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(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 \(2.87 \times 10^{-18} \mathrm{~J} ?\)

Short Answer

Expert verified
a) The energy of the photon is approximately \(4.47 \times 10^{-21} J\). b) The energy of the photon is approximately \(6.17 \times 10^{-19} J\). c) The wavelength of radiation with photons of energy \(2.87 \times 10^{-18} J\) is approximately \(69.3 nm\).

Step by step solution

01

a) Calculate the energy of a photon whose frequency is \(6.75 \times 10^{12} s^{-1}\)

Use the energy formula, \(E = h\nu\), where the frequency (\(\nu\)) is given as \(6.75 \times 10^{12} s^{-1}\). E = \((6.63 \times 10^{-34} Js)(6.75 \times 10^{12} s^{-1})\) E = \(4.47 \times 10^{-21} J\) The energy of the photon is approximately \(4.47 \times 10^{-21} J\).
02

b) Calculate the energy of a photon of radiation whose wavelength is \(322 nm\)

First, convert the wavelength to meters: \(\lambda = 322 nm = 322 \times 10^{-9} m\) Now, use the relationship between frequency and wavelength: \(\nu = \frac{c}{\lambda}\) \(\nu = \frac{3.00 \times 10^8 m/s}{322 \times 10^{-9} m} = 9.32 \times 10^{14} s^{-1}\) Calculate the energy using the energy formula: \(E = h\nu\) E = \((6.63 \times 10^{-34} Js)(9.32 \times 10^{14} s^{-1})\) E = \(6.17 \times 10^{-19} J\) The energy of the photon is approximately \(6.17 \times 10^{-19} J\).
03

c) What wavelength of radiation has photons of energy \(2.87 \times 10^{-18} J\)?

Given the energy of the photon (E), we can first find the frequency using the energy formula: \(E = h\nu\) \(\nu = \frac{E}{h} = \frac{2.87 \times 10^{-18} J}{6.63 \times 10^{-34} Js} = 4.33 \times 10^{15} s^{-1}\) Now, use the relationship between frequency and wavelength to find the wavelength: \(\nu = \frac{c}{\lambda}\) \(\lambda = \frac{c}{\nu} = \frac{3.00 \times 10^8 m/s}{4.33 \times 10^{15} s^{-1}} = 6.93 \times 10^{-8} m\) Convert the wavelength to nanometers: \(\lambda = 6.93 \times 10^{-8} m = 69.3 nm\) The wavelength of radiation with photons of energy \(2.87 \times 10^{-18} J\) is approximately \(69.3 nm\).

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Key Concepts

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

Electromagnetic Radiation
Electromagnetic radiation is a cornerstone of our understanding of physics and is integral to various technologies from radios to medical imaging. This form of energy travels through space as a combination of electric and magnetic waves. The spectrum of electromagnetic radiation includes a range of wavelengths and frequencies, from extremely long radio waves to very short gamma rays. What's fascinating about electromagnetic waves is that they do not require a medium to travel through; they can move through the vacuum of space, allowing us to receive light from the sun and stars.

When we talk about electromagnetic radiation in the context of photon energy, we're discussing how each photon carries a 'packet' of energy which relates directly to the radiation's frequency. This energy can be observed when radiation interacts with matter, possibly ejecting electrons from atoms, a phenomenon known as the photoelectric effect. Understanding this interaction is critical in fields such as quantum physics, chemistry, and in developing technologies like photovoltaic cells which convert light into electricity.
Planck's Constant (h)
Planck's constant, symbolized by the letter 'h', is a fundamental constant in physics that relates the energy of a single photon to its frequency. The value of Planck's constant is approximately \(6.63 \times 10^{-34} \) Joule seconds (Js), an incredibly small number reflecting the tiny energy bits carried by photons. This constant is named after Max Planck, the physicist who proposed that energy is quantized, meaning it can only exist in discrete amounts called 'quanta'.

In the equation \( E = hu \), Planck's constant bridges the world of the very small (quantum mechanics) with the energy we can observe and measure. Here, \( E \) represents the energy of the photon in Joules, and \( u \) (Greek letter nu) represents the frequency of the electromagnetic wave in Hertz (Hz, equivalent to s^{-1}). Planck's constant is foundational for the photoelectric effect and played a key role in the development of quantum theory, revolutionizing our understanding of how energy and matter interact at the microscopic level.
Wavelength to Frequency Conversion
Wavelength and frequency are two attributes of waves that are inversely related to each other. The wavelength (often represented by \( \lambda \)) is the distance between two consecutive peaks of a wave, while frequency (\( u \)), as mentioned before, is the number of waves that pass a given point per second. The relationship between the two is governed by the speed of light (\( c \)), which is a constant \(3.00 \times 10^8 \) meters per second in a vacuum.

To convert wavelength to frequency or vice versa, we use the equation \( u = \frac{c}{\lambda} \). When you know the wavelength, you can calculate the frequency by dividing the speed of light by the wavelength. The reverse is also true: to find the wavelength if you know the frequency, you divide the speed of light by the frequency. This conversion is not only crucial in photon energy calculation, where knowing either the wavelength or the frequency allows you to determine the energy of a photon using Planck's equation, but it's also used in understanding the properties of waves across all different regions of the electromagnetic spectrum.

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Most popular questions from this chapter

The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of \(8.95 \times 10^{6} \mathrm{~m} / \mathrm{s}\). What is the characteristic wavelength of this electron? Is the wavelength comparable to the size of atoms?

Molybdenum metal must absorb radiation with a minimum frequency of \(1.09 \times 10^{15} \mathrm{~s}^{-1}\) before it can eject an electron from its surface via the photoelectric effect. (a) What is the minimum energy needed to eject an electron? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of \(120 \mathrm{nm}\), what is the maximum possible kinetic energy of the emitted electrons?

(a) What is the frequency of radiation whose wavelength is \(5.0 \times 10^{-5} \mathrm{~m} ?\) (b) What is the wavelength of radiation that has a frequency of \(2.5 \times 10^{8} \mathrm{~s}^{-1} ?(\mathrm{c})\) Would the radiations in part (a) or part (b) be detected by an X-ray detector? (d) What distance does electromagnetic radiation travel in \(10.5 \mathrm{fs}\) ?

(a) In terms of the Bohr theory of the hydrogen atom, what process is occurring when excited hydrogen atoms emit radiant energy of certain wavelengths and only those wavelengths? (b) Does a hydrogen atom "expand" or "contract" as it moves from its ground state to an excited state?

(a) Why does the Bohr model of the hydrogen atom violate the uncertainty principle? (b) In what way is the description of the electron using a wave function consistent with de Broglie's hypothesis? (c) What is meant by the term probability density? Given the wave function, how do we find the probability density at a certain point in space?

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