300 000 000

1. Dec 30, 2004

DB

Looking at the electromagnetic spectrum, the product of wavelenght and frequency of a wave is constant. For example, Gamma Rays;
wavelenght 3e-14 m x frequency 10^22 Hz = 300 000 000. The same number is given for the product of all waves on the spectrum. My question is:

If:
$$c=\lambda\nu$$
and
$$\lambda\nu=3 e8$$

How does 300 000 000 come out to 299792 km/s???

2. Dec 30, 2004

Cyrus

Maybe one is in m/s and the other is in km/s?

3. Dec 30, 2004

brewnog

Because it's pretty damn close! Remember that you calculated c in m/s, divide by 1000 for km/s.

4. Dec 30, 2004

apchemstudent

Are you asking why it's not exactly 300 000 km/s? The reason for that is because the light of speed is not exactly 300 000km/s...

5. Dec 30, 2004

DB

Not exactly. I just came to realize that it could be 300 000 000 m/s, but it seems more of a coincidence to me. I am using exact measurements of wavelenght and frequency from a Mcgraw Hill Dictionary of physics. If $$c=\lambda\nu$$ then c should come out to exactly 299792458 m/s. I'm asking that when $$\lambda\nu=300 000 000$$ What has this done to obtain the exact speed of light? (But I very well could be wrong and that 300 000 000 is simply the speed of light in m/s, just a bit off.)

thnx

6. Dec 30, 2004

Cyrus

300 000 000 is in m/s and 299 792 458 is in m/s. You are looking up a table of values that were found through experimentation, nothing is perfectly exact. Look at the value you got and the recorded value, that's a percent error of 0.0692%! Still not close enough?

7. Dec 30, 2004

DB

I never knew that lol. Thnx, now it seems so simple that $$\lambda\nu=300 000 000$$

8. Dec 30, 2004

Cyrus

No, $$\lambda\nu=299 792 458$$ to be accurate. 299 792 458 m/s is the speed of light. You got 300 000 000 because of error in the values you looked up when calculating the speed of light. Look at the values you used:
. The first one has only 1 sig fig, the second has only two, how can you call those exact?

9. Dec 30, 2004

rachmaninoff

Gamma rays represent EM radiation from a whole band of frequencies; there is no 'exact measurment' involved. I am assuming you are reading from a table of representative values, or ranges, thereof?

10. Dec 30, 2004

Integral

Staff Emeritus
A better way to represent this calculation would be

3e-14 * 10e22 = 3e8

or $$3 x 10^{-14} m * 1 x 10^{22} hz= 3 x 10^8 \frac m s$$

This more accurately reflects the number of significant digits. To get the more precise result your frequency and wavelength numbers would need at least 8 digits of precision.

11. Dec 31, 2004

Tide

That depends entirely what you mean by "exact measurement" and "whole band of frequencies!" The gravitational red shift due to the Earth of photons travelling a mere 10's of meters vertically has been measured with gamma rays with Mossbauer spectroscopy.

12. Dec 31, 2004

DB

You're right, now I see that for a more detailed answer I would need more precise measurments.

13. Dec 31, 2004

dextercioby

There's a trick here and noone noticed.The speed of light in vacuum IS EXACTLY 299792458ms^{-1}.So it means no error from using it.So u can do a measurement of wavelength and try to get it as accurate as possible and then,to find the frequency,simply use the speed of light as it will bring you no further error.Backwards works the same way.Measure the frequency with greatest accuracy possible and use "c" to get the wave length with no further error.Exempli gratia:
$$\lambda=\frac{c}{\nu}$$
$$\Delta \lambda =\frac{c\Delta\nu}{\nu^{2}}$$
,where 'c' is fixed and given and \nu and \Delta nu are determined thoretically from measurements (mean value and standard mean deviation).

Daniel.

PS.There's no point in measuring both and get a value for "c",as i'm sure u'll find that $c'\neq 299792458ms^{-1}$ and moreover [itex] \Delta c'\neq 0 [/tex],which is a bunch of c***.

14. Dec 31, 2004

krab

Or:
$$3\times 10^{-14}\,\mbox{m}\,*\,1\times 10^{22}\,\mbox{Hz}= 3\times 10^8\,\mbox{m/s}$$
(Sorry, I can't resist fixing LaTeX..)