The energy levels of hydrogen atoms can be calculated using the formula:
E = -13.6 eV / n^2
where E is the energy, n is the principal quantum number of the energy level, and -13.6 eV is the ionization energy of hydrogen.
To calculate the energy difference between levels 143 and 142, we can subtract the energy of level 142 from the energy of level 143:
ΔE = E(143) - E(142) = (-13.6 eV / 143^2) - (-13.6 eV / 142^2)
Now, let's calculate the energy difference:
ΔE ≈ (-13.6 eV / 143^2) - (-13.6 eV / 142^2) ≈ 0.000008432 eV
The energy of the photon emitted during this transition is equal to the energy difference:
E_photon = ΔE ≈ 0.000008432 eV
To find the wavelength of the spectral line, we can use the equation:
E_photon = (hc) / λ
where E_photon is the energy of the photon, h is 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.
First, let's convert the energy of the photon from electron volts (eV) to joules (J):
E_photon ≈ 0.000008432 eV ≈ 0.000008432 x 1.6 x 10^-19 J ≈ 1.34 x 10^-26 J
Now, let's calculate the wavelength:
λ = (hc) / E_photon = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (1.34 x 10^-26 J) ≈ 1.47 x 10^-7 m
The wavelength of the spectral line is approximately 1.47 x 10^-7 meters.
To determine the part of the electromagnetic spectrum this wavelength corresponds to, we can refer to the electromagnetic spectrum diagram. This wavelength falls within the range of the ultraviolet (UV) spectrum.
Therefore, the spectral line corresponding to the transition from level 143 to level 142 in hydrogen emits a photon with an energy of approximately 0.000008432 eV, a wavelength of approximately 1.47 x 10^-7 meters, and it belongs to the ultraviolet (UV) part of the electromagnetic spectrum.