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.
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.
4.7E-15
WRONG EXPALIN PLEASE
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
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
To determine the frequency of the interacting photon, we can use the de Broglie wavelength formula, which relates the wavelength of a particle to its momentum:
λ = h / p
where λ is the de Broglie wavelength, h is the Planck's constant (6.626 x 10^-34 J*s), and p is the momentum of the particle.
In this case, the wavelength of the ejected electron is given as 3.188 x 10^-10 m. Since the electron is ejected, we can assume that its momentum is equal to the momentum of the absorbed photon, as their interaction occurs in an isolated system. Therefore, we can rewrite the equation as:
λ = h / (m * v)
where m is the mass of the electron and v is its velocity.
The mass of an electron is approximately 9.11 x 10^-31 kg.
Now we can rearrange the equation to solve for the velocity of the electron:
v = h / (m * λ)
Plugging in the values:
v = (6.626 x 10^-34 J*s) / (9.11 x 10^-31 kg * 3.188 x 10^-10 m)
v ≈ 6.944 x 10^6 m/s
Since we now have the velocity of the electron, we can use it to calculate the frequency of the photon using the equation:
f = v / λ
Plugging in the values:
f = (6.944 x 10^6 m/s) / (3.188 x 10^-10 m)
f ≈ 2.177 x 10^16 Hz
Therefore, the frequency of the interacting photon is approximately 2.177 x 10^16 Hz.