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(a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the \(n\) = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels. (c) The average lifetime of the first excited level of a hydrogen atom is 1.0 \(\times\) 10\(^{-8}\) s. In the Bohr model, how many orbits does an electron in the \(n\) = 2 level complete before returning to the ground level?

Short Answer

Expert verified
The speeds are 2.18, 1.09, and 0.73 million m/s for n=1, 2, 3. Periods are 1.52x10^-16, 1.22x10^-15, and 4.10x10^-15 s. ~8.2 million orbits before dropping to n=1.

Step by step solution

01

Understanding the Bohr Model

The Bohr model provides a way to calculate the speed of an electron in an orbit around the nucleus of a hydrogen atom. The formula for the speed is given by \( v_n = \frac{e^2}{2\varepsilon_0 h} \cdot \frac{1}{n} \), where \( e \) is the electron charge, \( \varepsilon_0 \) is the permittivity of free space, \( h \) is Planck's constant, and \( n \) is the principal quantum number.
02

Calculating Speed for Different Energy Levels

Substitute the known constants and calculate the speed for each energy level \( n = 1, 2, 3 \):- \( n = 1 \): \( v_1 = 2.18 \times 10^6 \) m/s- \( n = 2 \): \( v_2 = \frac{v_1}{2} = 1.09 \times 10^6 \) m/s- \( n = 3 \): \( v_3 = \frac{v_1}{3} = 0.73 \times 10^6 \) m/s
03

Formula for Orbital Period

The orbital period \( T \) can be calculated using the formula \( T = \frac{2\pi r_n}{v_n} \), where \( r_n = n^2 a_0 \) with \( a_0 \) as the Bohr radius \( 5.29 \times 10^{-11} \) m. This gives an expression for the period as \( T_n = \frac{2\pi n^3 \cdot a_0}{v_1} \).
04

Calculating Orbital Period for Each Level

Calculate \( T_n \) for levels \( n = 1, 2, 3 \):- \( n = 1 \): \( T_1 = \frac{2\pi \cdot 5.29 \times 10^{-11}}{2.18 \times 10^6} \approx 1.52 \times 10^{-16} \) s- \( n = 2 \): \( T_2 = 2^3 \cdot T_1 = 1.22 \times 10^{-15} \) s- \( n = 3 \): \( T_3 = 3^3 \cdot T_1 = 4.10 \times 10^{-15} \) s
05

Orbits in First Excited Level

Calculate the number of orbits in \( n=2 \) before returning to \( n=1 \). The number of orbits \( N \) is given by \( N = \frac{L}{T} \) where \( L = 1.0 \times 10^{-8} \) s and \( T = 1.22 \times 10^{-15} \) s from Step 4. Compute \( N = \frac{1.0 \times 10^{-8}}{1.22 \times 10^{-15}} \approx 8.2 \times 10^6 \) orbits.

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Key Concepts

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

Hydrogen Atom
The hydrogen atom is the simplest atom and plays a critical role in understanding broader atomic structure in physics. It consists of a single proton as the nucleus around which a single electron orbits. In the Bohr model, this setup provides a way to analyze the electron's energy levels and behavior when it transitions between them.
The Bohr model was revolutionary because it introduced the idea of quantized energy levels, meaning the electron can only occupy certain orbits with specific energy. This concept was integral in explaining the emission spectra of hydrogen, which are the light waves emitted when its electron transitions from a higher energy level to a lower one.
In essence, the hydrogen atom allows physicists to test theories about atomic energy levels and electron dynamics in a controlled and predictable manner, providing a foundation for more complex atomic models.
Principal Quantum Number
The principal quantum number, denoted by \( n \), is a key component of quantum mechanics and the Bohr model, helping to describe the energy levels of an electron in an atom. In terms of the hydrogen atom, the principal quantum number determines the possible orbits an electron can inhabit.
  • For \( n = 1 \), the electron occupies the closest orbit to the nucleus, known as the ground state.
  • As \( n \) increases (e.g., \( n = 2, 3 \)), the electron resides in higher energy orbits, further from the nucleus.
Each principal quantum number corresponds to a different amount of energy, with higher values of \( n \) indicating higher energy levels due to increased distance from the nucleus. This quantum number is crucial for calculating other properties like the electron's speed and the orbital period, providing insight into how electrons behave within atoms.
Orbital Period
The orbital period refers to the time it takes for the electron to complete one full orbit around the nucleus. In the Bohr model for the hydrogen atom, the orbital period depends on the principal quantum number \( n \) and can be determined using the formula:
\[T_n = \frac{2\pi n^3 \cdot a_0}{v_1}\]
Where \( a_0 \) is the Bohr radius, and \( v_1 \) is the speed of the electron at the ground state.
For different levels:
  • At \( n = 1 \), the orbital period is very short, about \( 1.52 \times 10^{-16} \) seconds.
  • Higher levels have longer periods; for example, \( n = 2 \) has \( T_2 = 1.22 \times 10^{-15} \) seconds.
  • An even higher level like \( n = 3 \) results in a period of \( T_3 = 4.10 \times 10^{-15} \) seconds.
The period increases with higher \( n \) because the electron travels a larger orbit at slower speeds. This concept helps quantify how energetic transitions occur in atoms.
Electron Speed
In the Bohr model of the hydrogen atom, the speed of an electron is dictated by its orbit or principal quantum number. The speed decreases as the electron's orbit becomes larger, or as \( n \) increases.
The speed \( v_n \) of an electron in level \( n \) is given by:
\[v_n = \frac{e^2}{2\varepsilon_0 h} \cdot \frac{1}{n}\]
Where \( e \) is the electron charge, \( \varepsilon_0 \) is the permittivity of free space, and \( h \) is Planck's constant.
  • At \( n = 1 \), the electron's speed is around \( 2.18 \times 10^6 \) m/s.
  • For \( n = 2 \), it slows to approximately half, or \( 1.09 \times 10^6 \) m/s.
  • At \( n = 3 \), the speed further decreases to \( 0.73 \times 10^6 \) m/s.
This relationship shows how an electron's dynamics change with its energy level, affecting the atom's interaction with light and other particles. Understanding this electron speed is essential for deeper studies in quantum mechanics and atomic physics.

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Most popular questions from this chapter

A triply ionized beryllium ion, Be\(^{3+}\) (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom except that the nuclear charge is four times as great. (a) What is the ground-level energy of Be\(^{3+}\)? How does this compare to the ground-level energy of the hydrogen atom? (b) What is the ionization energy of Be\(^{3+}\)? How does this compare to the ionization energy of the hydrogen atom? (c) For the hydrogen atom, the wavelength of the photon emitted in the \(n\) = 2 to \(n\) = 1 transition is 122 nm (see Example 39.6). What is the wavelength of the photon emitted when a Be\(^{3+}\) ion undergoes this transition? (d) For a given value of \(n\), how does the radius of an orbit in Be\(^{3+}\) compare to that for hydrogen?

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?

An atom in a metastable state has a lifetime of 5.2 ms. What is the uncertainty in energy of the metastable state?

If your wavelength were 1.0 m, you would undergo considerable diffraction in moving through a doorway. (a) What must your speed be for you to have this wavelength? (Assume that your mass is 60.0 kg.) (b) At the speed calculated in part (a), how many years would it take you to move 0.80 m (one step)? Will you notice diffraction effects as you walk through doorways?

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?

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