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Chapter 6: Problem 46

Place the following transitions of the hydrogen atom in order from shortest to longest wavelength of the photon emitted: \(n=5\) to \(n=2, n=4\) to \(n=3, n=8\) to \(n=4,\) and \(n=4\) to \(n=2\).

Short Answer

Expert verified
The transitions in order from shortest to longest wavelength of the photon emitted are: n=4 to n=3, n=4 to n=2, n=8 to n=4, and n=5 to n=2.

Step by step solution

01

Calculate the wavelength of each transition

Using the Rydberg formula for the hydrogen atom, we can calculate the wavelength of the emitted photon for each transition. For the transition n=5 to n=2: \[ \frac{1}{\lambda_{5 \to 2}} = R_H \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \] For the transition n=4 to n=3: \[ \frac{1}{\lambda_{4 \to 3}} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \] For the transition n=8 to n=4: \[ \frac{1}{\lambda_{8 \to 4}} = R_H \left( \frac{1}{4^2} - \frac{1}{8^2} \right) \] For the transition n=4 to n=2: \[ \frac{1}{\lambda_{4 \to 2}} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
02

Arrange the transitions in order of wavelength

Once we have calculated the wavelengths of the emitted photons for each transition, we can easily arrange them in order from shortest to longest wavelength: 1. \( \lambda_{4 \to 3} \), n=4 to n=3. 2. \( \lambda_{4 \to 2} \), n=4 to n=2. 3. \( \lambda_{8 \to 4} \), n=8 to n=4. 4. \( \lambda_{5 \to 2} \), n=5 to n=2. Thus, the transitions in order from shortest to longest wavelength of the photon emitted are n=4 to n=3, n=4 to n=2, n=8 to n=4, and n=5 to n=2.

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Most popular questions from this chapter

Consider a transition in which the electron of a hydrogen atom is excited from \(n=1\) to \(n=\infty\). (a) What is the end result of this transition? (b) What is the wavelength of light that must be absorbed to accomplish this process? (c) What will occur if light with a shorter wavelength than that in part (b) is used to excite the hydrogen atom? (d) How are the results of parts \((\mathrm{b})\) and \((\mathrm{c})\) related to the plot shown in Exercise \(6.88 ?\)

The speed of sound in dry air at \(20^{\circ} \mathrm{C}\) is $343 \mathrm{~m} / \mathrm{s}$ and the lowest frequency sound wave that the human ear can detect is approximately \(20 \mathrm{~Hz}\) (a) What is the wavelength of such a sound wave? (b) What would be the frequency of electromagnetic radiation with the same wavelength? (c) What type of electromagnetic radiation would that correspond to? [Section 6.1]

The first 25 years of the twentieth century were momentous for the rapid pace of change in scientists' understanding of the nature of matter. (a) How did Rutherford's experiments on the scattering of \(\alpha\) particles by a gold foil set the stage for Bohr's theory of the hydrogen atom? (b) In what ways is de Broglie's hypothesis, as it applies to electrons, consistent with J. J. Thomson's conclusion that the electron has mass? In what sense is it consistent with proposals preceding Thomson's work that the cathode rays are a wave phenomenon?

(a) Calculate the energy of a photon of electromagnetic radiation whose frequency is \(2.94 \times 10^{14} \mathrm{~s}^{-1}\). (b) Calculate the energy of a photon of radiation whose wavelength is 413 \(\mathrm{nm} .\) (c) What wavelength of radiation has photons of energy $6.06 \times 10^{-19} \mathrm{~J} ?$

(a) A green laser pointer emits light with a wavelength of \(532 \mathrm{nm}\). What is the frequency of this light? (b) What is the energy of one of these photons? (c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of \(532-\mathrm{nm}\) photons. What is the energy gap between the ground state and excited state in the laser material?
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