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 the excited atoms, we need to use the formula for calculating the wavelength of light:
λ = c / ν
Where:
λ is the wavelength in meters,
c is the speed of light (approximately 3 × 10^8 meters per second),
ν is the frequency of light in hertz.
The frequency of light can be related to the energy of the excited state using Planck's equation:
E = h * ν
Where:
E is the energy in joules,
h is Planck's constant (approximately 6.626 × 10^-34 joule-seconds),
ν is the frequency of light in hertz.
First, let's calculate the frequency:
E = h * ν
ν = E / h
Given the energy E = 3.211 × 10^−19 J, and Planck's constant h = 6.626 × 10^−34 J s, we can substitute these values into the equation:
ν = (3.211 × 10^−19 J) / (6.626 × 10^−34 J s)
ν ≈ 4.858 × 10^14 Hz
Now that we have the frequency, we can substitute it into the wavelength formula:
λ = c / ν
Substituting the value of the speed of light c ≈ 3 × 10^8 m/s and the frequency ν ≈ 4.858 × 10^14 Hz:
λ = (3 × 10^8 m/s) / (4.858 × 10^14 Hz)
λ ≈ 6.17 × 10^−7 meters
The wavelength is obtained in meters, but since the question asks for the wavelength in nanometers, we will convert it:
λ (nm) = λ (m) * 10^9
λ (nm) ≈ 617 nm
Therefore, the emission wavelength of these excited atoms is approximately 617 nanometers.
E = h * (c / λ)
energy = Planck's constant * speed of light / wavelength
λ = h * c / E
h = 6.63E-34 J·s
λ = 6.626E-34 * 2.998E8 / 3.211E-19
λ = 6.186E-7 m = 618.6 nm