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.

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

1.71*10^17

Right answer

To determine the frequency of the interacting photon, we can use the de Broglie equation, which relates the wavelength (λ) and the momentum (p) of a particle or a wave. The equation is given by:

λ = h / p

where λ is the wavelength, h is the Planck's constant (6.626 × 10^(-34) J·s), and p is the momentum of the particle.

In this case, the given wavelength (λ) is 0.468 × 10^(-10) m. We can use this to calculate the momentum (p) of the electron:

p = h / λ

p = (6.626 × 10^(-34) J·s) / (0.468 × 10^(-10) m)

p = 1.414 × 10^(-24) kg·m/s

Now, we know that the momentum (p) of a photon is related to its frequency (f) using the equation:

p = hf

where f is the frequency of the photon.

Rearranging the equation, we get:

f = p / h

f = (1.414 × 10^(-24) kg·m/s) / (6.626 × 10^(-34) J·s)

f ≈ 2.133 × 10^9 Hz

Therefore, the frequency of the interacting photon is approximately 2.133 × 10^9 Hz.