Asked by schoolgirl

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 3.188*10^-10m Determine the frequency (in hz) of the interacting photon.
Answered by drwls
The momentum of the electron is h /(wavelength)
h is Planck's constant. Use that to compute the electron energy.

Add the electron energy to the ionization energy (13.6 eV) for the photon energy.
Answered by helper
4.7E-15
Answered by Anonymous
WRONG EXPALIN PLEASE
Answered by qwerty
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
Answered by a
i put the numbers into the formula but didn't worked..where's my fault? my E=5.796*10^-31 ; my f=874.8039
maybe someone can tell me please what's wrong here, thx
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