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 3.188×10−10 m. Determine the frequency (in hz) of the interacting photon.
Yes! I foun the correct answer. It's 7.157e15
can you please help to explain the formula
I have tried everything but don't know how the formulas go if you could please give a formula i would be very thankful
Selly could you please give the formula
First you find the Momentum (P) from the de Broglie eqn: P = h/lambda, where h is the Plancks constant.
Then you find the velocity (v) from the eqn: P = m*v, where m is the Electron Mass. Then substitute v in the KE eqn: KE = 0.5*m*v^2.
Add this KE to the Ionization Energy of H from the periodic table to find the Photon Energy.
Then Photon Energy = h*nu, where nu is the desired frequency ..
P.S: Don't forget to convert the Ionization Energy from eV to J.
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:
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 electron.
The momentum of an object can be calculated using the equation:
p = m * v
where p is the momentum, m is the mass of the object (in this case, the mass of the electron), and v is the velocity of the object.
In this scenario, the electron is ejected from the atom, so it is no longer bound to the atom and does not have a velocity. However, we know from the question that the electron exhibits a de Broglie wavelength of 3.188 × 10^-10 m.
Using the de Broglie wavelength equation and rearranging it, we can find the momentum of the electron:
p = h / λ
p = (6.626 × 10^-34 J·s) / (3.188 × 10^-10 m)
p ≈ 2.080 × 10^-24 kg·m/s
Now, we can use the momentum obtained to calculate the velocity of the electron. Since the mass of an electron is approximately 9.11 × 10^-31 kg, we can substitute these values into the equation:
p = m * v
(2.080 × 10^-24 kg·m/s) = (9.11 × 10^-31 kg) * v
v ≈ 2.280 × 10^6 m/s
Now that we have the velocity of the electron, we can determine the frequency of the interacting photon by using the relationship between the energy of a photon and its frequency:
E = h * f
where E is the energy of the photon, h is Planck's constant, and f is the frequency.
The energy of a photon can be calculated using the equation:
E = 1/2 * m * v^2
Substituting the mass and velocity values, we have:
E = 1/2 * (9.11 × 10^-31 kg) * (2.280 × 10^6 m/s)^2
E ≈ 2.521 × 10^-18 J
Now we can find the frequency by rearranging the energy equation:
f = E / h
f ≈ (2.521 × 10^-18 J) / (6.626 × 10^-34 J·s)
f ≈ 3.810 × 10^15 Hz
Therefore, the frequency of the interacting photon is approximately 3.810 × 10^15 Hz.