light & the atomic model unit test study guide:
name:
period: 3 date:
section #1: the electromagnetic spectrum, wave, & frequency
1. what is the speed of all forms of electromagnetic radiation (emr) in a vacuum?
2. how are wavelength and frequency related?
3. how are frequency and energy related?
4. using your electromagnetic spectrum, list the colors of the visible light region in order of
5. what is the frequency of an electromagnetic radiation with a wavelength of 4.257 x 10⁹ m what type of emr is this?
6. radar signals are electromagnetic radiation in the microwave region. the wavelength o is 3.19 cm. what is the frequency of this radiation?
what is the wavelength, in nm, of a visible light radiation that has a frequency of 7.0 wavelength?

Question

Accepted Answer

### Question 1
# Explanation:
## Step1: Recall EMR speed in vacuum
All electromagnetic radiation travels at the speed of light in a vacuum. The speed of light $ c $ is a constant.
$ c = 3.0\times10^{8}\, \text{m/s} $
# Answer:
The speed of all forms of electromagnetic radiation (EMR) in a vacuum is $ 3.0\times10^{8}\, \text{m/s} $ (or $ 3.0\times10^{5}\, \text{km/s} $).

### Question 2
# Brief Explanations:
The relationship between wavelength ($ \lambda $) and frequency ($
u $) is given by the wave equation $ c=\lambda
u $ (where $ c $ is the speed of light, constant in a given medium like vacuum). Since $ c $ is constant, wavelength and frequency are inversely proportional. Mathematically, $ \lambda\propto\frac{1}{
u} $ (as $ c $ is constant, if wavelength increases, frequency decreases and vice - versa).
# Answer:
Wavelength and frequency are inversely proportional (mathematically, from $ c = \lambda
u $, $ \lambda=\frac{c}{
u} $ or $
u=\frac{c}{\lambda} $, so as one increases, the other decreases when $ c $ is constant).

### Question 3
# Brief Explanations:
The relationship between frequency ($
u $) and energy ($ E $) of a photon of electromagnetic radiation is given by Planck's equation $ E = h
u $, where $ h $ is Planck's constant ($ h = 6.626\times 10^{-34}\, \text{J}\cdot\text{s} $). Since $ h $ is a constant, energy and frequency are directly proportional. That is, as the frequency of the radiation increases, the energy of the photon also increases.
# Answer:
Frequency and energy are directly proportional (from $ E=h
u $, as $
u $ increases, $ E $ increases, with $ h $ constant).

### Question 4
# Brief Explanations:
The visible light region of the electromagnetic spectrum consists of colors in the order of decreasing wavelength (or increasing frequency) as: Red, Orange, Yellow, Green, Blue, Indigo, Violet (ROYGBIV). If the question was about increasing wavelength, the order would be Violet, Indigo, Blue, Green, Yellow, Orange, Red. Assuming the "order of d" (probably decreasing wavelength or increasing frequency), the order is Red, Orange, Yellow, Green, Blue, Indigo, Violet.
# Answer:
The colors of the visible light region (in order of decreasing wavelength, for example) are Red, Orange, Yellow, Green, Blue, Indigo, Violet (ROYGBIV).

### Question 5
# Explanation:
## Step1: Recall the wave equation
The wave equation is $ c=\lambda
u $, where $ c = 3.0\times10^{8}\, \text{m/s} $, $ \lambda $ is the wavelength, and $
u $ is the frequency. We need to solve for $
u $, so $
u=\frac{c}{\lambda} $.
## Step2: Substitute the values
Given $ \lambda = 4.257\times 10^{-9}\, \text{m} $ (assuming the exponent was a typo, maybe $ 10^{-9}\, \text{m} $ as $ 4.257\times 10^{9}\, \text{m} $ would be too large for EMR wavelength in most cases).
$
u=\frac{3.0\times 10^{8}\, \text{m/s}}{4.257\times 10^{-9}\, \text{m}} \approx 7.05\times 10^{16}\, \text{Hz} $
## Step3: Determine the type of EMR
The electromagnetic spectrum: X - rays have wavelengths in the range of $ 10^{-12}\, \text{m} $ to $ 10^{-8}\, \text{m} $ and frequencies in the range of $ 10^{16}\, \text{Hz} $ to $ 10^{20}\, \text{Hz} $. Since our calculated frequency is $ \approx7.05\times 10^{16}\, \text{Hz} $, this EMR is likely X - ray (if $ \lambda = 4.257\times 10^{-9}\, \text{m}=4.257\, \text{nm} $, which is in the X - ray region).
# Answer:
The frequency is approximately $ 7.05\times 10^{16}\, \text{Hz} $ and the type of EMR is X - ray (assuming $ \lambda = 4.257\times 10^{-9}\, \text{m} $).

### Question 6
# Explanation:
## Step1: Convert wavelength to meters
Given $ \lambda = 3.19\, \text{cm} $. Since $ 1\, \text{m}=100\, \text{cm} $, $ \lambda=\frac{3.19}{100}\, \text{m}=0.0319\, \text{m} $
## Step2: Use the wave equation to find frequency
From $ c = \lambda
u $, we can solve for $
u $: $
u=\frac{c}{\lambda} $
Substitute $ c = 3.0\times 10^{8}\, \text{m/s} $ and $ \lambda = 0.0319\, \text{m} $
$
u=\frac{3.0\times 10^{8}\, \text{m/s}}{0.0319\, \text{m}}\approx9.40\times 10^{9}\, \text{Hz} $
# Answer:
The frequency of the radar (microwave) radiation is approximately $ 9.40\times 10^{9}\, \text{Hz} $ (or $ 9.40\, \text{GHz} $).

### Question (last one, partially visible)
# Explanation:
## Step1: Recall the wave equation
We know that $ c=\lambda
u $, so $ \lambda=\frac{c}{
u} $. We need to find the wavelength in nanometers ($ \text{nm} $), where $ 1\, \text{nm}=10^{-9}\, \text{m} $.
## Step2: Assume the frequency value (since it's partially visible, let's assume the frequency $
u = 7.0\times 10^{14}\, \text{Hz} $ which is a typical visible light frequency)
Substitute $ c = 3.0\times 10^{8}\, \text{m/s} $ and $
u = 7.0\times 10^{14}\, \text{Hz} $ into the formula:
$ \lambda=\frac{3.0\times 10^{8}\, \text{m/s}}{7.0\times 10^{14}\, \text{Hz}}\approx4.29\times 10^{-7}\, \text{m} $
## Step3: Convert to nanometers
Since $ 1\, \text{m}=10^{9}\, \text{nm} $, $ \lambda = 4.29\times 10^{-7}\, \text{m}\times10^{9}\, \text{nm/m}=429\, \text{nm} $
# Answer:
If the frequency is $ 7.0\times 10^{14}\, \text{Hz} $, the wavelength is approximately $ 429\, \text{nm} $ (the answer depends on the exact frequency value, but the method is as above).

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Question 1

Step1: Recall EMR speed in vacuum

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Question 2