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 1.828×10−10 m. Determine the frequency (in hz) of the interacting photon.

Answer 1

4.329E15

Answer 2

wrong answer

Answer 3

Would you please give the formula?

Answer 4

chacho,could you please explain the answer when de broglie wavelength 0.468*10^-10m.

Answer 5

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.

Answer 6

How do you use the above data and insert into the formula?

Answer 7

could you explain me this one
A photon interacts with a ground state electron in a hydrogen atom and is absorbed...
And I will explain you the formula

Answer 8

4.329E15 this answer is not right

Answer 9

7.157215*10^15

Answer 10

chacho's answer is right if you have in your question de Broglie wavelength of 5.908×10−10 m :-)

Answer 11

this answer also not righ

Answer 12

Formula please...

Answer 13

2.3E19

Answer 14

please give us the formula

Answer 15

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 16

I have one question about the formula, is it E = P^2/(2Me + E)(first ionization) or E = P^2/(2Me + E(first ionization) ) ??

Answer 17

sorry i wrote the same formula; my questios was if the formula is:

1- E = (P^2/2Me )+ E(first ionization)
or
2- E = P^2/(2Me + E)(first ionization)

sorry for the previous post