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 with an initial configuration of 7p1, we need to calculate the energy difference between the final and initial energy levels of the electron.
In the electron configuration 7p1, the electron is in the 7th principal energy level and occupies the p-subshell with a magnetic quantum number (m) of +1.
To find the energy level of the electron, we can use the formula:
E = -13.6 eV / n^2
where E is the energy in electron volts (eV) and n is the principal quantum number.
In this case, the initial energy level (ni) is 7. Therefore, the energy Ei of the initial energy level is calculated as:
Ei = -13.6 eV / (7)^2 = -0.27 eV
Next, we need to find the final energy level (nf) after the electron transitions to a lower energy level. Since we are interested in finding the shortest wavelength, the final energy level will be the ground state of hydrogen, which is n = 1.
Hence, the final energy level (nf) is 1, and the energy Ef of the final energy level is calculated as:
Ef = -13.6 eV / (1)^2 = -13.6 eV
The energy difference (ΔE) between the initial and final energy levels is given by:
ΔE = Ef - Ei
ΔE = -13.6 eV - (-0.27 eV) = -13.33 eV
To convert the energy difference into the wavelength (λ) of the emitted light, we can use the equation:
λ = hc / ΔE
where 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 ΔE is the energy difference in joules (J).
To convert -13.33 eV into joules (J), we can multiply it by the conversion factor 1.602 x 10^-19 J/eV:
ΔE (J) = -13.33 eV * (1.602 x 10^-19 J/eV) = -2.13 x 10^-18 J
Now, we can calculate the wavelength:
λ = (6.626 x 10^-34 J.s * 3.0 x 10^8 m/s) / (-2.13 x 10^-18 J) = 9.28 x 10^-7 m
Therefore, the shortest wavelength of light emitted by a hydrogen atom with an initial configuration of 7p1 is approximately 9.28 x 10^-7 meters (or 928 nm).