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Order the following transitions in the hydrogen atom from smallest to largest frequency of light absorbed: \(n=3\) to \(n=6, n=4\) to \(n=9, n=2\) to \(n=3\), and \(n=1\) to \(n=2\).

Short Answer

Expert verified
The order of the transitions from the smallest to the largest frequency of light absorbed is: \(n=4\) to \(n=9\), \(n=3\) to \(n=6\), \(n=2\) to \(n=3\), and \(n=1\) to \(n=2\).

Step by step solution

01

Apply the Rydberg formula for hydrogen atoms

The formula that relates the energy difference between two energy levels in a hydrogen atom and the frequency of light absorbed is as follows: \(ν = R_H ( \frac{1}{n_1^2} - \frac{1}{n_2^2} )\) where \(ν\) is the frequency of light absorbed, \(R_H\) is the Rydberg constant for hydrogen, \(n_1\) and \(n_2\) are the principal quantum numbers of the initial and final states of the electron, respectively.
02

Calculate the frequency of light absorbed for each transition

Using the given transitions and the Rydberg formula, we can calculate the frequency of light absorbed for each transition. 1. For \(n=3\) to \(n=6\): \(ν_1 = R_H ( \frac{1}{3^2} - \frac{1}{6^2} ) = \frac{5 R_H}{36}\) 2. For \(n=4\) to \(n=9\): \(ν_2 = R_H ( \frac{1}{4^2} - \frac{1}{9^2} ) = \frac{65 R_H}{324}\) 3. For \(n=2\) to \(n=3\): \(ν_3 = R_H ( \frac{1}{2^2} - \frac{1}{3^2} ) = \frac{5 R_H}{36}\) 4. For \(n=1\) to \(n=2\): \(ν_4 = R_H ( \frac{1}{1^2} - \frac{1}{2^2} ) = \frac{3 R_H}{4}\) Now, let's compare the frequencies.
03

Order the transitions from smallest to largest frequency of light absorbed

We can order the frequencies (\(ν_1\), \(ν_2\), \(ν_3\), and \(ν_4\)) obtained in Step 2, from the smallest to the largest. Comparing the frequencies, we have: \(ν_2 < ν_1 = ν_3 < ν_4\) Therefore, the order of the transitions from the smallest to the largest frequency of light absorbed is: \(n=4\) to \(n=9\), \(n=3\) to \(n=6\), \(n=2\) to \(n=3\), and \(n=1\) to \(n=2\).

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Key Concepts

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

Hydrogen Atom Transitions
In a hydrogen atom, electrons occupy specific energy levels, also known as orbits, which are defined by quantum numbers. These energy levels are not fixed; electrons can move, or "transition," from one energy level to another. Such transitions are crucial because they either absorb or emit light, which is how we can study these processes experimentally.
When moving from a lower energy level to a higher one, an electron absorbs energy. The difference in energy between these two levels will determine the frequency of light absorbed.
Some common types of transitions in a hydrogen atom are:
  • *Lyman series:* Transitions where the electron ends at the n=1 level.
  • *Balmer series:* Transitions where the electron ends at the n=2 level.
  • *Paschen series:* Transitions where the electron ends at the n=3 level.
This energy absorption results in different wavelengths or colors of light, which are characteristic of the transition series.
Understanding these transitions helps us order the frequency of light absorbed in different circumstances.
Frequency of Light Absorbed
The frequency of light absorbed during a hydrogen atom transition is determined by the energy difference between initial and final quantum states. Frequency, denoted as \( ν \), is directly related to the amount of energy absorbed during this transition.
To calculate this frequency, we use the Rydberg formula:\[ ν = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where \( R_H \) is the Rydberg constant, \( n_1 \) is the initial energy level, and \( n_2 \) is the final energy level. This formula expresses how frequency depends on the square of the inverse of these quantum numbers.
When comparing transitions, a larger energy gap (where \( n_2 \) is much larger than \( n_1 \)) will result in a higher frequency of light absorbed. This relationship confirms that larger transitions involve more energy absorption, resulting in higher frequencies.
Quantum Numbers
Quantum numbers are sets of numbers that describe various properties of an electron in an atom. For hydrogen atom transitions, the principal quantum number, often represented as \( n \), is the most relevant. It indicates the energy level or shell in which an electron resides. Lower values of \( n \) correspond to lower energy levels closer to the nucleus.
The principal quantum number can take positive integer values such as 1, 2, 3, etc. When an electron transitions between these levels, it either absorbs or emits energy. This movement is associated with distinct changes in the electron's quantum state.
The relationship between quantum numbers and electron transitions is fundamental in calculating the frequency of light absorbed during these transitions. The differences in the reciprocal of the squares of the initial and final quantum numbers are central to the Rydberg formula. In essence, quantum numbers are key to understanding atomic behavior, energy absorption, and consequent light frequencies observed in hydrogen atom transitions.

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Most popular questions from this chapter

The Lyman series of emission lines of the hydrogen atom are those for which \(n_{\mathrm{f}}=1\). (a) Determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed. (b) Calculate the wavelengths of the first three lines in the Lyman series-those for which \(n_{\mathrm{i}}=2,3\), and 4 .

In the experiment shown schematically below, a beam of neutral atoms is passed through a magnetic field. Atoms that have unpaired electrons are deflected in different directions in the magnetic field depending on the value of the electron spin quantum number. In the experiment illustrated, we envision that a beam of hydrogen atoms splits into two beams. (a) What is the significance of the observation that the single beam splits into two beams? (b) What do you think would happen if the strength of the magnet were increased? (c) What do you think would happen if the beam of hydrogen atoms were replaced with a beam of helium atoms? Why? (d) The relevant experiment was first performed by Otto Stern and Walter Gerlach in 1921 . They used a beam of Ag atoms in the experiment. By considering the electron configuration of a silver atom, explain why the single beam splits into two beams.

Identify the specific element that corresponds to each of the following electron configurations and indicate the number of unpaired electrons for each: (a) \(1 s^{2} 2 s^{2}\), (b) \(1 s^{2} 2 s^{2} 2 p^{4}\), (c) \([\mathrm{Ar}] 4 s^{1} 3 d^{5}\), (d) \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}\).

A hydrogen atom orbital has \(n=5\) and \(m_{l}=-2\). (a) What are the possible values of \(l\) for this orbital? (b) What are the possible values of \(m_{s}\) for the orbital?

(a) The average distance from the nucleus of a \(3 s\) electron in a chlorine atom is smaller than that for a \(3 p\) electron. In light of this fact, which orbital is higher in energy? (b) Would you expect it to require more or less energy to remove a \(3 s\) electron from the chlorine atom, as compared with a \(2 p\) electron?

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