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 4.548 x 10^-10m. Determine the frequency (in hz) of the interacting photon.
i got 1.45691908531e-24 but it is wrong :(
9.14E18
9.14E18 is wrong
can you please explain how you got this number
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
Yup, it was meant to be wrong.
I don't like cheats.
when you gave the people wrong answer and you know that it is wrong, this is the pure cheating, stupid..!!
5.043661*10^15
To determine the frequency (in Hz) of the interacting photon, we can use the de Broglie wavelength equation:
λ = h / p
where λ is the wavelength, h is the Planck's constant (6.626 x 10^-34 J·s), and p is the momentum of the particle.
In this case, the de Broglie wavelength of the ejected electron is given as 4.548 x 10^-10 m. We can find the momentum of the electron by using its mass (m) and velocity (v). Since the electron is ejected, its velocity can be determined using the classical equation for kinetic energy:
K.E. = (1/2)mv^2
The total energy (E) of the ejected electron is equal to the energy of the absorbed photon:
E = hf
where h is Planck's constant and f is the frequency of the photon.
The kinetic energy of the ejected electron is given by the energy difference between the ground state (-13.6 eV) and the ejected state (0 eV):
K.E. = Ef - Ei
where Ef is the energy in the ejected state and Ei is the energy in the ground state.
Now, we can relate the momentum (p) to kinetic energy (K.E.):
K.E. = p^2 / (2m)
Rearranging this equation gives us:
p = √(2mK.E.)
Now, we have all the necessary equations to find the frequency (f) of the interacting photon. Let's substitute the known values:
1. Find the momentum of the ejected electron:
λ = h / p
4.548 x 10^-10 m = (6.626 x 10^-34 J·s) / p
Rearrange the equation to find p:
p = (6.626 x 10^-34 J·s) / (4.548 x 10^-10 m)
p ≈ 1.455 x 10^-24 kg·m/s
2. Find the mass of the electron:
m = 9.10938356 x 10^-31 kg
3. Find the kinetic energy of the ejected electron:
K.E. = Ef - Ei
K.E. = 0 eV - (-13.6 eV) = 13.6 eV
Convert eV to Joules:
1 eV = 1.602 x 10^-19 J
K.E. = 13.6 eV * (1.602 x 10^-19 J / 1 eV) = 2.18 x 10^-18 J
4. Calculate the frequency of the interacting photon:
Using the relation between momentum (p) and kinetic energy (K.E.):
K.E. = p^2 / (2m)
2.18 x 10^-18 J = (1.455 x 10^-24 kg·m/s)^2 / (2 * 9.10938356 x 10^-31 kg)
Simplify the equation:
(1.455 x 10^-24 kg·m/s)^2 = 2.18 x 10^-18 J * (2 * 9.10938356 x 10^-31 kg)
Take the square root:
1.455 x 10^-24 kg·m/s = √(2.18 x 10^-18 J * (2 * 9.10938356 x 10^-31 kg))
Calculate:
1.455 x 10^-24 kg·m/s ≈ 6.593 x 10^-24 kg·m/s
Finally, calculate the frequency of the interacting photon:
E = hf
hf = 6.593 x 10^-24 kg·m/s
f = (6.593 x 10^-24 kg·m/s) / (6.626 x 10^-34 J·s)
f ≈ 9.954 x 10^9 Hz
Hence, the frequency of the interacting photon is approximately 9.954 x 10^9 Hz.