Ch 38: Quantization

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 38: Quantization

Problem 27b

Chapter 38, Problem 27b

The allowed energies of a simple atom are 0.00 eV, 4.00 eV, and 6.00 eV. What wavelengths appear in the atom’s emission spectrum?

Verified step by step guidance

Step 1: Understand the problem. The atom has discrete energy levels: 0.00 eV, 4.00 eV, and 6.00 eV. When the atom transitions from a higher energy level to a lower one, it emits a photon. The energy of the photon corresponds to the difference between the two energy levels.

Step 2: Use the relationship between photon energy and wavelength. The energy of a photon is given by the equation:

E = h c / λ

, where

E

is the energy of the photon,

h

is Planck's constant (

6.63 \times 10 ⁻ 34

J·s),

c

is the speed of light (

3.00 \times 10 ⁸

m/s), and

λ

is the wavelength of the photon.

Step 3: Calculate the energy differences between the allowed energy levels. The possible transitions are: (6.00 eV to 4.00 eV), (6.00 eV to 0.00 eV), and (4.00 eV to 0.00 eV). The energy differences are:

2.00

eV,

6.00

eV, and

4.00

eV respectively.

Step 4: Convert the energy differences from electron volts (eV) to joules (J). Use the conversion factor:

1 eV = 1.60 \times 10 ⁻ 19

J. For example,

2.00 eV = 3.20 \times 10 ⁻ 19

Step 5: Solve for the wavelength

λ

using the formula

λ = h c / E

. Substitute the values of

h

c

, and the energy differences (converted to joules) to calculate the wavelengths for each transition. Ensure the units are consistent throughout the calculation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Energy Levels

In quantum mechanics, energy levels refer to the specific energies that an electron in an atom can occupy. Electrons can only exist in these discrete energy states, and transitions between them result in the absorption or emission of energy in the form of photons. The allowed energies for the atom in the question are 0.00 eV, 4.00 eV, and 6.00 eV, indicating the quantized nature of electron states.

Photon Emission

Photon emission occurs when an electron transitions from a higher energy level to a lower one, releasing energy in the form of a photon. The energy of the emitted photon corresponds to the difference in energy between the two levels. This process is fundamental in understanding atomic spectra, as the wavelengths of emitted light can be calculated from these energy differences.

Wavelength and Energy Relationship

The relationship between the energy of a photon and its wavelength is described by the equation E = hc/λ, where E is energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. This inverse relationship means that higher energy transitions produce shorter wavelengths. By calculating the energy differences for the allowed transitions, one can determine the corresponding wavelengths in the emission spectrum.

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