Question

(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

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

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 \(4.47 \times 10^{-21} J\).

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 \(6.17 \times 10^{-19} J\).

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 \(69.3 nm\).

Key concepts

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.