An electron begins in the n = 7 orbital of a hydrogen atom. It then emits a photon of wavelength λ and falls to a lower energy level. Before it has a chance to change levels again, it absorbs a 10 eV photon and is ejected from the atom with a kinetic energy of 6.6 eV. What is the wavelength, λ, of the first photon emitted?
First, let's find the initial energy of the electron in the n = 7 orbital. The energy levels in a hydrogen atom can be found using the formula:
E_n = -13.6 eV / n^2
where E_n is the energy of the electron in the nth orbital.
So for n = 7, the initial energy is:
E_7 = -13.6 eV / 7^2 ≈ -0.277 eV
Now let's find the energy of the electron after it absorbs the 10 eV photon and is ejected from the atom with a kinetic energy of 6.6 eV.
E_final = E_kinetic + 10 eV
E_final = 6.6 eV + 10 eV = 16.6 eV
In order for the electron to change energy levels and eventually be ejected, it must lose energy by emitting a photon with a certain wavelength. The energy of the emitted photon can be found using the change in energy of the electron:
E_photon = E_final - E_7
E_photon = 16.6 eV - (-0.277 eV) ≈ 16.877 eV
Now we can use the Planck-Einstein relation to find the wavelength of the emitted photon:
E_photon = (hc) / λ
where h is Planck's constant (4.136 x 10^-15 eV*s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength.
Rearranging the equation to find λ:
λ = (hc) / E_photon
Plugging in the values:
λ = (4.136 x 10^-15 eV*s * 3 x 10^8 m/s) / 16.877 eV ≈ 7.34 x 10^-8 m
So the wavelength of the emitted photon is approximately λ ≈ 734 nm, which falls in the near-infrared (NIR) region of the electromagnetic spectrum.
To solve this problem, we need to use the concepts of energy levels in a hydrogen atom and the relationship between the energy of a photon and its wavelength.
Step 1: Find the initial energy of the electron in the n = 7 orbital.
The energy of an electron in the nth energy level of a hydrogen atom is given by the equation: E = -13.6 eV/n^2, where n is the principal quantum number.
For n = 7, the initial energy (E_initial) is given by:
E_initial = -13.6 eV / (7^2)
E_initial = -13.6 eV / 49
E_initial = -0.2776 eV
Step 2: Determine the change in energy when the electron falls to a lower energy level and emits a photon.
The energy change (ΔE) can be calculated using the equation: ΔE = Ef - Ei, where Ef is the final energy and Ei is the initial energy.
Given that the electron is falling to a lower energy level, the final energy (Ef) is higher than the initial energy (E_initial).
Step 3: Calculate the energy of the emitted photon using the energy change.
The energy of a photon is given by the equation: E_photon = h*c/λ, where h is Planck's constant (h = 4.136 × 10^-15 eV s) and c is the speed of light (c = 3.0 × 10^8 m/s).
From the energy change (ΔE), we can find the energy of the emitted photon (E_photon).
Step 4: Convert the energy of the emitted photon to wavelength using the energy-wavelength relationship.
The energy-wavelength relationship is given by the equation: E_photon = hc/λ.
We will rearrange the equation to solve for the wavelength (λ).
Now, let's perform the calculations:
Step 1:
Initial energy (E_initial) = -0.2776 eV
Step 2:
ΔE = Ef - E_initial
ΔE = -10 eV (given)
Step 3:
E_photon = ΔE = -10 eV
Step 4:
E_photon = hc/λ
-10 eV = (4.136 × 10^-15 eV s) * (3.0 × 10^8 m/s) / λ
Simplifying the equation, we find:
λ = (4.136 × 10^-15 eV s) * (3.0 × 10^8 m/s) / (-10 eV)
Calculating, we get:
λ = 1.2412 × 10^-6 m
Therefore, the wavelength (λ) of the first photon emitted is approximately 1.2412 × 10^-6 meters.
To find the wavelength of the first emitted photon, we can use the formula:
λ = hc / E
where λ is the wavelength, h is the Planck's constant (6.626 x 10^-34 J⋅s), c is the speed of light (3.00 x 10^8 m/s), and E is the energy difference between the two levels.
First, let's calculate the energy difference between the n=7 and the lower energy level. Since this is a hydrogen atom, we can use the formula to calculate energy levels:
E = -13.6 eV / n^2
For n=7, the energy will be:
E_7 = -13.6 eV / 7^2 = -2.04 eV
Now, let's calculate the energy of the first emitted photon. The energy difference between the two levels is given by:
ΔE = - E_7
Substituting the values:
ΔE = - (-2.04 eV) = 2.04 eV
Next, we can substitute these values into the formula for wavelength:
λ = hc / E
λ = (6.626 x 10^-34 J⋅s) * (3.00 x 10^8 m/s) / (2.04 eV * 1.60 x 10^-19 J/eV)
Calculating the value:
λ ≈ 9.12 x 10^-8 m
Therefore, the wavelength of the first emitted photon is approximately 9.12 x 10^-8 meters.