How Do Atoms of Element Q Absorb or Emit Specific Photon Frequencies?

[steven10137]

Homework Statement

An atom of element Q can only absorb two photons of frequency 6.40E14 Hz and 9.05E14 Hz.

Explain how atoms of element Q could absorb or emit photons of particular frequencies.2. The attempt at a solution

OK I think I am missing something in my understanding of absorbtion and emission ...
I don't really understand what the question is asking.

But, perhaps;

The element Q has set energy levels; n=1, n=2 ...
Hence the element can only absorb or emit photons with energies equal to the energies of these energy levels.
These energies are 6.40E14 Hz and 9.05E14 respectively.As I said, I'm having trouble. If someone could please explain the concept of the problem, it would be greatly appreciated.

Thanks
Steven

 

[Astronuc]

One more or less has is correct. An atom of a given element will absorb or emit light (photons) at discrete energies because the electrons in the atom occupy particular energy levels.

The energy of the light (photon) is related to the frequency by Planck's constant.

http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html#c3

Absorption or emission - http://hyperphysics.phy-astr.gsu.edu/hbase/mod5.html#c2

Hydrogen model - http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html

 

[steven10137]

Astronuc, thanks for your reply :)

For the second part of the question,
I am required to draw the energy level diagram of the element, showing the known levels.
Then, calculate the longest wavelength photon that the element could emit.

http://img257.imageshack.us/img257/9219/photondiagramkd9.jpg

ok so the longest possible photon will be that of the smallest frequency photon.
From the question, it says the element can absorb photons of the the specifc frequencies.

Now from here
http://hyperphysics.phy-astr.gsu.edu/hbase/mod5.html#c2
I can see that the downward transition gives an emission of E2 - E1.

So we are looking for the downward transition with the smallest possible frequency, which is;
9.05E14 - 6.04E14 = 3.01E14 Hz
<br /> \begin{array}{l}<br /> \lambda = \frac{c}{f} \\ <br /> \lambda _{\max } = \frac{{3 \times 10^8 }}{{3.01 \times 10^{14} }} = 9.97 \times 10^{ - 7} m \\ <br /> \end{array}<br />

How am I looking?

Thanks for your help,
Steven

 

[Astronuc]

The energy level at n=2 is closer to n=1 than n=3 is to n=2.

Refer to the last link I cited -
http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html

If 9.05E14 Hz represents n=3 to n=1 and 6.04E14 n=2 to n=1, then the n=3 to n=2 transition would be the difference. This is based on the statement "An atom of element Q can only absorb two photons of frequency 6.40E14 Hz and 9.05E14 Hz." It would be unlikely that a photon would absorbed by an electron in the excited n=2 state.

So we are looking for the downward transition with the smallest possible frequency, which is;
9.05E14 - 6.04E14 = 3.01E14 Hz

Looks OK.

 

[steven10137]

OK for the diagram, the 2.5eV should be less than the 3.75eV (elementary mistake)

If 9.05E14 Hz represents n=3 to n=1 and 6.04E14 n=2 to n=1, then the n=3 to n=2 transition would be the difference. This is based on the statement "An atom of element Q can only absorb two photons of frequency 6.40E14 Hz and 9.05E14 Hz." It would be unlikely that a photon would absorbed by an electron in the excited n=2 state.

I'm not really fully understanding what you mean.
But are you trying to say that those frequencies do not represent those transitions?
Am I sort of on the right track?

Thanks for your patience.
Steven.

 

[Astronuc]

Note the different groups and emission lines in this -
http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html#c4


I was saying those energies do represent those transitions.

The two absorption energy would correspond to n=3 to n=1 (more energetic) and n=2 to n=1 less energetic, and these would be approximately equal to emission lines. There is also a possible n=3 to n=2 emission, which is the difference between 3-1 and 2-1.

The 2->3 would be an unlikely absorption because the transition 2->1 would occur very quickly.