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The spectrum of the hydrogen atom reflects the splitting of the \(1 s^{2} \mathrm{S}\) and \(2 p^{2} \mathrm{P}\) terms into levels. The energy difference between the levels in each term is much smaller than the difference in energy between the terms. Given this information, how many spectral lines are observed in the \(1 s^{2} \mathrm{S} \rightarrow 2 p^{2} \mathrm{P}\) transition? Are the frequencies of these transitions very similar or quite different?

Short Answer

Expert verified
There is only one allowed transition between the 1s²S and 2p²P terms, which means there will be one spectral line observed in the 1s²S → 2p²P transition. As there is only one transition, the question about the frequencies being very similar or quite different doesn't apply. There is only one frequency associated with the energy difference of the 1s²S → 2p²P transition.

Step by step solution

01

Identify the possible transitions

First, we need to identify the possible transitions between the 1s²S and 2p²P terms. We can use the selection rules for allowed transitions in the hydrogen atom, which state that a transition is allowed if: 1. The principal quantum number n changes (Δn ≠ 0). 2. The total angular momentum quantum number J changes by 0 or ±1 (ΔJ = 0, ±1), but J = 0 to J = 0 transition is forbidden. For the 1s²S term, the principal quantum number n = 1, and for the 2p²P term, n = 2. Since Δn ≠ 0, the transition is allowed. Now we need to look at the angular momentum quantum number J.
02

Determine the values of J for the initial and final terms

For the 1s²S term, since it's an S state (l = 0), the total angular momentum quantum number J can only be 0. For the 2p²P term, since it's a P state (l = 1), the total angular momentum quantum number J can be 0, 1, or 2.
03

Apply the selection rules for J

Now, we can apply the selection rules for J and find the possible transitions: 1. Initial J = 0 to Final J = 1 (ΔJ = +1), which is allowed. There is only one allowed transition, which means there will be one spectral line observed in the 1s²S → 2p²P transition.
04

Comment on the frequencies of the transitions

Since there is only one transition, the question about the frequencies being very similar or quite different doesn't apply. There is only one frequency associated with the energy difference of the 1s²S → 2p²P transition.

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Key Concepts

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

Spectral Lines
Spectral lines are essentially patterns formed in the spectrum from the emission or absorption of light by atoms. When an electron in a hydrogen atom transitions between different energy levels, it results in either the absorption or emission of light at specific wavelengths. Each set of spectral lines corresponds to a specific series of emissions or absorptions. In the hydrogen atom, spectral lines arise due to transitions between various energy levels. These lines are observed in the spectrum and are distinctive because each line represents a difference in energy levels. For example, the transition in the original exercise from the 1s²S level to the 2p²P level creates a distinct spectral line due to the specific energy difference between these two states.
Because these energy differences are unique, each spectral line is like a fingerprint for the transition occurring in the atom. Students might be interested to know that spectral lines are the key behind the understanding of atomic structure and can tell us much about an element when analyzed through spectroscopy.
Transition Selection Rules
Transition selection rules are a set of principles that determine which electron transitions between energy levels are allowed in a quantum mechanical system, such as an atom like hydrogen. These rules need to be satisfied for a transition to occur and for a spectral line to be observed. Consider the main transitional selection rules:
  • The principal quantum number n must change (Δn ≠ 0). This means that the initial and final states of the electron must be in different shells.
  • The total angular momentum quantum number J must change by 0 or ±1. However, transitions where J = 0 in the initial state to J = 0 in the final state are strictly forbidden.
For example, in our hydrogen atom scenario, the transition from 1s²S to 2p²P leverages these rules to conclude that the Δn is non-zero and ΔJ equals +1 (from 0 to 1), allowing the transition. Understanding these selection rules is crucial for predicting which transitions will be allowed and consequently, which spectral lines will appear, helping students grasp how quantum mechanics governs atomic behavior.
Angular Momentum Quantum Number
The angular momentum quantum number, often denoted as L or J in quantum mechanics, plays a vital role in defining the electron's state in an atom and determining its allowed transitions. Angular momentum comes in two forms: orbital and spin angular momentum. However, in the context of transitions and spectral lines, we usually focus on the total angular momentum quantum number J which combines these. For our hydrogen atom:
  • The initial 1s²S level has a J value of 0 because it is in an S state (L=0), adding up to J = 0.
  • The final 2p²P state has possible J values of 0, 1, or 2 because it is a P state (L=1), offering more complexity.
These values matter because they help determine which transitions are possible under the aforementioned selection rules. Only transitions where the change in J adheres to the rules are allowed, thus impacting how many spectral lines we observe. This quantum number aids in explaining not just theoretical principles but also observable phenomena in spectroscopy, making it an essential concept for students to understand.

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Most popular questions from this chapter

The first ionization potential of ground-state He is \(24.6 \mathrm{eV} .\) The wavelength of light associated with the \(1 s 2 p^{1} \mathrm{P}\) term is \(58.44 \mathrm{nm} .\) What is the ionization energy of the He atom in this excited state?

Derive the ground-state term symbols for the following atoms or ions: a. \(\mathrm{H}\) b. \(\mathrm{F}\) c. \(\mathrm{Na}^{+}\) d. \(\mathrm{Sc}\)

Calculate the wavelengths of the first three lines of the Lyman, Balmer, and Paschen series, and the series limit (the shortest wavelength) for each series.

The ground-state level for the phosphorus atom is \(^{4} \mathrm{S}_{3 / 2}\). List the possible values of \(L, M_{b} S, M_{S}, J,\) and \(M_{J}\) consistent with this level.

The inelastic mean free path of electrons in a solid, \(\lambda,\) governs the surface sensitivity of techniques such as AES and XPS. The electrons generated below the surface must make their way to the surface without losing energy in order to give elemental and chemical shift information. An empirical expression for elements that give \(\lambda\) as a function of the kinetic energy of the electron generated in AES or XPS is \(\lambda=538 E^{-2}+0.41(l E)^{0.5} .\) The units of \(\lambda\) are monolayers, \(E\) is the kinetic energy of the electron in eV, and \(l\) is the monolayer thickness in nanometers. On the basis of this equation, what kinetic energy maximizes the surface sensitivity for a monolayer thickness of \(0.3 \mathrm{nm} ?\) An equation solver would be helpful in obtaining the answer.

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