.07 nm as identified in the exercise. The hydrogen emission spectra are significant because they provide a deep understanding of atomic structure and quantum mechanics. Each wavelength corresponds to a photon's energy emitted as an electron transitions between energy levels within an atom.

Understanding this spectrum not only allows us to identify the type of light by its wavelength but also assists in discovering the properties of atoms and molecules. Applications of this knowledge range from astronomical studies, where analyzing the light from stars can tell us their composition, to the development of various technologies in the field of communications and medicine.
Rydberg Constant
The Rydberg constant is a fundamental figure in atomic physics related to the energy differences of electron transitions in hydrogen and other simple, one-electron systems. Represented by the symbol RH, it quantifies the limiting value of the highest wavenumber (the inverse of the wavelength) of any photon that can be emitted from an atom, as an electron moves from higher energy levels down to its lowest energy level.

Using the Rydberg constant, which has a value of approximately 1.097 x 10^7 m-1, scientists calculate the wavelengths of the spectral lines of hydrogen. The formula emerging from this constant, known as the Rydberg formula, is vital in spectroscopy. In the context of the exercise, when an electron transitions between two energy levels with quantum numbers n1 and n2, the wavelength of the emitted photon can be calculated. This has been crucial in understanding the structure of atoms and the interplay of energy, frequency, and wavelength in atomic emissions.
Quantum Number
Quantum numbers are essentially the addresses of electrons within an atom. They describe characteristics of the orbital where an electron resides, including its energy level, shape, and orientation. There are four types of quantum numbers: principal (n), azimuthal (l), magnetic (m), and spin (s).

The principal quantum number n, which can take any positive integer value, is mainly responsible for determining the energy of an electron in an atom and the size of the orbital. In the case of the hydrogen emission spectrums, when an electron makes a transition from a higher to a lower energy level, represented by a change in n values, it emits energy in the form of light. This light corresponds to a specific wavelength, which is observed in the emission spectrum.

As shown in the exercise, determining the initial and final principal quantum numbers (n1 and n2) is key in identifying the specific electronic transition and thus the wavelength of the emitted light. The larger the difference between these quantum numbers, the greater the energy release, and vice versa. This concept is a cornerstone of quantum mechanics and highlights the particle-like and wave-like duality of electrons.

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Most popular questions from this chapter

The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of \(9.47 \times 10^{6} \mathrm{~m} / \mathrm{s}\). What is the characteristic wavelength of this electron? Is the wavelength comparable to the size of atoms?

The hydrogen atom can absorb light of wavelength \(1094 \mathrm{~nm}\). (a) In what region of the electromagnetic spectrum is this absorption found? (b) Determine the initial and final values of \(n\) associated with this absorption.

(a) Why does the Bohr model of the hydrogen atom violate the uncertainty principle? (b) In what way is the description of the electron using a wave function consistent with de Broglie's hypothesis? (c) What is meant by the term probability density? Given the wave function, how do we find the probability density at a certain point in space?

Consider a transition of the electron in the hydrogen atom from \(n=4\) to \(n=9\). (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)?

Calculate the uncertainty in the position of (a) an electron moving at a speed of \((3.00 \pm 0.01) \times 10^{5} \mathrm{~m} / \mathrm{s},(\mathbf{b})\) a neutron moving at this same speed. (The masses of an electron and a neutron are given in the table of fundamental constants in the inside cover of the text.) (c) Based on your answers to parts (a) and (b), which can we know with greater precision, the position of the electron or of the neutron?

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