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.
frequency*wavelength=speed of light
not correct answer - this is a de Broglie wave ;)
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
5.032*10^15
1.71*10^17
It is wrong rex.
To find the frequency of the interacting photon, we can use the de Broglie wavelength equation and relate it to the energy of the photon.
The de Broglie wavelength (λ) is given by:
λ = h / p,
where λ is the wavelength, h is Planck's constant (6.626 × 10^(-34) J·s), and p is the momentum of the particle.
In this case, we have an electron that exhibits a de Broglie wavelength. The momentum of the electron (p) can be related to its kinetic energy (K) using the equation:
p = sqrt(2 * m * K),
where m is the mass of the electron (9.109 × 10^(-31) kg).
Since the electron is ejected from the hydrogen atom, it gained energy in the process. This energy can be given by the energy of the photon (E) using the equation:
E = h * f,
where E is the energy, h is Planck's constant, and f is the frequency of the photon.
Now, we can find the frequency by relating all the given quantities.
1. Firstly, convert the de Broglie wavelength from meters to kilograms using scientific notation:
λ = 0.468 × 10^(-10) m = 4.68 × 10^(-11) m
2. Calculate the momentum of the electron using the de Broglie wavelength:
p = h / λ = 6.626 × 10^(-34) J·s / 4.68 × 10^(-11) m = 1.415 × 10^(-23) kg·m/s
3. Calculate the kinetic energy (K) of the electron using the momentum:
K = p^2 / (2 * m) = (1.415 × 10^(-23) kg·m/s)^2 / (2 * 9.109 × 10^(-31) kg) = 1.993 × 10^(-17) J
4. Now, we can relate the energy (E) to the frequency (f) using Planck's constant:
E = h * f
f = E / h = (1.993 × 10^(-17) J) / (6.626 × 10^(-34) J·s) = 3.009 × 10^16 Hz
Therefore, the frequency of the interacting photon is approximately 3.009 × 10^16 Hz.