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 5.908×10−10 m. Determine the frequency (in hz) of the interacting photon.
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
Chemgam, can you please clarify what is (P^2)? do you mean (Momentum) X (Momentum)
I made it like this, but it gave me wrong answer
P = h/ BroglieWavelength
Like he says. It is correct
To determine the frequency of the interacting photon, we can use the de Broglie wavelength equation, which relates the wavelength (λ) and the momentum (p) of a particle:
λ = h / p
where λ is the wavelength, h is the Planck's constant (6.62607015 × 10^(-34) m²·kg/s), and p is the momentum.
The momentum of a particle is given by:
p = mv
where m is the mass of the particle, and v is its velocity.
In this case, the electron is ejected from the atom, so it is no longer bound. We assume that it is initially at rest, as it is in the ground state. Therefore, the velocity of the electron is the final velocity after it is ejected.
Now, the energy of a photon can be given by:
E = hf
where E is the energy, h is Planck's constant, and f is the frequency of the photon.
The energy of the photon can also be related to the energy required to ionize the electron from the ground state to its final velocity:
E = (1/2)mv^2
Setting these two energy equations equal to each other:
hf = (1/2)mv^2
From this equation, we can solve for v:
v = √(2hf / m)
Now, since we know the de Broglie wavelength (λ) and the mass (m) of the electron, we can solve for the momentum (p) using the equation:
λ = h / p
Solving for p:
p = h / λ
Finally, we can find the frequency (f) of the interacting photon by using the equation:
f = p / m
Now we can substitute the values provided into the equations to calculate the frequency of the interacting photon:
λ = 5.908×10^(-10) m
h = 6.62607015 × 10^(-34) m²·kg/s
m (mass of electron) = 9.10938356 × 10^(-31) kg
First, let's calculate the momentum (p):
p = h / λ = (6.62607015 × 10^(-34) m²·kg/s) / (5.908×10^(-10) m) ≈ 1.120912 × 10^(-24) kg·m/s
Then, we can calculate the frequency (f):
f = p / m = (1.120912 × 10^(-24) kg·m/s) / (9.10938356 × 10^(-31) kg) ≈ 1.228123 × 10^6 Hz
Therefore, the frequency of the interacting photon is approximately 1.228123 × 10^6 Hz.