Wavelength, Frequency, and Planck's Constant

AI Thread Summary
The discussion focuses on calculating the wavelength and energy of electromagnetic waves using the formulas LAMBDA = c/v and E = hv. Participants verify calculations for frequencies of 2812571875.00MHz and 7084.00MHz, noting the importance of significant digits in constants like the speed of light. There is a debate about using a more precise value for the speed of light versus a rounded version from textbooks. Additionally, there are corrections regarding unit conversions and the proper application of formulas for energy calculations. Overall, the conversation emphasizes accuracy in scientific calculations and the importance of adhering to given significant figures.
Soaring Crane
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1) What is the wavelength (in meters) of an electromagnetic wave of frequency 2812571875.00MHz?
Example: 1.11e-5
LAMBDA = c/v
=(3.00*10^8 m/s)/(2812571875.00MHz*10^6 Hz/MHz) = 1.07E-7 m??

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2) What is the wavelength (in meters) of an electromagnetic wave of frequency 7084.00MHz?
LAMBDA = c/v
=(3.00*10^8 m/s)/(7084.00MHz*10^6 Hz/MHz) = 4.23E-1 m??

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3) How much energy is carried by a mole of photons with frequency 704.00MHz?
Give your answer in kilojoules per mole, Example: 1.11e-5
Careful with your conventions.
E = hv
=(6.626*10^-34)*(704.00MHz*10^6 Hz/MHz) = 4.665E-25 J
4.665E-25 J(1 kJ/10^-3 J)* (6.02*10^23 photons/1mol) = 2.81E-4 kJ/mol??
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Did I express my answers with the right significant digits and calculations?
Thanks.
 
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Yes, the significant digits look correct and the calculations seem correct although I don't have a calculator around at the moment.
 
You are given a lot of signifiant digits in your problem, why did you only use 3 for c? Is that c to 3 digits? There are about 8 digits of c available, why not use them?
 
Integral said:
You are given a lot of signifiant digits in your problem, why did you only use 3 for c? Is that c to 3 digits? There are about 8 digits of c available, why not use them?

I suppose its because he's using a textbook and I bet he might get the answers wrong (as far as grading is concerned) if he starts using different sources for the numbers.
 
Soaring Crane said:
1) What is the wavelength (in meters) of an electromagnetic wave of frequency 2812571875.00MHz?
Example: 1.11e-5
LAMBDA = c/v
=(3.00*10^8 m/s)/(2812571875.00MHz*10^6 Hz/MHz) = 1.07E-7 m??
If your text does give constant c to more significant digits, I would use them as Integral suggests. If you are not given speed of light in your text, you can use any valid source such as NIST (national institute of standards & technology). An equally correct answer to this part is 107nm. If your question does not specify units for your answers, both solutions are good.

2) What is the wavelength (in meters) of an electromagnetic wave of frequency 7084.00MHz?
LAMBDA = c/v
=(3.00*10^8 m/s)/(7084.00MHz*10^6 Hz/MHz) = 4.23E-1 m??
I would check this again, especially your decimal place.

3) How much energy is carried by a mole of photons with frequency 704.00MHz?
Give your answer in kilojoules per mole, Example: 1.11e-5
Careful with your conventions.
E = hv
=(6.626*10^-34)*(704.00MHz*10^6 Hz/MHz) = 4.665E-25 J
4.665E-25 J(1 kJ/10^-3 J)* (6.02*10^23 photons/1mol) = 2.81E-4 kJ/mol??
Your solution is correct but your equation is not...
You should have (1 kJ/10^3J), using your equation as written, you would get 281 kJ/mol
 
For #2, it is 4.23E-2 m?
For the constant c, my textbook gives the value in 8 decimal places, but it says that it is mostly rounded off to 3.00*10^8 and it uses this rounded off version in examples.
 
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Nice job on #2 ! Try solving your questions, more than one way. Then you will be able to catch things like this on your own. :smile:

If they use the rounded off version of c in their examples, for practical purposes, you probably can use it as well. If you're in doubt, ask your teacher first before handing in those questions..
 
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