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.
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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
To determine the frequency of the interacting photon, we can use the de Broglie wavelength equation. The equation relates the wavelength of a particle to its momentum and mass.
λ = h / (mv)
where λ is the wavelength, h is Planck's constant (6.626 × 10^-34 J·s), m is the mass of the particle, and v is its velocity.
In this case, the particle is an electron. The final wavelength (λ) is given as 1.828 × 10^-10 m.
We can rearrange the equation to solve for v:
v = h / (mλ)
The mass of an electron (m) is approximately 9.109 × 10^-31 kg.
Now, we can calculate the velocity of the electron.
v = (6.626 × 10^-34 J·s) / (9.109 × 10^-31 kg × 1.828 × 10^-10 m)
Simplifying the expression:
v ≈ 3.598 × 10^6 m/s
Now that we know the velocity of the electron, we can calculate the frequency (f) of the photon using the equation:
f = v / λ
f ≈ (3.598 × 10^6 m/s) / (1.828 × 10^-10 m)
Simplifying the expression:
f ≈ 1.970 × 10^16 Hz
Therefore, the frequency of the interacting photon is approximately 1.970 × 10^16 Hz.