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

Well, if we're talking about wavelengths, we could describe it using some H-umor! The shortest wavelength of light emitted by a hydrogen atom in the 7p1 configuration would be like the tiniest clown car at a circus parade – it's really small! In fact, it would be in the ultraviolet (UV) region, with a wavelength so short that it would make even the tiniest clown nose blush! So, be careful when you're dealing with those 7p1 hydrogen atoms – they may look harmless, but they can emit some shockingly short wavelengths!

To determine the shortest wavelength of light emitted by a hydrogen atom with an initial configuration of 7p1, we need to consider the energy levels and transitions within the atom.

In this case, the initial configuration 7p1 refers to the electron being in the 7p orbital. The "7" denotes the principal quantum number, while the "p" represents the orbital type.

Hydrogen's energy levels can be described by the equation: E = -13.6/n² eV, where E is the energy level and n is the principal quantum number.

To calculate the energy difference between two energy levels, we can use the equation: ΔE = E_final - E_initial.

Given that the initial configuration is 7p1, we can determine the initial energy level:
E_initial = -13.6/7² = -0.281 eV

To find the final energy level, we need to determine the configuration to which the electron transitions. Since the question does not provide this information, let's assume the electron transitions to the ground state, which is the configuration with n=1.

E_final = -13.6/1² = -13.6 eV

Now, we can calculate the energy difference:
ΔE = -13.6 - (-0.281) = -13.319 eV

To convert this energy difference to a wavelength, we can use the equation: ΔE = hc/λ, where h is Planck's constant (6.626 x 10^-34 J⋅s) and c is the speed of light (3 x 10^8 m/s).

Let's convert energy from eV to Joules:
1 eV = 1.6 x 10^-19 J

ΔE = -13.319 eV x 1.6 x 10^-19 J/eV = -2.13104 x 10^-18 J

Now, we can solve for the wavelength:
-2.13104 x 10^-18 J = (6.626 x 10^-34 J⋅s)(3 x 10^8 m/s)/λ

Rearranging the equation: λ = (6.626 x 10^-34 J⋅s)(3 x 10^8 m/s)/(-2.13104 x 10^-18 J)

λ = 9.381 x 10^-7 meters

Therefore, the shortest wavelength of light emitted by the hydrogen atom with an initial configuration of 7p1 is approximately 9.381 x 10^-7 meters (938.1 nm).

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

In hydrogen, the energy levels are labeled using a combination of the principal quantum number (n), the azimuthal quantum number (l), and the magnetic quantum number (ml). For the given configuration of 7p1, n = 7 and l = 1.

The energy of an electron in a hydrogen atom can be calculated using the formula:

E = -13.6 eV / n^2

where E is the energy in electron volts (eV) and n is the principal quantum number. Plugging in the value of n = 7, we can calculate the energy of the electron:

E(7p1) = -13.6 eV / 7^2
= -13.6 eV / 49
≈ -0.2776 eV

Now, we need to find the energy difference between the initial configuration and the final configuration, which will determine the frequency (and hence, the wavelength) of the emitted light.

For a hydrogen atom, the energy difference between two energy levels is given by the formula:

ΔE = 13.6 eV * [(1 / nf^2) - (1 / ni^2)]

where ΔE is the energy difference in electron volts, nf is the final principal quantum number, and ni is the initial principal quantum number.

For the initial configuration 7p1 (ni = 7), let's calculate the energy difference for various final configurations (nf). We will calculate the energy difference up to nf = 1, as the question asks for the shortest wavelength.

ΔE(7p1 → 1s1) = 13.6 eV * [(1 / 1^2) - (1 / 7^2)]
≈ 13.52 eV

Now, to find the wavelength of the emitted light, we need to use the equation:

E = hc / λ

where E is the energy in joules, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.0 x 10^8 m/s), and λ is the wavelength in meters.

Converting the energy difference into joules from electron volts and substituting the values into the equation, we can find the shortest wavelength:

13.52 eV * (1.6 x 10^-19 J)/(eV) = 2.1632 x 10^-18 J

Plugging this value into the wavelength equation:

2.1632 x 10^-18 J = (6.626 x 10^-34 J·s)(3.0 x 10^8 m/s) / λ

Rearranging the equation to solve for λ:

λ = (6.626 x 10^-34 J·s)(3.0 x 10^8 m/s) / (2.1632 x 10^-18 J)

Calculating this expression will give us the shortest wavelength of light emitted by the hydrogen atom with the given initial configuration.

the lowest energy transition that can occur is to the next lower level from 7p1. For Hydrogen like atoms, the energy livel depends on n alone. So n goes from 7 to 6, the formula is here:

https://en.wikipedia.org/wiki/Energy_level#Orbital_state_energy_level:_atom.2Fion_with_nucleus_.2B_one_electron