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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 Ankur
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.
Answers
Answered by
bobpursley
frequency*wavelength=speed of light
Answered by
Peter
not correct answer - this is a de Broglie wave ;)
Answered by
Ling
First find: 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
Once E is found the find frequency:
frequency(hz)= E / h
P= h / BroglieWavelength
E(first ionization)=21.7*10^-19
Me= 9.1*10^-31
h= 6.626*10^-34
Once E is found the find frequency:
frequency(hz)= E / h
Answered by
rex
5.032*10^15
Answered by
Anonymous
1.71*10^17
Answered by
LI WU
It is wrong rex.
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