Consider a sample of excited atoms that lie 3.211 × 10–19 J above the ground state. Determine the emission wavelength (in nanometers) of these atoms.
To determine the emission wavelength of these excited atoms, we can use the equation:
E = (hc) / λ
where E is the energy difference between the excited state and the ground state of the atom, h is Planck's constant, c is the speed of light, and λ is the wavelength of the emitted light.
First, let's convert the energy difference E from joules (J) to electron volts (eV) using the conversion factor 1 eV = 1.6 × 10^-19 J:
E = 3.211 × 10^-19 J × (1 eV / 1.6 × 10^-19 J) = 2.0075 eV
Next, we can use the equation:
E = hc / λ
Rearranging the equation, we can solve for λ:
λ = hc / E
Plugging in the values of h (Planck's constant = 6.626 × 10^-34 J·s) and c (speed of light = 3.0 × 10^8 m/s), we get:
λ = (6.626 × 10^-34 J·s × 3.0 × 10^8 m/s) / (2.0075 eV)
To convert eV to joules, we use the conversion factor 1 eV = 1.6 × 10^-19 J:
λ = (6.626 × 10^-34 J·s × 3.0 × 10^8 m/s) / (2.0075 eV × 1.6 × 10^-19 J/eV)
Simplifying the equation, we find:
λ ≈ 656 nm
Therefore, the emission wavelength of these excited atoms is approximately 656 nanometers.