Find energy of each of the photons which : (a) correspond to light of frequency \(3 \times 10^{15} \mathrm{~Hz}\) (b) have wavelength of \(0.50 \mathrm{~A}\).

Short Answer

Expert verified
For (a) the energy is \(1.988 \times 10^{-18} \mathrm{J}\) and for (b) the energy is \(3.976 \times 10^{-15} \mathrm{J}\).

Step by step solution

01

Understand the Concept

The energy of a photon can be calculated using the equation: \(E = h \cdot f\), where \(E\) is the energy of the photon, \(h\) is Planck's constant which is approximately \(6.626 \times 10^{-34} \mathrm{J \cdot s}\), and \(f\) is the frequency of the light.
02

Calculate the Energy for Part (a)

Substitute the given frequency value \(f = 3 \times 10^{15} \mathrm{Hz}\) into the equation \(E = h \cdot f\). This gives us \(E = (6.626 \times 10^{-34} \mathrm{J \cdot s}) \cdot (3 \times 10^{15} \mathrm{Hz})\).
03

Compute the Result for Part (a)

Perform the multiplication to find the energy of the photon: \(E = (6.626 \times 10^{-34} \mathrm{J \cdot s}) \cdot (3 \times 10^{15} \mathrm{Hz}) = 1.988 \times 10^{-18} \mathrm{J}\).
04

Understand the Relationship between Wavelength and Frequency

The relationship between the wavelength (\(\lambda\)) and the frequency (\(f\)) of light is given by the equation: \(c = \lambda \cdot f\), where \(c\) is the speed of light in a vacuum (approximately \(3 \times 10^{8} \mathrm{m/s}\)).
05

Calculate the Frequency for Part (b)

Rearrange the equation to solve for frequency: \(f = \frac{c}{\lambda}\). Substitute the given wavelength \(\lambda = 0.50 \mathrm{\AA} = 0.50 \times 10^{-10} \mathrm{m}\) into the equation to find the frequency: \(f = \frac{3 \times 10^{8} \mathrm{m/s}}{0.50 \times 10^{-10} \mathrm{m}}\).
06

Compute the Frequency for Part (b)

Carry out the division to find the frequency: \(f = \frac{3 \times 10^{8} \mathrm{m/s}}{0.50 \times 10^{-10} \mathrm{m}} = 6 \times 10^{18} \mathrm{Hz}\).
07

Calculate the Energy for Part (b)

Using the equation for photon energy \(E = h \cdot f\), substitute the calculated frequency back into the equation: \(E = (6.626 \times 10^{-34} \mathrm{J \cdot s}) \cdot (6 \times 10^{18} \mathrm{Hz})\).
08

Compute the Result for Part (b)

Perform the multiplication to find the energy of the photon: \(E = (6.626 \times 10^{-34} \mathrm{J \cdot s}) \cdot (6 \times 10^{18} \mathrm{Hz}) = 3.976 \times 10^{-15} \mathrm{J}\).

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Key Concepts

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

Planck's Constant
A fundamental aspect of quantum mechanics, Planck's constant serves as a cornerstone in the calculation of photon energy. Named after Max Planck, the constant (\( h \)) has a value of approximately \( 6.626 \times 10^{-34} \text{Joule seconds} (J \times s) \) and plays a pivotal role in the relationship between energy and frequency for a photon. When calculating the energy of a photon, the formula we use is \( E = h \times f \) where \( E \) is the energy in Joules, \( h \) is Planck's constant, and \( f \) is the frequency in Hertz (Hz).

This constant is not just a number; it represents the quantization of energy levels. In other words, energy is not continuous but occurs in discrete 'packets' called quanta. A photon is a quantum of electromagnetic radiation and its energy is directly proportional to its frequency, which brings us to the concept of frequency of light.
Frequency of Light
The frequency of light (\( f \)) is an essential aspect when determining a photon's energy. Frequency refers to the number of wave cycles that pass a fixed point in one second, and it's measured in Hertz (\( Hz \) or cycles per second). High-frequency light, such as ultraviolet rays, has more energy than low-frequency light, such as infrared.

To illustrate, in the context of the given exercise, the frequency provided for part (a) is \( 3 \times 10^{15} \) Hz. Applying this value into the energy formula alongside Planck's constant, you can determine that a photon of this frequency carries significant energy. Understanding the frequency allows us to unlock the energy stored in a single photon of light. The direct proportionality between energy and frequency means that as one increases, so does the other.
Wavelength and Frequency Relationship
Delving deeper into the properties of light, we encounter the intrinsic connection between wavelength (\(\text{denoted as } \( \) \( \(lambda\) \)\)) and frequency. These two properties of a wave are inversely related through the speed of light (\( c \)), expressed in the fundamental equation \( c = \( \(lambda\) \) \times f \) where \( c = 3 \times 10^8 m/s \) - the constant speed of light in a vacuum.

For instance, a photon with a lower wavelength has a higher frequency, leading to greater energy. Conversely, a greater wavelength implies a lower frequency and therefore less energy. This is why different wavelengths of light (such as red versus violet) have different energies, which we perceive as colors. In part (b) of the exercise, once you've determined the wavelength, you can rearrange this equation to solve for frequency, allowing you to use the energy formula covered earlier. The intimate relationship between wavelength and frequency highlights the beautiful harmony within the electromagnetic spectrum and is key to understanding phenomena across physics.

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