Chemistry
posted by kaken on .
A photon interacts with a ground state electron in a hydrogen atom and is absorbed. The electron is ejected from the atom and exhibits a de Broglie wavelength of 1.828×10−10 m. Determine the frequency (in hz) of the interacting photon.

4.329E15

wrong answer

Would you please give the formula?

chacho,could you please explain the answer when de broglie wavelength 0.468*10^10m.

A photon interacts with a ground state electron in a hydrogen atom and is absorbed. The electron is ejected from the atom and exhibits a de Broglie wavelength of 0.468×10−10 m. Determine the frequency (in hz) of the interacting photon.

How do you use the above data and insert into the formula?

could you explain me this one
A photon interacts with a ground state electron in a hydrogen atom and is absorbed...
And I will explain you the formula 
4.329E15 this answer is not right

7.157215*10^15

chacho's answer is right if you have in your question de Broglie wavelength of 5.908×10−10 m :)

this answer also not righ

Formula please...

2.3E19

please give us the formula

E = P^2/2Me + E(first ionization)
P= h / BroglieWavelength
E(first ionization)=21.7*10^19
Me= 9.1*10^31
h= 6.626*10^34
frecuency(hz)= E / h 
I have one question about the formula, is it E = P^2/(2Me + E)(first ionization) or E = P^2/(2Me + E(first ionization) ) ??

sorry i wrote the same formula; my questios was if the formula is:
1 E = (P^2/2Me )+ E(first ionization)
or
2 E = P^2/(2Me + E)(first ionization)
sorry for the previous post