Relate wavelength and energy scale

[senobim]

Homework Statement

Light intensity is measured by monochromator and is given by $$I(\lambda ) = I_{0}\lambda ^{3}$$ How to change it to the energy scale $$I(h\nu ) = ?$$

Homework Equations

Photons energy $$E = h\nu, E =\frac{hc}{\lambda }$$

The Attempt at a Solution

It's kind of strange to relate energy with wavelenght

 

[mfb]

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

 

[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

 

[mfb]

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

 

[senobim]

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

any hints? ;]

 

[mfb]

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

 

[senobim]

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

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

 

[mfb]

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

That is not right.

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

You'll need that formula once you found the intensity in the wavelength range.

 

[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?

 

[mfb]

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 $\lambda$?

 

[senobim]

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

 

[mfb]

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.

 

[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?

 

[mfb]

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?

 

[senobim]

i seems that i don't get the concept at all..

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

P.S. Thanks for your patiece

 

[mfb]

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

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

 

[senobim]

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

 

[mfb]

That's what I suggested in post #6.

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

 

[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?

 

[mfb]

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.

 

[senobim]

Could I let $$\delta\lambda \rightarrow 0$$?

 

[mfb]

Well, if you set it exactly to zero then the integrated intensity is zero, so you want to keep the first order in $\delta \lambda$.

 

[senobim]

still struggling on the approximation, what do you mean by that? Do i need to use trapezoidal rule for the definite integral approximation? Or something else?

 

[mfb]

Even easier. Approximate it as rectangle.

 

[senobim]

In order to perform approximation how do I find y-cordinate for this function?

 

[mfb]

What do you expect $I(\lambda)$ to be?

 

[senobim]

I(λ) = I₀λ³

 

[senobim]

$$I=I_{0}\int_{\lambda }^{\lambda +\delta \lambda }\lambda' d\lambda = \lambda +\delta \lambda -\lambda * f(\lambda') = I_{0} \delta \lambda \lambda^3$$

does that make any sense?

 

[mfb]

That is the intensity between $\lambda$ and $\lambda + \delta \lambda$ right (not exactly, but we let $\delta \lambda$ go to zero later so that approximation works).

Now you have to find the frequencies that correspond to $\lambda$ and $\lambda + \delta \lambda$.

 

[senobim]

λ=c/ν

$$I=I_{0}\int_{\nu }^{\nu +\delta \nu }\left ( \frac{c}{\nu } \right )^3d\nu \approx I_{0}\delta \nu \left ( \frac{c}{\nu } \right )^3$$

 

[mfb]

$\delta \lambda$ and $\delta \nu$ are different things, you cannot simply replace them.

 

[senobim]

of course.

λ = c/ν, ν=c/λ --> do I nead to integrate over this range this time? c/λ and c/λ + c/δλ

 

[mfb]

c/λ and c/λ + c/δλ

a/(b+c) and a/b + a/c are not the same.

 

[senobim]

c/λ and c/λ + δ(c/λ)?

 

[mfb]

What does δ(c/λ) mean?