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(a) What is the relationship between the wavelength and the frequency of radiant energy? (b) Ozone in the upper atmosphere absorbs energy in the \(210-230-\mathrm{nm}\) range of the spectrum. In what region of the electromagnetic spectrum does this radiation occur?

Short Answer

Expert verified
(a) Wavelength (λ) and frequency (v) of radiant energy are inversely proportional to each other, as expressed by the equations \(λ = \frac{c}{v}\) and \(v = \frac{c}{λ}\), where 'c' is the speed of light in a vacuum (\(3 × 10^8\ \mathrm{m/s}\)). As the wavelength increases, the frequency decreases, and vice versa. (b) The given wavelength range of \(210-230\ \mathrm{nm}\) falls under the category of Ultraviolet radiation within the electromagnetic spectrum.

Step by step solution

01

(a) Wavelength and Frequency Relationship

The relationship between the wavelength (denoted as 'λ', in meters) and frequency (denoted as 'v' or 'f', in Hertz) of radiant energy can be described using the equation: \(c = λv\) Here, 'c' represents the speed of light in a vacuum, which is approximately equal to \(3 × 10^8\ \mathrm{m/s}\). To find the relationship between wavelength and frequency of radiant energy, isolate the variable of interest in this equation.
02

(a) Explain Wavelength and Frequency

Divide both sides of the equation \(c = λv\) by the frequency 'v' to isolate the wavelength 'λ': \(λ = \frac{c}{v}\) Now, divide both sides of the equation \(c = λv\) by the wavelength 'λ' to isolate the frequency 'v': \(v = \frac{c}{λ}\) These equations show that wavelength and frequency are inversely proportional to each other while the speed of light remains constant. As the wavelength of radiant energy increases, its frequency decreases, and vice versa.
03

(b) Finding Electromagnetic Spectrum Region

The given range of the spectrum, where ozone in the upper atmosphere absorbs energy, is \(210-230\ \mathrm{nm}\). To identify the region of the electromagnetic spectrum that this radiation belongs to, refer to the classification of the electromagnetic spectrum based on wavelengths. Here is the classification: - Gamma Rays: < 0.01 nm - X-Rays: 0.01 nm - 10 nm - Ultraviolet: 10 nm - 400 nm - Visible Light: 400 nm - 700 nm - Infrared: 700 nm - 1 mm - Microwaves: 1 mm - 1 m - Radio Waves: > 1 m
04

(b) Determine the Radiation Region

As the given wavelength range is \(210-230\ \mathrm{nm}\), it can be observed that this range falls under the category of Ultraviolet radiation within the electromagnetic spectrum.

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Most popular questions from this chapter

(a) Calculate the energy of a photon of electromagnetic radiation whose frequency is \(6.75 \times 10^{12} \mathrm{~s}^{-1}\). (b) Calculate the energy of a photon of radiation whose wavelength is \(322 \mathrm{nm} .\) (c) What wavelength of radiation has photons of energy \(2.87 \times 10^{-18} \mathrm{~J} ?\)

What is the maximum number of electrons in an atom that can have the following quantum numbers: (a) \(n=2\), \(m_{s}=-\frac{1}{2},\) (b) \(n=5, l=3 ;\) (c) \(n=4, l=3, m_{l}=-3\) (d) \(n=4, l=0, m_{l}=0 ?\)

(a) Calculate the energies of an electron in the hydrogen atom for \(n=1\) and for \(n=\infty .\) How much energy does it require to move the electron out of the atom completely (from \(n=1\) to \(n=\infty),\) according to Bohr? Put your answer in \(\mathrm{kJ} / \mathrm{mol}\). (b) The energy for the process \(\mathrm{H}+\) energy \(\rightarrow \mathrm{H}^{+}+\mathrm{e}^{-}\) is called the ionization energy of hydrogen. The experimentally determined value for the ionization energy of hydrogen is \(1310 \mathrm{~kJ} / \mathrm{mol}\). How does this compare to your calculation?

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Einstein's 1905 paper on the photoelectric effect was the first important application of Planck's quantum hypothesis. Describe Planck's original hypothesis, and explain how Einstein made use of it in his theory of the photoelectric effect.

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