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Chapter 7: Problem 37

The laser in an audio CD player uses light with a wavelength of \(7.80 \times 10^{2} \mathrm{~nm} .\) Calculate the frequency of this light.

Short Answer

Expert verified
The frequency of the light used by the laser in the audio CD player is approximately \(3.846 \times 10^{14} \mathrm{~Hz}\).

Step by step solution

01

Write down the speed of light, wavelength, and frequency relationship formula

The formula that relates the speed of light (c), wavelength (λ), and frequency (υ) is given by the equation: \[c = \lambda \times \nu \] where c = speed of light \( (3.00 \times 10^8 \mathrm{~m/s}) \) λ = wavelength of light ν = frequency of light
02

Convert the wavelength from nm to m

The wavelength given in the problem is in nm (nanometers). We need to convert this to meters in order to use this formula. Use the following conversion factor: \(1 \mathrm{~m} = 10^9 \mathrm{~nm} \) So, \(7.80 \times 10^2 \mathrm{~nm} = 7.80 \times 10^2 \times 10^{-9} \mathrm{~m} = 7.80 \times 10^{-7} \mathrm{~m} \).
03

Rearrange the equation to solve for the frequency

We need to find the frequency (ν), so we need to rearrange the speed of light equation to solve for the frequency: \[\nu = \frac{c}{\lambda}\]
04

Calculate the frequency of the light

Now, we can plug in the given values for the speed of light (c) and the wavelength (λ) to find the frequency: \[\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \mathrm{~m/s}}{7.80 \times 10^{-7} \mathrm{~m}}\] \[\nu = 3.846 \times 10^{14} \mathrm{~Hz}\] The frequency of the light used by the laser in the audio CD player is approximately \(3.846 \times 10^{14} \mathrm{~Hz}\).

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Most popular questions from this chapter

One type of electromagnetic radiation has a frequency of \(107.1\) MHz, another type has a wavelength of \(2.12 \times 10^{-10} \mathrm{~m}\), and another type of electromagnetic radiation has photons with energy equal to \(3.97 \times 10^{-19} \mathrm{~J} /\) photon. Identify each type of electromagnetic radiation and place them in order of increasing photon energy and increasing frequency.

Assume that we are in another universe with different physical laws. Electrons in this universe are described by four quantum numbers with meanings similar to those we use. We will call these quantum numbers \(p, q, r\), and \(s .\) The rules for these quantum numbers are as follows: \(p=1,2,3,4,5, \ldots\) \(q\) takes on positive odd integers and \(q \leq p\) \(r\) takes on all even integer values from \(-q\) to \(+q\). (Zero is considered an even number.) \(s=+\frac{1}{2}\) or \(-\frac{1}{2}\) a. Sketch what the first four periods of the periodic table will look like in this universe. b. Wh?t are the atomic numbers of the first four elements you would expect to be least reactive? c. Give an example, using elements in the forst four rows, of ionic compounds with the formulas \(\mathrm{XY}, \mathrm{XY}_{2}, \mathrm{X}_{2} \mathrm{Y}, \mathrm{XY}_{3}\), and \(\mathrm{X}_{2} \mathrm{Y}_{3}\) d. How many electrons can have \(p=4, q=3 ?\) e. How many electrons can have \(p=3, q=0, r=0 ?\) f. How many electrons can have \(p=6\) ?

An electron is excited from the \(n=1\) ground state to the \(n=3\) state in a hydrogen atom. Which of the following statements are true? Correct the false statements to make them true. a. It takes more energy to ionize (completely remove) the electron from \(n=3\) than from the ground state. b. The electron is farther from the nucleus on average in the \(n=3\) state than in the \(n=1\) state. c. The wavelength of light emitted if the electron drops from \(n=3\) to \(n=2\) will be shorter than the wavelength of light emitted if the electron falls from \(n=3\) to \(n=1\). d. The wavelength of light emitted when the electron returns to the ground state from \(n=3\) will be the same as the wavelength of light absorbed to go from \(n=1\) to \(n=3\). e. For \(n=3\), the electron is in the first excited state.

It takes \(208.4 \mathrm{~kJ}\) of energy to remove 1 mole of electrons from an atom on the surface of rubidium metal. How much energy does it take to remove a single electron from an atom on the surface of solid rubidium? What is the maximum wavelength of light capable of doing this?

Consider an electron for a hydrogen atom in an excited state. The maximum wavelength of electromagnetic radiation that can completely remove (ionize) the electron from the \(\mathrm{H}\) atom is \(1460 \mathrm{~nm} .\) What is the initial excited state for the electron \((n=?) ?\)
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