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.

5.26E23

5.26E23 is incorrect

1.4169*10^16

To determine the frequency of the interacting photon, we can use the formula that relates the energy of a photon to its frequency:

E = hf

Where:
E is the energy of the photon,
h is Planck's constant (6.626 x 10^-34 J·s),
and f is the frequency of the photon.

We know that the de Broglie wavelength (λ) of the ejected electron can be related to its momentum (p):

λ = h / p

Where:
λ is the wavelength,
h is Planck's constant (6.626 x 10^-34 J·s),
and p is the momentum of the electron.

We can rearrange the equation to solve for p:

p = h / λ

Given that the de Broglie wavelength (λ) of the ejected electron is 1.828×10^-10 m, we can calculate its momentum (p):

p = (6.626 x 10^-34 J·s) / (1.828×10^-10 m)

Now, since momentum (p) is equal to the product of mass (m) and velocity (v):

p = mv

For an electron, we know its mass (m) is approximately 9.109 x 10^-31 kg, so we can rearrange the equation to solve for the velocity (v):

v = p / m

Using the calculated momentum (p) and the electron's mass (m), we can find its velocity (v). Then, since the ejected electron absorbs the photon's energy, the energy of the photon will be equal to the kinetic energy of the electron:

E = ½ mv^2

Now we can substitute the calculated velocity (v) into the equation:

E = ½ (9.109 x 10^-31 kg) (v^2)

Finally, substituting the calculated energy (E) into the equation E = hf, we can solve for the frequency (f) of the interacting photon.