# 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 3.188×10−10 m. Determine the frequency (in hz) of the interacting photon.

• Chemistry -

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

• Chemistry -

• Chemistry -

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

• Chemistry -

Selly could you please give the formula

• Chemistry -

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.

• Chemistry -

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

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