Chemistry
posted by selly .
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 3.188×10−10 m. Determine the frequency (in hz) of the interacting photon.

Yes! I foun the correct answer. It's 7.157e15

can you please help to explain the formula

I have tried everything but don't know how the formulas go if you could please give a formula i would be very thankful

Selly could you please give the formula

First you find the Momentum (P) from the de Broglie eqn: P = h/lambda, where h is the Plancks constant.
Then you find the velocity (v) from the eqn: P = m*v, where m is the Electron Mass. Then substitute v in the KE eqn: KE = 0.5*m*v^2.
Add this KE to the Ionization Energy of H from the periodic table to find the Photon Energy.
Then Photon Energy = h*nu, where nu is the desired frequency ..
P.S: Don't forget to convert the Ionization Energy from eV to J. 
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