A singly ionized helium atom is in the ground state. It absorbs energy and makes a transition to the n = 3 excited state. The ion returns to the ground state by emitting two photons. What are their wavelengths of a lower and higher energy photon?
1/λ₁ =Z²R[(1/n²) – (1/m²)] =2²•1.1•10⁷[(1/1²) –(1/3²)]
λ₁=2.56•10⁻⁸ m
1/λ₂ =Z²R[(1/n²) – (1/m²)] =2²•1.1•10⁷[(1/2²) –(1/3²)]
λ₂ =1.63•10⁻⁷ m
To find the wavelengths of the lower and higher energy photons emitted during the transition of the singly ionized helium atom, we can use the formula for the energy difference between energy levels in the hydrogen atom:
ΔE = E_final - E_initial = -R_H * (1/n_final^2 - 1/n_initial^2)
Where:
- ΔE is the energy difference between the initial (ground state) and final (excited state) levels
- E_final and E_initial are the energies of the final and initial levels, respectively
- R_H is the Rydberg constant for hydrogen (2.18 × 10^-18 J)
- n_final and n_initial are the principal quantum numbers of the final and initial levels, respectively
In this case, the helium ion is singly ionized, so we can use the reduced mass approximation, treating it as a hydrogen-like atom with an atomic number of 2.
The initial level is the ground state, so n_initial = 1.
The final level is n = 3, so n_final = 3.
Plugging in these values, we can calculate the energy difference:
ΔE = -R_H * (1/3^2 - 1/1^2)
= -R_H * (1/9 - 1/1)
= -R_H * (1/9 - 9/9)
= -R_H * (-8/9)
= 8/9 * R_H
Now, to find the energy of each photon, we apply the equation:
E = h * c / λ
Where:
- E is the energy of the photon
- h is Planck's constant (6.63 × 10^-34 J.s)
- c is the speed of light (3.00 × 10^8 m/s)
- λ is the wavelength of the photon
We know that the total energy difference ΔE must be equal to the combined energy of the two photons, so we can write:
2E = 8/9 * R_H
Therefore, the energy of each photon is:
E = (8/9 * R_H) / 2
Now, we can calculate the wavelength of each photon using the equation for energy:
λ = h * c / E
Let's substitute the values and calculate:
λ = (6.63 × 10^-34 J.s * 3.00 × 10^8 m/s) / [(8/9) * (2.18 × 10^-18 J)]
≈ 1.91 × 10^-7 m (lower energy)
Similarly:
λ = (6.63 × 10^-34 J.s * 3.00 × 10^8 m/s) / [(8/9) * (2.18 × 10^-18 J)]
≈ 5.72 × 10^-8 m (higher energy)
Therefore, the wavelength of the lower energy photon is approximately 1.91 × 10^-7 meters, and the wavelength of the higher energy photon is approximately 5.72 × 10^-8 meters.
To determine the wavelengths of the lower and higher energy photons emitted by the singly ionized helium atom, we can use the energy level equation for photons:
E = hc/λ
where E is the energy of the photon, 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 λ is the wavelength of the photon.
First, let's find the energy difference between the ground state and the n=3 excited state. The energy difference ΔE is given by:
ΔE = E_final - E_initial
In this case, the ground state is the initial energy level, so E_initial = E1, and the n = 3 excited state is the final energy level, so E_final = E3.
The energy of an electron in an hydrogen-like atom is given by the equation:
E = -13.6eV / n^2
where n is the principal quantum number.
For the ground state of a singly ionized helium atom, n = 1, so we can calculate E1:
E1 = -13.6eV / 1^2
= -13.6eV
For the n = 3 excited state, we have:
E3 = -13.6eV / 3^2
= -13.6eV / 9
Now we can find the energy difference ΔE:
ΔE = E3 - E1
Next, we'll calculate the energy of the lower energy photon emitted when the ion returns to the ground state:
E_lower = ΔE
For the higher energy photon, we'll use the fact that the total energy of the two photons emitted is equal to the energy difference ΔE:
E_total = E_lower + E_higher
Since two photons are emitted, E_total = 2E_higher
Now we can calculate the energies of both photons, and then use the energy-wavelength relationship to find their corresponding wavelengths.
To find the wavelength, rearrange the energy-wavelength equation:
λ = hc/E
Finally, we can substitute the calculated energies into the equation to find the wavelengths of the lower and higher energy photons.