Asked by Anonymous
The meter has been defined as 1,650,763.73 wavelengths of the orange-red line of the emission spectrum of ^86Kr. Calculate the frequency of this radiation. In what proportion of the electromagnetic spectrum does the radiation fall?
Answers
Answered by
DrBob222
What is the wavelength of the Kr line? That is 1 m = 1,650,763.73 x wavelength. Solve for wavelength. Then
c = freq*wavelength
wavelength is from above.
c = 3E8 m/s
Solve for freq.
Obviously if it is an orange-red line it must be in the visible part of the spectrum.
By the way, I think you meant, "In what PORTION of .....".
c = freq*wavelength
wavelength is from above.
c = 3E8 m/s
Solve for freq.
Obviously if it is an orange-red line it must be in the visible part of the spectrum.
By the way, I think you meant, "In what PORTION of .....".
Answered by
Anonymous
I got 5.506x10^-3 when I divide the wavelength by the speed of light but I don't think that's right because Kr is an isotope. Could you please explain how this works too?
Answered by
DrBob222
Not what I get. What is the wavelength you have. Post your work and I'll find the error.
Answered by
Anonymous
Oh is it 1.816x10^2? I divide the speed of light by the wavelength: 2.998x10^8 / 1,650,763.73
I did the reverse the first time. Is this correct now?
I did the reverse the first time. Is this correct now?
Answered by
DrBob222
No. 1,650,763.73 is not the wavelength. Look at my first response again.
1 m = 1,650,763.73 x wavelength and you calculate the wavelength from that. Frequency is in the order of 10^14 Hz unless I punched te wrong buttons.
1 m = 1,650,763.73 x wavelength and you calculate the wavelength from that. Frequency is in the order of 10^14 Hz unless I punched te wrong buttons.
Answered by
Anonymous
ok so for wavelength I get 6.058x10^-7
then I divide: 2.998x10^8 / 6.058x10^-7 and get 4.949x10^14 Hz
Is this correct?
then I divide: 2.998x10^8 / 6.058x10^-7 and get 4.949x10^14 Hz
Is this correct?
Answered by
DrBob222
Looks ok to me.
Answered by
Anonymous
Thank you for your help
Answered by
DrBob222
I didn't think to post the earlier but since you have so many significant figures in the wavelength calculation, you could use that many places in the speed of light. I don't know that many places but I know it's available on the web.
Answered by
Johne163
I do not even know how I stopped up here, however I assumed this put up was good. I do not recognise who you are however definitely you're going to a wellknown blogger for those who are not already bfdedacekdbd
Answered by
bleh
how did you get 6.058x10^-7?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.