# Chemistry

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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 exhibits a de Broglie wavelength of 1.828×10−10 m. Determine the frequency (in hz) of the interacting photon.

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4.329E15

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Would you please give the formula?

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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 exhibits a de Broglie wavelength of 0.468×10−10 m. Determine the frequency (in hz) of the interacting photon.

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

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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

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4.329E15 this answer is not right

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7.157215*10^15

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

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2.3E19

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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

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I have one question about the formula, is it E = P^2/(2Me + E)(first ionization) or E = P^2/(2Me + E(first ionization) ) ??

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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

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