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 given configuration, we need to calculate the energy difference between the initial and final states.
The configuration "7p1" implies that the electron is in the 7th energy level (n=7) and the p subshell (l=1). In order to find the energy of the initial state, we can use the formula for the energy levels of hydrogen:
En = -13.6eV / n^2
Substituting n=7 into the formula, we find:
Einitial = -13.6eV / (7^2) = -13.6eV / 49
Now, let's consider the emission of light due to a transition to a lower energy level. The shortest wavelength corresponds to the largest energy difference. In this case, the final state will be the lowest-energy state, which is the ground state with n=1.
The energy of the ground state can be calculated using the same formula:
Efinal = -13.6eV / (1^2) = -13.6eV
The energy difference is given by:
ΔE = Efinal - Einitial = -13.6eV - (-13.6eV / 49) = -13.6eV + 0.2776eV
Next, we can use the equation for the energy of a photon:
E = hc/λ
Where:
E is the energy difference (given by ΔE)
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)
λ is the wavelength of light we want to find
Rearranging the equation, we have:
λ = hc / E
Substituting the values, we get:
λ = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (-(13.6eV + 0.2776eV))
The negative sign in the denominator is because the energy is being released (emitted) by the atom.
Using the conversion factor 1eV = 1.602 x 10^-19 J, we can convert the energy from electron volts (eV) to joules (J):
λ = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (-(13.6 * 1.602 x 10^-19 J + 0.2776 * 1.602 x 10^-19 J))
Calculating this expression will give us the shortest wavelength of light (λ) emitted by the hydrogen atom in the given configuration.