What Frequency Photon Ionizes an Atom from n=2 State?

[sskk221]

Homework Statement

The following diagram represents the allowed electronic energy levels of a hypothetical atom. What frequency photon would be required to ionize this atom if it were already in the n = 2 state?

[PLAIN]http://img32.imageshack.us/img32/667/picture2vt.png

Homework Equations

E = hf

The Attempt at a Solution

Ionizing the atom would require emission of a photon. From the picure, E = 4eV for an emission of a photon:

4eV = hf
f= 9.65e14 Hz but this is not the answer. Any suggestions??

 

[rock.freak667]

You used the correct formula. Did you change eV to J? That is the only way I could see you getting it wrong.

 

[sskk221]

yeah I did 4eV*1.6e-19 to convert it to Joules. Then I divided by Planck's constant to get the frequency. This problem seems so easy, but I'm missing something... :\

 

[ehild]

What does ionization mean? What is the energy of the electron when the atom is ionized?

ehild

 

[rdl]

Your attempt at a solution is on the right track. However, it is important to note that the energy levels in the diagram are not to scale, so using the value of 4eV as the energy of the photon may not be accurate. Also, the energy of the photon required for ionization is actually equal to the difference in energy between the n = 2 state and the ionization energy level (n = infinity).

To find the frequency of the photon, we can use the equation E = hf, where E is the energy of the photon, h is Planck's constant, and f is the frequency. So we can rewrite the equation as f = E/h.

The energy of the photon required for ionization can be calculated by taking the difference between the energy of the ionization level (n = infinity) and the energy of the n = 2 state. In this case, the energy of the ionization level is equal to 10eV (again, not to scale), and the energy of the n = 2 state is equal to 4eV. So the energy of the photon required for ionization is 10eV - 4eV = 6eV.

Now we can plug this value into our equation for frequency:

f = (6eV)/(4.14x10^-15 eVs) = 1.45x10^15 Hz

This would be the frequency of the photon required to ionize the atom from the n = 2 state. I hope this helps!