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A photon interacts with a ground state electron in a hydrogen atom and is absorbed. The electron is ejected from the atom and e...Asked by Blair
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 5.908×10−10 m. Determine the frequency (in hz) of the interacting photon.
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Answered by
Chemgam
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
P= h / BroglieWavelength
E(first ionization)=21.7*10^-19
Me= 9.1*10^-31
h= 6.626*10^-34
frecuency(hz)= E / h
Answered by
Anonymous
Chemgam, can you please clarify what is (P^2)? do you mean (Momentum) X (Momentum)
I made it like this, but it gave me wrong answer
I made it like this, but it gave me wrong answer
Answered by
Anonymous
P = h/ BroglieWavelength
Like he says. It is correct
Like he says. It is correct
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