The energies, E, for the first few states of an unknown element are shown here in arbitrary units:

n= 1, E=-21
n=2, E=-9
n=3, E=-4
n=4, E=-3
...,...
infinity, 0

A gaseous sample of this element is bombarded by photons of various energies (in these same units). Match each photon to the result of its absorption (or lack thereof) by an n=1 electron.

Photon/Energy:
A/21
B/17
C/15
D/12

Possible answers:
n=1 to n=2
n=1 to n=3
n=1 to n=4
electron ejected
not absorbed

I think the A/21 is electron ejected, but I don't know how to figure the rest...

A= ejected

B=1->3
C=not absorbed
D=1->2

A= ejected

B=1-3
C= not absorbed
D=1-2

Well, let's shed some light on this situation, shall we?

For the photon with an energy of A/21, since the energy of the n=1 state is -21, it's exactly equal to the energy of the photon. Therefore, the electron would have enough energy to be completely ejected from the atom. So, you're absolutely correct, the electron is ejected!

Now, let's figure out the rest without being too negative:

For the photon with an energy of B/17, we can see that the energy difference between the n=1 state and n=2 state is -21 - (-9) = -12. Since B/17 is less than this energy difference, the photon will not be absorbed and no transition will take place.

For the photon with an energy of C/15, the energy difference between the n=1 state and n=3 state is -21 - (-4) = -17. Since C/15 is greater than this energy difference, the photon will be absorbed and cause a transition from n=1 to n=3.

Lastly, for the photon with an energy of D/12, the energy difference between the n=1 state and n=4 state is -21 - (-3) = -18. Since D/12 is less than this energy difference, the photon will not be absorbed and no transition will occur.

So, to summarize:

A/21 - Electron ejected
B/17 - Not absorbed
C/15 - n=1 to n=3
D/12 - Not absorbed

Remember, it's electrifying to talk about atoms and photons, even if it's a bit shocking!

To determine the result of absorption for each photon, we need to compare the energy of the photon to the energy differences between different states of the electron.

The energy difference between two energy states is given by the equation:

ΔE = E(final state) - E(initial state)

Let's calculate the energy differences for each possible absorption transition:

1. For photon A/21:
ΔE = E(n=2) - E(n=1)
ΔE = (-9) - (-21)
ΔE = 12

Since the energy difference between n=2 and n=1 is 12, which matches the energy of photon A/21, the photon is absorbed and causes an excitation from n=1 to n=2. So the result is n=1 to n=2.

2. For photon B/17:
ΔE = E(n=3) - E(n=1)
ΔE = (-4) - (-21)
ΔE = 17

The energy difference between n=3 and n=1 is 17, which matches the energy of photon B/17, so the photon is absorbed and causes an excitation from n=1 to n=3. Therefore, the result is n=1 to n=3.

3. For photon C/15:
ΔE = E(n=4) - E(n=1)
ΔE = (-3) - (-21)
ΔE = 18

The energy difference between n=4 and n=1 is 18, which does not match the energy of photon C/15. Therefore, the photon is not absorbed.

4. For photon D/12:
ΔE = E(n=4) - E(n=1)
ΔE = (-3) - (-21)
ΔE = 18

Similar to photon C/15, the energy difference between n=4 and n=1 is 18, which does not match the energy of photon D/12. Hence, the photon is not absorbed.

To summarize:

- Photon A/21 is absorbed, resulting in an excitation from n=1 to n=2.
- Photon B/17 is absorbed, resulting in an excitation from n=1 to n=3.
- Photons C/15 and D/12 are not absorbed.

Thus, the results are as follows:

A/21: n=1 to n=2
B/17: n=1 to n=3
C/15: not absorbed
D/12: not absorbed

To determine the result of the absorption of each photon by an n=1 electron, we need to compare the energy of each photon to the given energy levels of the unknown element.

The energy levels provided for the unknown element are given as negative numbers, and they increase in magnitude as the energy level increases. For example, the energy of the first excited state (n=2) is -9, while the energy of the second excited state (n=3) is -4. The energy of the ground state (n=1) is -21, and the energy increases towards 0 as the energy level approaches infinity.

Now let's analyze each photon and its absorption:

A/21:
The energy of this photon is equal to the energy of the ground state (n=1) of the unknown element. This means that the photon has exactly the right energy to promote an electron from the ground state to a higher energy level. Since the energy of the ground state is -21, which is more negative than the energy of any other state, we can conclude that the absorption of this photon will result in an electron being ejected from the atom (electron ejected).

B/17:
The energy of this photon is greater than the energy of the ground state (-21) but less than the energy of the first excited state (-9). Therefore, this photon does not have enough energy to promote an electron from the ground state to the first excited state. As a result, the photon will not be absorbed by an n=1 electron (not absorbed).

C/15:
Similar to the previous case, the energy of this photon is greater than the ground state energy (-21) but less than the energy of the first excited state (-9). Hence, this photon will also not be absorbed by an n=1 electron (not absorbed).

D/12:
The energy of this photon is greater than the energy of the ground state (-21) but less than the energy of the first excited state (-9). Just like the previous two cases, this photon lacks sufficient energy to promote an electron from the n=1 ground state. Therefore, this photon will also not be absorbed by an n=1 electron (not absorbed).

In summary, we have determined that:
- For photon A/21, the result of its absorption is electron ejected.
- For photons B/17, C/15, and D/12, the result of their absorption is not absorbed.

Remember, we reached these conclusions by comparing the energy of each photon with the energy levels provided for the unknown element.

Name the following compounds:

CH3COONa

(NH4)2SO4

Mg(H2PO4)2

If I could draw a diagram but I can't. Let me try to use periods as spaces.

E = k/N2

.....1.....2.....3.....4.....inf
E = -21...-9....-4....-3......0

In going from n = 1 to n = 2 we have dE = 13
Going from 1 to 3 we have dE = 17
Going from 1 to 4 we have dE = 18
Going form 1 to inf we have dE = 21
Match them up.
A is 21 so that must be the one ejected; i.e., ionized.
B is 17 so that must be 1 to 3
C and D are not absorbed.

Check my thinking