What is the shortest wavelength of light that can be emitted by a hydrogen atom that has an initial configuration of 7p1?

To determine the shortest wavelength of light emitted by a hydrogen atom in the 7p1 configuration, we need to first calculate the final configuration after the emission of a photon.

In the electron configuration of the hydrogen atom, the energy level is represented by a number, and the subshell is represented by a letter. The energy levels are labeled as n=1,2,3,..., and the subshells are labeled as s,p,d,f,..., with s having the lowest energy.

In the 7p1 configuration, the "7" represents the energy level, and the "p1" represents the subshell. The 7p subshell has 6 electrons, and the 7p1 configuration means there is 1 electron in the 7p subshell.

To calculate the final configuration after the emission of the photon, we need to determine the new energy level and subshell.

When an electron transitions from a higher energy level to a lower energy level, it emits a photon. The energy of the emitted photon can be calculated using the Rydberg formula:

1/λ = R * (1/n1^2 - 1/n2^2)

Where λ is the wavelength of the emitted light, R is the Rydberg constant (approximately 1.097 x 10^7 m^-1), n1 is the initial energy level, and n2 is the final energy level.

In this case, the initial energy level is 7, and the final energy level will be a lower value. We can start by assuming n1 = 7 and calculating the wavelength for different values of n2.

Let's calculate the wavelength for the transition to the 6p subshell (n2 = 6):

1/λ = R * (1/7^2 - 1/6^2)
1/λ = R * (1/49 - 1/36)
1/λ = R * (36/1764 - 49/1764)
1/λ = R * (-13/1764)
λ = -1764/13R ≈ -135.69 nm

The calculated wavelength is a negative value, which indicates that the transition from the 7p1 configuration to the 6p subshell is not possible.

Let's calculate the wavelength for the transition to the 7s subshell (n2 = 7):

1/λ = R * (1/7^2 - 1/7^2)
1/λ = R * (1/49 - 1/49)
1/λ = R * (0)
λ = ∞

The calculated wavelength is infinity, indicating that the transition from the 7p1 configuration to the 7s subshell does not produce any emitted light.

Since there is no emitted light in the transitions considered so far, it means that there is no shorter wavelength of light that can be emitted by a hydrogen atom in the 7p1 configuration.

To determine the shortest wavelength of light emitted by a hydrogen atom with an initial configuration of 7p1, we first need to understand the electronic configuration of hydrogen atoms.

The electronic configuration of hydrogen atoms can be represented as nℓm, where:
- n represents the principal quantum number (1, 2, 3, ...)
- ℓ represents the azimuthal quantum number (0 to n-1)
- m represents the magnetic quantum number (-ℓ to +ℓ)

In the given initial configuration of 7p1, the principal quantum number (n) is 7, the azimuthal quantum number (ℓ) is p (ℓ = 1), and the magnetic quantum number (m) is 1.

The energy levels for hydrogen atoms can be calculated using the Rydberg formula:
1/λ = R * (1/n1^2 - 1/n2^2)

Where:
- λ represents the wavelength of light emitted
- R is the Rydberg constant (approximately 1.097 × 10^7 m^-1)
- n1 and n2 are the principal quantum numbers of the initial and final states, respectively.

In this case, the initial state is 7p1. Since the azimuthal quantum number (ℓ) for p is 1, we can determine the total number of energy sublevels using the formula 2ℓ + 1. Therefore, there are (2*1 + 1) = 3 energy sublevels for the p configuration.

Now, to find the shortest wavelength, we need to determine the final state. The final state can be determined based on the selection rule for electric dipole transitions, which states that Δℓ = ±1. This means that the final state can be 6p0, 6p1, or 6p2.

To find the shortest wavelength, we need to find the largest energy difference between the initial and final energy levels. This occurs when the final state is 6p0, as the difference in principal quantum numbers is the largest (7 - 6 = 1) compared to the other possible final states.

Using the Rydberg formula, we can calculate the wavelength:
1/λ = R * (1/7^2 - 1/6^2)

Simplifying the equation further:
1/λ = R * (1/49 - 1/36)
1/λ = R * (36/1764 - 49/1764)
1/λ = R * (-13/1764)
1/λ = -R/136

Finally, we can calculate the wavelength:
λ = 136/R

Substituting the value of R (1.097 × 10^7 m^-1), we can calculate the wavelength:
λ = 136 / (1.097 × 10^7)
λ ≈ 1.24 × 10^-8 meters

Hence, the shortest wavelength of light emitted by a hydrogen atom with an initial configuration of 7p1 is approximately 1.24 × 10^-8 meters.