What is the frequency of the photon?

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The discussion focuses on calculating the frequency of a photon given its wavelength, with the correct method involving dividing the speed of light by the wavelength. Participants emphasize the importance of careful arithmetic, especially with exponents, to avoid errors. The energy of a photon can be calculated using the formula E = h f, where h is Planck's constant, and the discussion also touches on various physics problems related to mirrors and light travel time in different mediums. Overall, the conversation highlights the need for precision in calculations and understanding fundamental physics concepts.
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a photon of red light have a wavelength of 750 x 10^-9 meters. What is the frequency??

3x10^8 divided by 750x10^-9

=

4 x 10^-4

Would this be the correct way to do the problem??

Also how would I go about finding the energy of a photon when I am just given the frequency?
 
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helpm3pl3ase said:
Would this be the correct way to do the problem??
Right method but wrong answer. Check your arithmetic, especially how you handle the exponents.

Also how would I go about finding the energy of a photon when I am just given the frequency?
The energy of a photon is given by E = h f, where h is Plank's constant. (Look it up.)
 
When I punch it into the calculator that's the answer I recieve?
 
Do it again, more carefully. If your calculator allows parentheses, use them: (3x10^8)/(750x10^-9).

Or you can handle the exponents separately and just use the calculator for the other numbers. What's (10^8)/(10^-9)?
 
Finally figure it out.. 225.. Thank you for your help.

I have another question about mirrors. The question states:

When parallel light hits a concave mirror made from a piece of a sphere, all of the light rays converage at a point 3.8 cm from the mirror. What is the radius of the sphere??

Wouldnt it be 7.6?? 3.8 x 2 = 7.6??
 
That's right: The focal length of a spherical mirror is half the radius.
 
I love the faster responses :smile:

I have another question... I know I am annoying, but I don't understand this stuff...

How long does it take for light to go 300 meters in water? The index of refraction of water is 1.33.

I understand I have to use the formula N=C/V. Would I use 3 X 10^8 divided by 1.33??
 
helpm3pl3ase said:
I understand I have to use the formula N=C/V. Would I use 3 X 10^8 divided by 1.33??
That will give you the speed of light in water. Then use the usual "Distance = speed X time" to find the time.
 
3x10^8 divided by 1.33 = 225,563,909.8

Distance = speed x time

300 meters = 225,563,909.8 x time

300 meters divided by 225,563,909.8 = 1.33 X 10^-6

That look right??

Another Mirror Question:

The average distance from the surface of the Earth to the surface of the Moon is 3.80 X 10^5 kilometers. How many seconds does it take for a laser beam starting on the Earth and reflecting from a mirror on the moon to make the round trip??

I don't even know where to start on that problem :frown:
 
  • #10
helpm3pl3ase said:
3x10^8 divided by 1.33 = 225,563,909.8

Distance = speed x time

300 meters = 225,563,909.8 x time

300 meters divided by 225,563,909.8 = 1.33 X 10^-6

That look right??
Looks good to me. (Don't forget your units.)

Just a note: Whenever possible, what until the very last step before reaching for the calculator--that will save you time. For example, here's how I would solve that problem.
(1) V = c/n = 3x10^8/1.33 (leave it just like that for now)
(2) T = D/v = (300)/(3x10^8/1.33) = [(300)/(3x10^8)]x 1.33 = 1.33 10^-6.
(I don't even need a calculator!)

Another Mirror Question:

The average distance from the surface of the Earth to the surface of the Moon is 3.80 X 10^5 kilometers. How many seconds does it take for a laser beam starting on the Earth and reflecting from a mirror on the moon to make the round trip??
Just use "Distance = speed x time" again. (What's the round trip distance?)
 
  • #11
Thanks for the tip, it helped me a bunch and saved some time.

For the moon problem:

3.8x10^5/3x10^8 = 1.27 x 10^13

Look and seem right?

Mirror question 2:

A near sighted snake, 17 feet long, lies perpendicular to and facing, a flat mirror. The snake can't see anything farther away than 25 feet from its eyes. What is the largest distance from its face to the mirror at which it can see the image of its tail??

The answer is 8 correct?? No tricks involved??
 
  • #12
helpm3pl3ase said:
For the moon problem:

3.8x10^5/3x10^8 = 1.27 x 10^13

Look and seem right?
Not quite right. Use the total distance: Earth to Moon and back.

Mirror question 2:

A near sighted snake, 17 feet long, lies perpendicular to and facing, a flat mirror. The snake can't see anything farther away than 25 feet from its eyes. What is the largest distance from its face to the mirror at which it can see the image of its tail??

The answer is 8 correct?? No tricks involved??
Not quite correct. Hint: If you stand 5 feet from a mirror, how far away does your image appear?
 
  • #13
(1.27 x 10^13)^2 = 1.6129 X 10^26?? Hows that look??

For the mirror question:

If you stand 5 feet infront of a mirror your image will appear to be 10 ft away.

Answer: The snake can stand 12.5 feet away from the mirror?? Is that correct?
 
  • #14
helpm3pl3ase said:
(1.27 x 10^13)^2 = 1.6129 X 10^26?? Hows that look??
Pretty crazy! Why in the world did you square that answer? Hint: If the distance to the moon is D, what's the distance to the moon and back? (Or: If it takes x seconds to get to the moon, how many seconds does it take to get to the moon and back?)

For the mirror question:

If you stand 5 feet infront of a mirror your image will appear to be 10 ft away.
Right!

Answer: The snake can stand 12.5 feet away from the mirror?? Is that correct?
Nope. If he "stood" 12.5 feet from the mirror, the image of his head would appear 25 feet away. Which means he'll just barely see his head, never mind his tail! (If he were 12.5 feet away from the mirror, how far away would the image of his tail be?)
 
  • #15
I don't know why I squared it.. Iam suppose to double it, I see this now:rolleyes:

1.27x10^13 X 2 = 2.54x10^13


Ok so the snake would have to be 4 feet away from the mirror??


New problem: A proton is moving with a velocity of 2.2 X 10^4 meters per second. What is the deBroglie wavelength of this proton? The mass of the proton is 1.67x10^-27 kg and its charge is +1.6x10^-19 Coulombs.

I would use the formula: Wavelength = Planks constant/(Mass)(Velocity)

Is that correct??
 
  • #16
helpm3pl3ase said:
I don't know why I squared it.. Iam suppose to double it, I see this now

1.27x10^13 X 2 = 2.54x10^13
OK


Ok so the snake would have to be 4 feet away from the mirror??
Right!


New problem: A proton is moving with a velocity of 2.2 X 10^4 meters per second. What is the deBroglie wavelength of this proton? The mass of the proton is 1.67x10^-27 kg and its charge is +1.6x10^-19 Coulombs.

I would use the formula: Wavelength = Planks constant/(Mass)(Velocity)

Is that correct??
That is correct.
 
  • #17
Thanks for you help. I really appreciate it!
 
  • #18
Doc Al said:
Do it again, more carefully. If your calculator allows parentheses, use them: (3x10^8)/(750x10^-9).
Or you can handle the exponents separately and just use the calculator for the other numbers. What's (10^8)/(10^-9)?

Your calculator should have a way of entering numbers in scientific notation without entering the "x 10" explicitly. For example, on one of my calculators, this is done with the "EE" button, so that to enter the speed of light I would enter:

3 EE (-) 8

This calculator has separate keys for entering a negative number "(-)" versus the subtraction operation "-".

On another calculator, the sequence is

3 EXP CHS 8

where "EXP" is the same as "EE" on the first calculator, and "CHS" changes the sign of the following number (makes it negative).

Whatever method your calculator uses, it's a good idea to learn it. People use scientific notation a lot in physics!
 
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