Wavelength, Frequency, and Planck's Constant

In summary: Your solution is correct but your equation is not... The equation is correct, however, the problem is with the units. You should have (1 kJ/10^3J), using your equation as written, you would get 281 kJ/mol.
  • #1
Soaring Crane
469
0
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??

_________________________________________________________________
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??

_________________________________________________________________
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??
_________________________________________________________________

Did I express my answers with the right significant digits and calculations?
Thanks.
 
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  • #2
Yes, the significant digits look correct and the calculations seem correct although I don't have a calculator around at the moment.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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
 
  • #6
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.
 
Last edited:
  • #7
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..
 

1. What is the relationship between wavelength and frequency?

Wavelength and frequency are inversely proportional to each other. This means that as wavelength increases, frequency decreases, and vice versa. This relationship is described by the equation: wavelength = speed of light / frequency.

2. What is Planck's constant and why is it important?

Planck's constant, denoted by the symbol "h", is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. It is important because it allows us to understand the behavior of light at a microscopic level and has applications in fields such as electronics and energy production.

3. How is Planck's constant calculated?

Planck's constant is calculated by dividing the energy of a photon by its frequency. This can be represented by the equation: E = hf, where E is the energy, h is Planck's constant, and f is the frequency.

4. What are the units for wavelength, frequency, and Planck's constant?

Wavelength is typically measured in meters (m), frequency in hertz (Hz), and Planck's constant in joule-seconds (J*s). However, depending on the context, other units such as nanometers (nm) and electron volts (eV) may also be used.

5. How does Planck's constant relate to the wave-particle duality of light?

Planck's constant is a key factor in the understanding of the wave-particle duality of light. It shows that light can behave as both a wave and a particle, and the value of Planck's constant determines the energy and frequency of the particle-like behavior of light.

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