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In the Bohr model of the hydrogen atom, what is the de Broglie wavelength of the electron when it is in (a) the \(n\) = 1 level and (b) the \(n\) = 4 level? In both cases, compare the de Broglie wavelength to the circumference 2\(\pi{r_n}\) of the orbit.

Short Answer

Expert verified
The de Broglie wavelengths are 3.32 × 10⁻¹⁰ m for n=1 and 1.32 × 10⁻⁹ m for n=4. The wave matches the circumference for n=1, but is shorter for n=4.

Step by step solution

01

Understanding the Bohr Model

The Bohr model of the hydrogen atom describes electrons in quantized orbits around the nucleus. Each orbit corresponds to a specific energy level, denoted by the principal quantum number \(n\).
02

de Broglie Wavelength Formula

The de Broglie wavelength \(\lambda\) of a particle is given by \(\lambda = \frac{h}{mv}\), where \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ m}^2\text{kg/s}\), \(m\) is the mass of the electron \(9.109 \times 10^{-31} \text{ kg}\), and \(v\) is the velocity of the electron.
03

Calculating Electron Velocity for Orbit

For the Bohr model, the velocity \(v\) of an electron in orbit \(n\) is given by \(v = \frac{e^2}{2\varepsilon_0 h}\cdot\frac{1}{n}\), where \(e\) is the elementary charge \(1.602 \times 10^{-19} \text{ C}\), and \(\varepsilon_0\) is the vacuum permittivity \(8.854 \times 10^{-12} \text{ C}^2/\text{N} \cdot m^2\).
04

Electron Velocity at n=1

For \(n=1\), substitute the known values into the formula to calculate \(v\). \[v_{n=1} = \frac{1.602^2 \times 10^{-38}}{2 \times 8.854 \times 10^{-12} \times 6.626 \times 10^{-34}} = 2.18 \times 10^6 \text{ m/s}.\]
05

de Broglie Wavelength for n=1

Use the calculated velocity to find the de Broglie wavelength for \(n=1\). Substitute values in \(\lambda = \frac{h}{mv}\).\[\lambda_{n=1} = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 2.18 \times 10^6} \approx 3.32 \times 10^{-10} \text{ m}.\]
06

Circumference for n=1

The circumference for \(n=1\) is given by \(2\pi r_n\), where \(r_n = \frac{n^2 h^2 \varepsilon_0 }{\pi m e^2}\). Calculate value when \(n=1\).\[r_1 = \frac{1^2 \times 6.626^2 \times 10^{-68} \times 8.854 \times 10^{-12}}{\pi \times 9.109 \times 10^{-31} \times 1.602^2 \times 10^{-38}} \approx 5.29 \times 10^{-11} \text{ m}\]\[C_{n=1} = 2\pi \times 5.29 \times 10^{-11} = 3.32 \times 10^{-10} \text{ m}.\]
07

Electron Velocity at n=4

For \(n=4\), apply the velocity formula.\[v_{n=4} = \frac{1.602^2 \times 10^{-38}}{2 \times 8.854 \times 10^{-12} \times 6.626 \times 10^{-34} \times 4} = 5.45 \times 10^5 \text{ m/s}.\]
08

de Broglie Wavelength for n=4

Calculate the de Broglie wavelength for \(n=4\).\[\lambda_{n=4} = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 5.45 \times 10^5} \approx 1.32 \times 10^{-9} \text{ m}.\]
09

Circumference for n=4

Calculate the circumference for \(n=4\). \[r_4 = \frac{4^2 \times 6.626^2 \times 10^{-68} \times 8.854 \times 10^{-12}}{\pi \times 9.109 \times 10^{-31} \times 1.602^2 \times 10^{-38}} \approx 8.468 \times 10^{-10} \text{ m}\]\[C_{n=4} = 2\pi \times 8.468 \times 10^{-10} \approx 5.316 \times 10^{-9} \text{ m}.\]
10

Comparing Wavelength and Circumference

For both \(n=1\) and \(n=4\), compare the de Broglie wavelength to the orbit's circumference. For \(n=1\), the wavelength and circumference approximately match, as expected in the Bohr model. For \(n=4\), the wavelength is much smaller than the circumference.

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Key Concepts

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

de Broglie Wavelength
The de Broglie wavelength is a fundamental concept in quantum mechanics. It describes the wave-like nature of particles such as electrons. According to de Broglie, every moving particle or object has an associated wavelength, given by the formula \( \lambda = \frac{h}{mv} \). Here, \( \lambda \) represents the wavelength, \( h \) is Planck's constant \(6.626 \times 10^{-34} \text{ m}^2\text{kg/s}\), \( m \) is the mass of the particle, and \( v \) is its velocity.

This concept is crucial in understanding how particles can exhibit properties of both particles and waves. In the Bohr model, the de Broglie wavelength helps explain why electrons occupy specific orbital paths around the nucleus. These paths correspond to integer multiples of the electron's wavelength, ensuring the wave pattern is complete and stable. This integrative view accounts for the electron's stability and the discrete energy levels observed in atoms.
Principal Quantum Number
The principal quantum number, denoted by \( n \), is a key factor in the Bohr model and quantum mechanics. It represents the different energy levels available to an electron within an atom and essentially dictates the electron's orbit and energy.

The value of \( n \) can be a positive integer (1, 2, 3, etc.), with each number indicating a different energy level or "shell." The lower the value of \( n \), the lower the energy level, and the closer the electron is to the nucleus. The higher the \( n \), the higher the energy and the further the electron's orbit from the nucleus. For instance, when an electron is in the \( n = 1 \) level, it's in the ground state, which is its lowest energy state. Conversely, \( n = 4 \) represents a higher energy level.

The principal quantum number also influences the electron's velocity and the radius of its orbit. It is a critical component in understanding electron arrangement and behavior in atoms.
Electron Velocity
Electron velocity in the Bohr model is vital in determining the behavior of the electron within its orbit. The velocity of an electron orbiting the nucleus is influenced by the principal quantum number \( n \). It can be calculated using the formula:

\[ v = \frac{e^2}{2\varepsilon_0 h}\cdot\frac{1}{n} \]

Here, \( e \) is the elementary charge, \( \varepsilon_0 \) is the vacuum permittivity, and \( h \) is Planck's constant.

From this formula, we observe that the electron's velocity decreases as the principal quantum number \( n \) increases. In simpler terms, electrons in higher energy levels \( (n = 4) \) move slower compared to those in lower levels \( (n = 1) \). This change in speed reflects how the attractive force from the nucleus lessens with increased distance, affecting the electron's kinetic energy and stability within its orbit.
Electron Orbit
Electron orbit refers to the specific path or "shell" that an electron follows around the nucleus in the Bohr model. This model visualizes electrons as occupying fixed orbits, much like planets revolving around the sun. Each orbit corresponds to a quantized energy level, determined by the principal quantum number \( n \).

The electron orbit’s radius is not arbitrary but follows from the formula:

\[ r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \]

Here, \( r_n \) is the radius of the orbit for a given \( n \). It shows that the radius increases with the square of \( n \), meaning higher energy levels are further from the nucleus.

Moreover, the circumference of the electron's orbit at any energy level \( n \) is related to its de Broglie wavelength, reinforcing the concept that only specific orbits are allowed where the electron's wave nature results in constructive interference. This determines the stable orbits of electrons as seen in reality slightly collapsing into the expected circumference, such as at \( n = 1 \) and \( n = 4 \). The understanding of electron orbits is fundamental for explaining atomic structure and spectra.

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Most popular questions from this chapter

A CD-ROM is used instead of a crystal in an electrondiffraction experiment. The surface of the CD-ROM has tracks of tiny pits with a uniform spacing of 1.60 \(\mu{m}\). (a) If the speed of the electrons is 1.26 \(\times\) 10\(^4\) m/s, at which values of \(\theta\) will the \(m\) = 1 and \(m\) = 2 intensity maxima appear? (b) The scattered electrons in these maxima strike at normal incidence a piece of photographic film that is 50.0 cm from the CD-ROM. What is the spacing on the film between these maxima?

An atom with mass \(m\) emits a photon of wavelength \(\lambda\). (a) What is the recoil speed of the atom? (b) What is the kinetic energy \(K\) of the recoiling atom? (c) Find the ratio \(K/E\), where \(E\) is the energy of the emitted photon. If this ratio is much less than unity, the recoil of the atom can be neglected in the emission process. Is the recoil of the atom more important for small or large atomic masses? For long or short wavelengths? (d) Calculate \(K\) (in electron volts) and \(K/E\) for a hydrogen atom (mass 1.67 \(\times\) 10\(^{-27}\)kg) that emits an ultraviolet photon of energy 10.2 eV. Is recoil an important consideration in this emission process?

(a) In an electron microscope, what accelerating voltage is needed to produce electrons with wavelength 0.0600 nm? (b) If protons are used instead of electrons, what accelerating voltage is needed to produce protons with wavelength 0.0600 nm? (\(Hint\): In each case the initial kinetic energy is negligible.)

To investigate the structure of extremely small objects, such as viruses, the wavelength of the probing wave should be about one-tenth the size of the object for sharp images. But as the wavelength gets shorter, the energy of a photon of light gets greater and could damage or destroy the object being studied. One alternative is to use electron matter waves instead of light. Viruses vary considerably in size, but 50 nm is not unusual. Suppose you want to study such a virus, using a wave of wavelength 5.00 nm. (a) If you use light of this wavelength, what would be the energy (in eV) of a single photon? (b) If you use an electron of this wavelength, what would be its kinetic energy (in eV)? Is it now clear why matter waves (such as in the electron microscope) are often preferable to electromagnetic waves for studying microscopic objects?

High-speed electrons are used to probe the interior structure of the atomic nucleus. For such electrons the expression \(\lambda = h/p\) still holds, but we must use the relativistic expression for momentum, \(p = mv/\sqrt{1 - v^2/c^2}\). (a) Show that the speed of an electron that has de Broglie wavelength \(\lambda\) is $$v = \frac{c}{\sqrt1+(mc\lambda/h)^2} $$ (b) The quantity \(h/mc\) equals 2.426 \(\times\) 10\(^{-12}\) m. (As we saw in Section 38.3, this same quantity appears in Eq. (38.7), the expression for Compton scattering of photons by electrons.) If \(\lambda\) is small compared to \(h/mc\), the denominator in the expression found in part (a) is close to unity and the speed \(v\) is very close to c. In this case it is convenient to write \(v = (1 - \Delta)c\) and express the speed of the electron in terms of rather than v. Find an expression for \(\delta\) valid when \(\lambda \ll h mc\). [\(Hint:\) Use the binomial expansion (1 + \(z)^n = 1 + nz + [n(n - 1)z^2/2] + \cdots\) g, valid for the case 0 z 0 6 1.4 (c) How fast must an electron move for its de Broglie wavelength to be 1.00 \(\times\) 10\(^{-15}\) m, comparable to the size of a proton? Express your answer in the form \(v =(1 - \Delta)c\), and state the value of \(\Delta\)

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