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A hydrogen atom undergoes a transition from the \(n=5\) to the \(n=2\) state. (a) What are the energy and wavelength of the photon that is emitted? (b) If the angular momentum is conserved and if the Bohr model is used to describe the atom, what must the angular momentum be of the photon that is emitted? (As we will see in Chapter \(41,\) the modern quantum-mechanical description of the hydrogen atom gives a different result.)

Short Answer

Expert verified
The photon has an energy of 2.856 eV and wavelength of 435 nm; its angular momentum is \( \hbar \).

Step by step solution

01

Determine Energy Levels in Bohr Model

In the Bohr model of the hydrogen atom, the energy of an electron in the nth orbit is given by \[ E_n = -\frac{13.6 \text{ eV}}{n^2} \]where \( E_n \) is the energy of the nth level and \( n \) is the principal quantum number. For levels \( n=5 \) and \( n=2 \), calculate these energies:\[ E_5 = -\frac{13.6 \text{ eV}}{5^2} = -0.544 \text{ eV} \]\[ E_2 = -\frac{13.6 \text{ eV}}{2^2} = -3.4 \text{ eV} \]
02

Calculate the Energy of the Emitted Photon

The energy of the photon emitted during the transition is the difference between the initial and final energy levels:\[ \Delta E = E_2 - E_5 = -3.4 \text{ eV} + 0.544 \text{ eV} = -2.856 \text{ eV} \]Since it's the difference, the value is positive:\[ \Delta E = 2.856 \text{ eV} \]Convert this energy to joules using the conversion \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \):\[ \Delta E = 2.856 \times 1.602 \times 10^{-19} \text{ J} \approx 4.57 \times 10^{-19} \text{ J} \]
03

Calculate Wavelength of the Photon

Use the relation between energy and wavelength given by\[ E = \frac{hc}{\lambda} \]where \( h = 6.626 \times 10^{-34} \text{ J s} \) is Planck's constant and \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light. Solve for wavelength \( \lambda \):\[ \lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{4.57 \times 10^{-19}} \approx 4.35 \times 10^{-7} \text{ m} \]Convert meters to nanometers:\[ \lambda = 4.35 \times 10^{-7} \text{ m} \times 10^9 \text{ nm/m} \approx 435 \text{ nm} \]
04

Find Angular Momentum of Emitted Photon

In the Bohr model, each electron transition is associated with a change in angular momentum. The change in angular momentum \( \Delta L \) associated with the transition is given by\[ \Delta L = \hbar \]where \( \hbar = \frac{h}{2\pi} \) is the reduced Planck's constant. Using the conversion \[ \hbar = \frac{6.626 \times 10^{-34}}{2\pi} \approx 1.055 \times 10^{-34} \text{ Js} \],the angular momentum of the emitted photon is therefore \( \hbar \).

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Key Concepts

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

Energy Levels
The Bohr model introduces the idea of quantized energy levels in atoms, vital for understanding atomic behavior. For the hydrogen atom, which is the simplest case, the energy for any electron in an orbit (quantum state) is calculated using the formula: \[ E_n = -\frac{13.6 \text{ eV}}{n^2} \]Here, \(E_n\) represents the energy, while \(n\) is a positive integer known as the principal quantum number.
This principal quantum number identifies the specific energy levels, with lower values of \(n\) indicating more stable, lower energy states. In our specific case, the electron transitions from the \(n=5\) level, which has a higher energy (less negative), to the \(n=2\) level, a more stable lower energy level.
- **Energy Level Calculation**: - For \(n=5\): \(E_5 = -0.544 \text{ eV}\) - For \(n=2\): \(E_2 = -3.4 \text{ eV}\)The difference in these energy levels is crucial. The energy is released as a photon when the electron falls from a higher to a lower energy state. This is a key concept in understanding atomic transitions.
Photon Emission
Photon emission occurs when an electron transitions between different energy levels within an atom. In this process, the energy difference between the initial and final states is converted into a photon—a packet of electromagnetic radiation.
- **Energy of the Photon**: - Calculated as the difference between the energies of the two levels: \(\Delta E = 2.856 \text{ eV}\) for the transition from \(n=5\) to \(n=2\). - This energy is often converted from electron volts (eV) to joules for calculation purposes, using the conversion factor \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\).
- **Wavelength of the Photon**: - The relationship between energy \(E\), wavelength \(\lambda\), and the constants Planck's constant \(h\) and the speed of light \(c\) is given by the formula: \(E = \frac{hc}{\lambda}\). - Solving for \(\lambda\), this leads to a wavelength \(\lambda \approx 435 \text{ nm}\), placing the photon within the visible light spectrum.
Photon emission is integral to many applications, such as lasers and light bulbs, highlighting the practical importance of understanding photon energy and wavelength.
Angular Momentum Conservation
Angular momentum conservation is a fundamental principle in physics, stating that in a closed system, the total angular momentum remains constant unless acted upon by external forces. In the context of the Bohr model and atomic transitions, this principle also applies on the atomic scale where each electron is viewed as having an angular momentum quantized by the rule:\[ L = n\hbar \]Here, \(n\) is an integer and \(\hbar\) (the reduced Planck's constant) is a fundamental physical constant. When an electron transitions from an upper to a lower energy level, its angular momentum changes by a unit of \(\hbar\).
- **Angular Momentum of the Photon**: - The change in angular momentum \(\Delta L = \hbar\). - The transition from \(n=5\) to \(n=2\) results in a photon possessing angular momentum of \(\hbar\), conserving the overall angular momentum of the system.
Understanding angular momentum conservation helps explain numerous physical phenomena beyond just atomic physics, such as the behavior of celestial bodies and the mechanics of everyday rotating objects.

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Most popular questions from this chapter

Blue Supergiants. A typical blue supergiant star (the type that explode and leave behind black holes) has a surface temperature of \(30,000 \mathrm{K}\) and a visual luminosity \(100,000\) times that of our sun. Our sun radiates at the rate of \(3.86 \times 10^{26} \mathrm{W}\) . (Visual) luminosity is the total power radiated at visible wavelengths. (a) Assuming that this star behaves like an ideal blackbody, what is, the principal wavelength it radiates? Is this light visible? Use your answer to explain why these stars are blue. (b) If we assume that the power radiated by the star is also \(100,000\) times that of our sun, what is the radius of this star? Compare its size to that of our sun, which has a radius of \(6.96 \times 10^{5} \mathrm{km}\) . (c) Is it really correct to say, that the visual luminosity is proportional to the total power radiated? Explain.

A triply ionized beryllium ion, \(\mathrm{Be}^{3+}\) (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"t? How does this compare to the ground-level energy of the hydrogen atom? (b) What is the ionization energy of \(\mathrm{Be}^{3+} ?\) How does this compare to the ionization energy of the hydrogen atom?(c) For the hydrogen atom the wave-length of the photon emitted in the \(n=2\) to \(n=1\) transition is 122 \(\mathrm{nm}\) (see Example \(38.6 ) .\) What is the wavelength of the photon emitted when a \(B e^{3+}\) ion undergoes this transition? (d) For a given value of \(n\) , how does the radius of an orbit in \(B e^{3+}\) compare to that for hydrogen?

Two stars, both of which behave like ideal blackbodies, radiate the same total energy per second. The cooler one has a surface temperature \(T\) and 3.0 times the diameter of the hotter star. (a) What is the temperature of the hotter star in terms of \(T\) ? (b) What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star?

(a) Show that, as \(n\) gets very large, the energy levels of the hydrogen atom get closer and closer together in energy. (b) Do the radii of these energy levels also get closer together?

A photon has momentum of magnitude \(8.24 \times 10^{-28} \mathrm{kg}\) . \(\mathrm{m} / \mathrm{s}\) (a) What is the energy of this photon? Give your answer in joules and in electron volts. (b) What is the wavelength of this photon? In what region of the electromagnetic spectrum does it lie?

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