Question

Considering the three electronic transitions \(n=2 \rightarrow\) \(n=1, n=3 \rightarrow n=2\) and \(n=4 \rightarrow n=3\) for the hydrogen at which one of the following is true. (1) The photon emitted in the transition \(n=4\) to \(n=3\) would have the longest wavelength. (2) The photon emitted in the transition \(n=2\) to \(n=1\) would have the longest wavelength. (3) The transition from \(n=3\) to \(n=1\) is forbidden. (4) The electron does not experience any change in orbit radius for any of these transitions.

Short answer

The photon emitted in the transition \( n=4 \to n=3 \) would have the longest wavelength.

Step-by-step solution

Understand the Question

Identify what is being asked: comparing the wavelengths of photons emitted during specific transitions in the hydrogen atom.

Review Electron Transitions

Recall that an electron dropping to a lower energy level emits a photon. The energy difference between the levels determines the photon's wavelength.

Use the Rydberg Formula

For hydrogen, the energy difference between two levels is given by the Rydberg formula: \[ E = - R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \] where \( E \) is the energy, \( R_H \) is the Rydberg constant, and \( n_i \) and \( n_f \) are the initial and final energy levels, respectively.

Calculate Energies for Each Transition

For the transition \(n=2 \to n=1\): \[ E_1 = -R_H \left( 1 - \frac{1}{4} \right) = -R_H \left( \frac{3}{4} \right) \] For the transition \(n=3 \to n=2\): \[ E_2 = -R_H \left( \frac{1}{4} - \frac{1}{9} \right) = -R_H \left( \frac{5}{36} \right) \] For the transition \(n=4 \to n=3\): \[ E_3 = -R_H \left( \frac{1}{9} - \frac{1}{16} \right) = -R_H \left( \frac{7}{144} \right) \]

Determine Wavelengths

The wavelength \( \lambda \) is related to energy \( E \) by the formula: \[ E = \frac{hc}{\lambda} \] where \( h \) is Planck's constant and \( c \) is the speed of light. From this, \( \lambda = \frac{hc}{E} \). Since \( E_3 \) is the smallest amongst \( E_1 \), \( E_2 \), and \( E_3 \), the wavelength \( \lambda_3 \) of the transition \( n=4 \to n=3 \) will be longest.

Validate Other Statements

Option (2): The transition \( n=2 \to n=1 \) does not have the longest wavelength as \( E_1 > E_2 > E_3 \). Option (3): No rule explicitly forbids the transition from \( n=3 \to n=1 \). Option (4): Electron transitions must change orbit radius since different energy levels correspond to different orbits.

Conclusion

The correct statement is (1): The photon emitted in the transition \( n=4 \to n=3 \) would have the longest wavelength.

Key concepts

Rydberg Formula

The Rydberg formula is essential for understanding the energy levels of an electron in a hydrogen atom. It allows us to calculate the energy difference between two energy levels. The formula is given as: oindent E = - R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) Here,

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• \(E\) is the energy of the photon emitted or absorbed during the transition,
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• \(R_H\) is the Rydberg constant,
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• \(n_i\) is the initial energy level,
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• \(n_f\) is the final energy level.
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By plugging in the values of \(n_i\) and \(n_f\) for the transition, we can calculate the energy and further use it to determine the wavelength of the photon involved in the transition. Understanding this formula gives us a strong foundation to analyze transitions in hydrogen atoms and to predict the characteristics of emitted or absorbed photons.

Electron Energy Levels

Electrons in a hydrogen atom occupy specific energy levels, typically denoted by quantum numbers (\(n\)). The lower the energy level (lower \(n\)), the closer the electron is to the nucleus:

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• \(n = 1\) is the ground state, the closest electron shell,
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• Higher \(n\) values (\(n = 2, 3, 4, ...\)) represent excited states with electrons further from the nucleus.
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When an electron transitions between these levels, it either absorbs or emits a photon. The energy difference between these levels determines the photon's energy and hence its wavelength. This is crucial for understanding why different transitions emit photons of different wavelengths. The larger the energy difference (drop from higher \(n\) to lower \(n\)), the higher the energy (and shorter the wavelength) of the emitted photon.

Photon Wavelength Calculation

Calculating the wavelength of a photon emitted or absorbed during an electron transition involves a few key steps. Firstly, we use the Rydberg formula to determine the energy of the transition (E\text{}). Next, we relate this energy to wavelength using the formula: oindentE = \frac{hc}{\lambda} \textrm { or } \lambda = \frac{hc}{E} where:

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• \(h\) is Planck's constant (6.62607015 x 10^{-34} \textrm {Joule sec}),
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• \(c\) is the speed of light (3.00 x 10^8 \textrm {m/sec}), and
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• \(\lambda\) is the wavelength of the photon.
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Calculating \(\lambda\) from \(\textE\text\) involves simple algebra, and understanding this helps us link the concepts of energy levels and observed wavelengths. Shorter wavelengths correspond to higher energy transitions, and longer wavelengths to lower energy transitions.

Electron Transitions

Electron transitions occur when an electron moves from one energy level to another within an atom. These transitions result in the emission or absorption of a photon:

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• When an electron falls to a lower energy level, it emits a photon whose energy equals the energy difference between the levels.
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• Conversely, when an electron absorbs energy and jumps to a higher level, it absorbs a photon of that corresponding energy.
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In the context of the hydrogen atom, electrons can move between various levels (\(n = 1, 2, 3,...\)). Let's take specific examples:

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• \(n=2 \to n=1\) is a high energy transition resulting in the emission of a photon with short wavelength.
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• \(n=4 \to n=3\) is a lower energy transition compared to \(n=2 \to n=1\), thus emits a photon with a longer wavelength.
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Electron transitions are key to understanding many phenomena in atomic physics and spectroscopy, and mastering these concepts helps in predicting and explaining the behavior of atoms during energy changes.