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Chapter 7: Problem 33

In addition to continuous radiation, fluorescent lamps emit some visible lines from mercury. A prominent line has a wavelength of \(436 \mathrm{nm}\). What is the energy (in \(\mathrm{J}\) ) of one photon of it?

Short Answer

Expert verified
The energy of one photon is approximately \(4.56 \times 10^{-19} \, J\).

Step by step solution

01

- Understand the Relationship Between Wavelength and Energy

The energy of a photon is related to its wavelength via the equation: \[ E = \frac{hc}{\lambda} \]where- \(E\) is the energy of the photon,- \(h\) is Planck's constant (approximately \(6.626 \times 10^{-34} \, J \cdot s\)),- \(c\) is the speed of light in a vacuum (approximately \(3.00 \times 10^8 \, m/s\)),- \(\lambda\) is the wavelength of the photon. In this case, the wavelength is given as \(436 \mathrm{nm}\) which needs to be converted to meters.
02

- Convert Wavelength to Meters

Convert the given wavelength from nanometers to meters. We know that \(1 \, \mathrm{nm} = 10^{-9} \, \mathrm{m}\). Thus,\[ 436 \, \mathrm{nm} = 436 \times 10^{-9} \, \mathrm{m} \]
03

- Substitute Values into the Equation

Substitute the known values of Planck's constant \(h\), the speed of light \(c\), and the converted wavelength \(\lambda\) into the equation \[ E = \frac{hc}{\lambda} \]\[ E = \frac{(6.626 \times 10^{-34} \, J \cdot s)(3.00 \times 10^8 \, m/s)}{436 \times 10^{-9} \, m} \]
04

- Calculate the Energy

Perform the calculation to find the energy of one photon.\[ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{436 \times 10^{-9}} \]\[ E = \frac{1.9878 \times 10^{-25}}{436 \times 10^{-9}} \]\[ E \approx 4.56 \times 10^{-19} \, J \]

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Key Concepts

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

Planck's Constant
Planck's constant is a fundamental physical constant denoted by the letter \(h\). It is named after the German physicist Max Planck, who is considered to be the father of quantum theory. The value of Planck's constant is approximately \(6.626 \times 10^{-34} \text{J} \text{鈰卻}\).
This constant is incredibly small and plays a crucial role in the realm of quantum mechanics. It relates the energy of a photon to the frequency of its electromagnetic wave.
In an equation form, Energy (E) can be calculated using frequency (\(u\)) and Planck's constant as:
\[ E = h u \]
Understanding and using this constant is essential for various calculations in physics and chemistry, especially when dealing with particles at the atomic and subatomic levels.
Wavelength to Energy Conversion
Converting wavelength to energy is a key concept in understanding the behavior of photons and their interactions with matter. The energy of a photon is inversely related to its wavelength, which means that as the wavelength decreases, the energy increases.
This relationship is dictated by the equation: \( E = \frac{hc}{\lambda} \)
Here, \(E\) is the energy of the photon, \(h\) is Planck's constant, \(c\) is the speed of light, and \( \lambda\) is the wavelength of the photon.
For example, in the problem given, we have a wavelength of 436 nm. First, converting this wavelength to meters, we get: \(436 \text{ nm} = 436 \times 10^{-9} \text{ m}\)
Then, substituting the values into the equation, we calculate the energy of the photon:
\[ E = \frac{(6.626 \times 10^{-34} \text{ J} \text{ 鈰卻})(3.00 \times 10^8 \text{ m/s})}{436 \times 10^{-9} \text{ m}} \]
Therefore, understanding wavelength to energy conversion helps us determine the energy carried by photons, which is critical in fields like spectroscopy and quantum mechanics.
Speed of Light
The speed of light, denoted by \(c\), is a fundamental constant in physics. Its value is approximately \(3.00 \times 10^8 \text{ m/s}\). This is the speed at which light travels in a vacuum, and it is one of the most important constants in nature.
The speed of light is crucial not just for understanding optical phenomena, but also for calculations in both classical and quantum physics. In the formula to calculate photon energy, \(E = \frac{hc}{\lambda}\), \(c\) represents the speed of light.
This value is fixed and does not change, making it a reliable constant to use in various equations and scientific principles.
Understanding the speed of light allows us to grasp how quickly light and, by extension, all electromagnetic waves travel. This knowledge is essential for understanding phenomena such as the formation of rainbows, the behavior of fiber optics, and even the time it takes for sunlight to reach Earth.
Therefore, mastering the concept of the speed of light is fundamental for any student of physics or related sciences.

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Most popular questions from this chapter

Only certain transitions are allowed from one energy level to another. In one- electron species, the change in \(I\) for an allowed transition is \(\pm 1 .\) For example, a \(3 p\) electron can move to a \(2 s\) orbital but not to a \(2 p\). Thus, in the UV series, where \(n_{\text {final }}=1\), allowed transitions can start in a \(p\) orbital \((l=1)\) of \(n=2\) or higher, not in an \(s(l=0)\) or \(d(l=2)\) orbital of \(n=2\) or higher. From what orbital do each of the allowed transitions start for the first four emission lines in the visible series \(\left(n_{\text {final }}=2\right) ?\)

A ground-state \(\mathrm{H}\) atom absorbs a photon of wavelength \(94.91 \mathrm{nm},\) and its electron attains a higher energy level. The atom then emits two photons: one of wavelength \(1281 \mathrm{nm}\) to reach an intermediate energy level, and a second to return to the ground state. (a) What higher level did the electron reach? (b) What intermediate level did the electron reach? (c) What was the wavelength of the second photon emitted?

What feature of an orbital is related to each of the following? (a) Principal quantum number \((n)\) (b) Angular momentum quantum number \((l)\) (c) Magnetic quantum number \(\left(m_{i}\right)\)

One reason carbon monoxide (CO) is toxic is that it binds to the blood protein hemoglobin more strongly than oxygen does. The bond between hemoglobin and CO absorbs radiation of \(1953 \mathrm{~cm}^{-1}\). (The unit is the reciprocal of the wavelength in centimeters.) Calculate the wavelength (in \(\mathrm{nm}\) and \(\dot{\mathrm{A}}\) ) and the frequency (in \(\mathrm{Hz}\) ) of the absorbed radiation.

For each of the following sublevels, give the \(n\) and \(l\) values and the number of orbitals: (a) \(5 s ;\) (b) \(3 p ;\) (c) \(4 f\).
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