# Relate wavelength and energy scale

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1. Sep 15, 2016

### senobim

1. The problem statement, all variables and given/known data
Light intensity is measured by monochromator and is given by $$I(\lambda ) = I_{0}\lambda ^{3}$$ How to change it to the energy scale $$I(h\nu ) = ?$$

2. Relevant equations
Photons enery $$E = h\nu, E =\frac{hc}{\lambda }$$

3. The attempt at a solution
It's kind of strange to relate energy with wavelenght

2. Sep 15, 2016

### Staff: Mentor

How much power is there between the frequencies v and v+$\delta$v? What happens if you let $\delta$v go to zero?

3. Sep 16, 2016

### senobim

Power could be calculated from intensity and area P = I * A.
and It could be related in terms of photon energy P = N * E --> P = N * hv
N - number of photons

Last edited: Sep 16, 2016
4. Sep 16, 2016

### Staff: Mentor

Yes, but that is not the point. Can you calculate it with the given intensity profile? Approximations for small $\delta$v are fine.

5. Sep 16, 2016

### senobim

can't think of anything how to do that..

any hints? ;]

Last edited: Sep 16, 2016
6. Sep 16, 2016

### Staff: Mentor

How much intensity is there between two wavelengths $\lambda$ and $\lambda + \delta \lambda$? How is that related to my previous question?

7. Sep 16, 2016

### senobim

$$I = I_{0}(\lambda + (\lambda + \delta \lambda ))^{3}$$

and $$\nu = \frac{c}{\lambda }$$

Last edited: Sep 16, 2016
8. Sep 16, 2016

### Staff: Mentor

That is not right.
You'll need that formula once you found the intensity in the wavelength range.

9. Sep 16, 2016

### senobim

emm.. Intensity at some wavelenght is defind by $$I(\lambda)=I_{0}\lambda^{3}$$

at some different lamda $$I(\lambda_{2})=I_{0}(\lambda_{2})^{3}$$

where am I going wrong?

10. Sep 16, 2016

### Staff: Mentor

Right, and how is this related to what you wrote in #7?

Can you draw a sketch, $I(\lambda)$? How would you get the integrated intensity between $\lambda$ and $\lambda + \delta \lambda$? How can you approximate this for very small intervals of $\lambda$?

11. Sep 16, 2016

### senobim

you mean something like this? $$I = I_{0}\int \lambda ^{3}d\lambda$$

12. Sep 16, 2016

### Staff: Mentor

Yes.

There are two approaches: keep the integrals everywhere, or go via the integrands. The former is easier if you know how to do substitutions in integrals, otherwise the latter is easier.

13. Sep 16, 2016

### senobim

so my answer is $$I=I_{0}\int \lambda ^{3}d\lambda =I_{o}\frac{\lambda ^{4}}{4}$$

and $$I(h\nu )=I_{o}\frac{h}{4}\left ( \frac{c}{\nu } \right )^4$$

am I right?

14. Sep 16, 2016

### Staff: Mentor

No, that does not work.

The second "=" in the first line is wrong, and the transition between the first line and the second line does not make sense.

What is $\lambda$ at the very right of the first line? Which wavelength is that?

15. Sep 16, 2016

### senobim

i seems that i dont get the concept at all..

I just integrated the expression, what could be wrong by that

P.S. Thanks for your patiece

16. Sep 16, 2016

### Staff: Mentor

Okay, let me ask differently: what did you use as range for the integral, and why?

17. Sep 17, 2016

### senobim

I did indefinte integral and that's not the case here, maybe i should try something like this
$$I = I_{0}\int_{\lambda }^{\lambda +\delta \lambda }\lambda ^{3}d\lambda$$

18. Sep 17, 2016

### Staff: Mentor

That's what I suggested in post #6.

The integrand should use a different symbol (like $\lambda'$) to avoid mixing two different things.

19. Sep 19, 2016

### senobim

good, so now we have this $$I = I_{0}\int_{\lambda }^{\lambda +\delta \lambda }\lambda'd\lambda = I_{0}\frac{\lambda'^4}{4}|_{\lambda}^{\lambda + \delta \lambda} = I_{0}(\frac{(\lambda +\delta \lambda)^4 }{4} - \frac{\lambda^4}{4})$$

and how to relate this with energy?

20. Sep 19, 2016

### Staff: Mentor

You can find an approximation of this for small $\delta \lambda$. What is the approximate value of an integral if the function is (roughly) constant over the integration range?

This is also the integrated intensity between two specific frequencies, which you can get with the formula relating wavelengths and frequencies.

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