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