An energetically excited hydrogen atom has its electron in a 5f subshell. The electron drops down to the 3d subshell, releasing a photon in the process. What wavelength of light (in nm) is emitted by the process?

Use the same equation as in the previous problem.

1/wavelength in meters = R(1/n^2 - 1/n^2) where R is the Rydberg constant. You can look that up on the web for the EXACT number but it is about 1.097E7 but many more digits. The first n in the equation is nsubscript 1 and the second n is nsubscript 2. It works out as 1/wavelength = R(1/3^2 - 1/5^2) and I obtained about 1280 nm. By the way I found a spectral line calculator on the web. Here is the url. https://www.omnicalculator.com/physics/rydberg-equation
You shouldn't use this because it doesn't give you any practice on the math part that must be learned but it sure is nice to check your manual work. Post your work if you get stuck.

To determine the wavelength of light emitted by the process, we need to calculate the energy difference between the initial and final states of the hydrogen atom, and then use that energy to find the corresponding wavelength of light.

Step 1: Calculate the energy difference between the 5f and 3d subshells.
The energy of an electron in a hydrogen atom is given by the formula:
E = - 13.6 * (Z^2 / n^2) eV

For the initial 5f subshell, n = 5 and Z (atomic number) = 1, so the initial energy is:
E_initial = - 13.6 * (1^2 / 5^2) eV

For the final 3d subshell, n = 3, so the final energy is:
E_final = - 13.6 * (1^2 / 3^2) eV

Step 2: Calculate the energy difference.
ΔE = E_final - E_initial

Step 3: Convert the energy difference into units of Joules.
To convert from electron volts (eV) to Joules (J), we can use the conversion factor 1 eV = 1.6 x 10^-19 J.

Step 4: Calculate the wavelength of the emitted light.
The wavelength (λ) of light can be determined using the equation:
λ = (c * h) / ΔE

Where c is the speed of light (2.998 x 10^8 m/s) and h is Planck's constant (6.626 x 10^-34 J·s).

Now, let's do the calculations step by step.

Step 1:
E_initial = - 13.6 * (1^2 / 5^2) eV
E_initial = - 13.6 * (1 / 25) eV

Step 2:
E_final = - 13.6 * (1^2 / 3^2) eV
E_final = - 13.6 * (1 / 9) eV

ΔE = E_final - E_initial
ΔE = (- 13.6 * (1 / 9)) - (- 13.6 * (1 / 25)) eV

Step 3:
To convert from eV to J, we multiply by the conversion factor:
1 eV = 1.6 x 10^-19 J

ΔE_J = ΔE * (1.6 x 10^-19 J / 1 eV)

Step 4:
λ = (c * h) / ΔE_J

Substitute the known values and calculate:
λ = (2.998 x 10^8 m/s * 6.626 x 10^-34 J·s) / ΔE_J

By following these steps and performing the calculations, we can determine the wavelength of the emitted light in nm.

To calculate the wavelength of light emitted when an electron transitions between energy levels in a hydrogen atom, we can use the Rydberg formula, which gives the relationship between the wavelength of light emitted and the energy levels involved.

The Rydberg formula is given by:

1/λ = R * (1/n₁² - 1/n₂²),

where λ is 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 energy levels, respectively.

In this case, the electron drops from the 5f subshell to the 3d subshell. The principal quantum number n₁ for the 5f subshell is 5, and for the 3d subshell, it is 3.

Plugging these values into the Rydberg formula, we get:

1/λ = R * (1/5² - 1/3²),

1/λ = R * (1/25 - 1/9).

Simplifying further:

1/λ = R * (9/225 - 25/225),

1/λ = R * (-16/225).

Now, solve for the wavelength (λ):

λ = 1 / (R * (-16/225)).

Using the given value for the Rydberg constant (R ≈ 1.097 × 10^7 m⁻¹), we can calculate the wavelength:

λ = 1 / (1.097 × 10^7 m⁻¹ * (-16/225)).

λ ≈ 2.413 × 10⁻⁷ m.

Since the wavelength is usually expressed in nanometers (nm), we can convert the result to nm:

λ ≈ 241.3 nm.

Therefore, the wavelength of light emitted by the process is approximately 241.3 nm.