Electromagnetic Spectrum - Revision history

Croberts65 at 19:39, 5 December 2015

2015-12-05T19:39:28Z

← Older revision		Revision as of 15:39, 5 December 2015
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	===Simple===		===Simple===
			What type of electromagnetic wave has a higher frequency, ultraviolet or infrared?

			A. Ultraviolet

			===Middling===
	Calculate the frequency of violet light (450 nm).		Calculate the frequency of violet light (450 nm).

Croberts65 at 07:16, 5 December 2015

2015-12-05T07:16:54Z

← Older revision		Revision as of 03:16, 5 December 2015
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	*Infrared Waves - causes molecular vibration, basically, this is what heat lamps produce, which is why the air gets hot around the lamp		*Infrared Waves - causes molecular vibration, basically, this is what heat lamps produce, which is why the air gets hot around the lamp
	*Visible Light - can produce ionization, but really acts more like infrared waves than ultraviolet waves		*Visible Light - can produce ionization, but really acts more like infrared waves than ultraviolet waves
	*Ultraviolet Waves - causes molecular ionization, but doesn't move the molecule around; this is why you get sunburned, because the light is destroying your skin		*Ultraviolet Waves - causes molecular ionization, but doesn't move the molecule around; this is why you get sunburned, because the ultraviolet light is destroying your skin

	==History==		==History==

Croberts65 at 07:16, 5 December 2015

2015-12-05T07:16:21Z

@@ Line 53: / Line 53: @@
 ==Interaction With Matter==
-{|class="wikitable" style="margin: 1em auto 1em auto; width:70%; font-size:95%;
+*Micro Waves - cause forces on molecules that cause torsion and rotation, causing friction in more strongly interconnected molecules, like food and other solids; this is why food heats up in a microwave oven
-|+Electromagnetic radiation interaction with matter
+*Infrared Waves - causes molecular vibration, basically, this is what heat lamps produce, which is why the air gets hot around the lamp
-|- style="text-align:center; background:#BBB;"
+*Visible Light - can produce ionization, but really acts more like infrared waves than ultraviolet waves
-|-
+*Ultraviolet Waves - causes molecular ionization, but doesn't move the molecule around; this is why you get sunburned, because the light is destroying your skin
 ==History==

Croberts65 at 07:06, 5 December 2015

2015-12-05T07:06:20Z

← Older revision		Revision as of 03:06, 5 December 2015
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	==Examples==		==Examples==

	~~Be sure to show all steps in your solution and include diagrams whenever possible~~

	===Simple===		===Simple===
	Calculate the frequency of violet light (450 nm).		Calculate the frequency of violet light (450 nm).

	1. <math>~~450 \times 10^~~{~~-9} =~~ \frac {~~2.998 \times 10^8~~}{\lambda}</math>		1. <math>v = {\frac{c}{\lambda}}</math>

	2. <math>~~\frac {1}{\lambda} = \frac {~~450 \times 10^{-9}}{2.998 \times 10^8} ~~= 1.501~~ \~~times 10^{-15~~}</math>		2. <math>450 \times 10^{-9} = \frac {2.998 \times 10^8}{\lambda}</math>

	3. <math>\lambda = 6.662 \times 10^{14}</math> Hz		3. <math>\frac {1}{\lambda} = \frac {450 \times 10^{-9}}{2.998 \times 10^8} = 1.501 \times 10^{-15}</math>

			4. <math>\lambda = 6.662 \times 10^{14}</math> Hz

	===Difficult===		===Difficult===
	A small laser used as a pointer produces a beam of red light 2 mm in diameter, and has a power output of 4 milliwatts. What is the magnitude of the electric field in the laser beam?		A small laser used as a pointer produces a beam of red light 2 mm in diameter, and has a power output of 4 milliwatts. What is the magnitude of the electric field in the laser beam?

	1. <math>E = \sqrt { \frac{(4 \times 10^{-3}) / ((2/2) \pi \times 10^{-3 + -3})}{(2.998 \times 10^{8}) \times (8.85 \times 10^{-12})} }</math>		1. <math>E = \sqrt {\frac{P/A}{c \times \epsilon_{0}}}</math>

			2. <math>E = \sqrt { \frac{(4 \times 10^{-3}) / ((2/2) \pi \times 10^{-3 + -3})}{(2.998 \times 10^{8}) \times (8.85 \times 10^{-12})} }</math>

			3. <math>E = \sqrt { \frac{4 \times 10^{3}}{2.65 \pi \times 10^{-3}} }</math>

			4. <math>E = \sqrt {4.80 \times 10^5} = 6.93 \times 10^{2}</math>

	~~2. <math>E~~ = ~~\sqrt { \frac{4 \times 10^{3}}{2.65 \pi \times 10^{-3}} }</math>~~		==Interaction With Matter==

	3. ~~<math>E = \sqrt {4~~.~~80 \times 10^5} = 6~~.~~93 \times 10^{2~~}~~</math>~~		{\|class="wikitable" style="margin: 1em auto 1em auto; width:70%; font-size:95%;
			\|+Electromagnetic radiation interaction with matter
			\|- style="text-align:center; background:#BBB;"
			\|-
			! Region of the spectrum
			! Main interactions with matter
			\|-
			\|[[Radio wave\|Radio]]
			\|Collective oscillation of charge carriers in bulk material ([[plasma oscillation]]). An example would be the oscillatory travels of the electrons in an [[antenna (radio)\|antenna]].
			\|-
			\|nowrap\|[[Microwave]] through far [[infrared]]
			\|Plasma oscillation, molecular rotation
			\|-
			\|Near [[infrared]]
			\|Molecular vibration, plasma oscillation (in metals only)
			\|-
			\|[[Light\|Visible]]
			\|Molecular electron excitation (including pigment molecules found in the human retina), plasma oscillations (in metals only)
			\|-
			\|[[Ultraviolet]]
			\|Excitation of molecular and atomic valence electrons, including ejection of the electrons ([[photoelectric effect]])
			\|-
			\|[[Xray\|X-rays]]
			\|Excitation and ejection of core atomic electrons, [[Compton scattering]] (for low atomic numbers)
			\|-
			\|[[Gamma ray]]s
			\|Energetic ejection of core electrons in heavy elements, [[Compton scattering]] (for all atomic numbers), excitation of atomic nuclei, including dissociation of nuclei
			\|-
			\|nowrap\|High-energy [[gamma ray]]s
			\|Creation of [[Virtual pair\|particle-antiparticle pairs]]. At very high energies a single photon can create a shower of high-energy particles and antiparticles upon interaction with matter.
			\|}

	==History==		==History==
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	==Connectedness==		==Connectedness==
	~~We can use this~~ for a variety of things including		The electromagnetic spectrum is used for a variety of things including
	*Cooking (microwaves)		*Cooking (microwaves)
	*Radio and Communication		*Radio and Communication
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	*Medical Treatments (gamma rays, x-rays)		*Medical Treatments (gamma rays, x-rays)
	*Nuclear Physics		*Nuclear Physics
			*The sense of sight

	==References & Further Reading==		==References & Further Reading==

Croberts65 at 06:58, 5 December 2015

2015-12-05T06:58:40Z

← Older revision		Revision as of 02:58, 5 December 2015
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	2. <math>E = \sqrt { \frac{4 \times 10^{3}}{2.65 \pi \times 10^{-3}} }</math>		2. <math>E = \sqrt { \frac{4 \times 10^{3}}{2.65 \pi \times 10^{-3}} }</math>

	3. <math>E = \sqrt {4.80 \times 10^5} = 6.93 \times 10^{2}		3. <math>E = \sqrt {4.80 \times 10^5} = 6.93 \times 10^{2}</math>

	==History==		==History==

Croberts65 at 06:58, 5 December 2015

2015-12-05T06:58:24Z

← Older revision		Revision as of 02:58, 5 December 2015
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	A small laser used as a pointer produces a beam of red light 2 mm in diameter, and has a power output of 4 milliwatts. What is the magnitude of the electric field in the laser beam?		A small laser used as a pointer produces a beam of red light 2 mm in diameter, and has a power output of 4 milliwatts. What is the magnitude of the electric field in the laser beam?

	1. <math>E = \sqrt { \frac{(4 \times 10^{-3}) / ((2/2) \pi \times 10^{-3 + -3})}{(2.998 \times 10^{8}) \times (1.~~501~~ \times 10^{-15})} }</math>		1. <math>E = \sqrt { \frac{(4 \times 10^{-3}) / ((2/2) \pi \times 10^{-3 + -3})}{(2.998 \times 10^{8}) \times (8.85 \times 10^{-12})} }</math>

	2. <math>E = \sqrt { \frac{4 ~~\pi~~ \times 10^{3}}{4.50 \times 10^{-7}} }</math>		2. <math>E = \sqrt { \frac{4 \times 10^{3}}{2.65 \pi \times 10^{-3}} }</math>

	3. <math>E =		3. <math>E = \sqrt {4.80 \times 10^5} = 6.93 \times 10^{2}

	==History==		==History==

Croberts65 at 06:55, 5 December 2015

2015-12-05T06:55:48Z

← Older revision		Revision as of 02:55, 5 December 2015
Line 43:		Line 43:
	A small laser used as a pointer produces a beam of red light 2 mm in diameter, and has a power output of 4 milliwatts. What is the magnitude of the electric field in the laser beam?		A small laser used as a pointer produces a beam of red light 2 mm in diameter, and has a power output of 4 milliwatts. What is the magnitude of the electric field in the laser beam?

	1. <math>E = \sqrt { \frac{(4 \times 10^{-3}) ~~\divide~~ ((2/2) \pi \times 10^{-3 + -3})}{(2.998 \times 10^{8}) \times (1.501 \times 10^{-15})} }</math>		1. <math>E = \sqrt { \frac{(4 \times 10^{-3}) / ((2/2) \pi \times 10^{-3 + -3})}{(2.998 \times 10^{8}) \times (1.501 \times 10^{-15})} }</math>

	2. <math>E = \sqrt { \frac{4 \pi \times 10^{3}}{4.50 \times 10^{-7}} }</math>		2. <math>E = \sqrt { \frac{4 \pi \times 10^{3}}{4.50 \times 10^{-7}} }</math>

Croberts65 at 06:52, 5 December 2015

2015-12-05T06:52:58Z

← Older revision		Revision as of 02:52, 5 December 2015
Line 43:		Line 43:
	A small laser used as a pointer produces a beam of red light 2 mm in diameter, and has a power output of 4 milliwatts. What is the magnitude of the electric field in the laser beam?		A small laser used as a pointer produces a beam of red light 2 mm in diameter, and has a power output of 4 milliwatts. What is the magnitude of the electric field in the laser beam?

	1. <math>E = \sqrt { \frac{(4 \times 10^{-3}) / ((2/2) \pi \times 10^{-3 + -3})}{(2.998 \times 10^{8}) \times (1.501 \times 10^{-15})} }</math>		1. <math>E = \sqrt { \frac{(4 \times 10^{-3}) \divide ((2/2) \pi \times 10^{-3 + -3})}{(2.998 \times 10^{8}) \times (1.501 \times 10^{-15})} }</math>

	2. <math>E = \sqrt { \frac{4 \pi \times 10^{3}}{4.50 \times 10^{-7}} }</math>		2. <math>E = \sqrt { \frac{4 \pi \times 10^{3}}{4.50 \times 10^{-7}} }</math>

Croberts65 at 06:52, 5 December 2015

2015-12-05T06:52:31Z

← Older revision		Revision as of 02:52, 5 December 2015
Line 43:		Line 43:
	A small laser used as a pointer produces a beam of red light 2 mm in diameter, and has a power output of 4 milliwatts. What is the magnitude of the electric field in the laser beam?		A small laser used as a pointer produces a beam of red light 2 mm in diameter, and has a power output of 4 milliwatts. What is the magnitude of the electric field in the laser beam?

	1. <math>E = \sqrt { \frac{(4 \times 10^{-3}) / ((2/2) \times 10^{-3 + -3})}{(2.998 \times 10^{8}) \times (1.501 \times 10^{-15})} }</math>		1. <math>E = \sqrt { \frac{(4 \times 10^{-3}) / ((2/2) \pi \times 10^{-3 + -3})}{(2.998 \times 10^{8}) \times (1.501 \times 10^{-15})} }</math>

	2. <math>E = \sqrt { \frac{4 *10^{3}}{4.50 \times 10^{-7}} }</math>		2. <math>E = \sqrt { \frac{4 \pi \times 10^{3}}{4.50 \times 10^{-7}} }</math>

			3. <math>E =

	==History==		==History==

Croberts65 at 06:50, 5 December 2015

2015-12-05T06:50:58Z

← Older revision		Revision as of 02:50, 5 December 2015
Line 45:		Line 45:
	1. <math>E = \sqrt { \frac{(4 \times 10^{-3}) / ((2/2) \times 10^{-3 + -3})}{(2.998 \times 10^{8}) \times (1.501 \times 10^{-15})} }</math>		1. <math>E = \sqrt { \frac{(4 \times 10^{-3}) / ((2/2) \times 10^{-3 + -3})}{(2.998 \times 10^{8}) \times (1.501 \times 10^{-15})} }</math>

	2. <math>E = \sqrt { \frac{4 ~~\times~~ 10^{3}}{4.50 \times 10^{-7}} }</math>		2. <math>E = \sqrt { \frac{4 *10^{3}}{4.50 \times 10^{-7}} }</math>

	==History==		==History==