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.

Answer 1

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.

Answer 2

4.7E-15

Answer 3

WRONG EXPALIN PLEASE

Answer 4

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

Answer 5

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