When table salt is put in a candle flame, it burns bright yellow because of emission from the sodium atoms. This emission is at 590 nm. (a) What is the natural resonant frequency of the
sodium atom that produces this emission (i.e. what is the frequency of this light)? (b) What is the difference in energy between the two levels of the sodium atom that are responsible for this emission (i.e. what is the energy of a photon
To find the natural resonant frequency of the sodium atom that produces the 590 nm emission, we can use the relationship between wavelength, frequency, and speed of light. The formula is as follows:
c = λ * f
where c is the speed of light (approximately 3.00 x 10^8 meters per second), λ is the wavelength in meters, and f is the frequency in hertz.
(a) First, let's convert the wavelength of 590 nm to meters. We know that 1 nm (nanometer) is equal to 1 x 10^-9 meters. So:
590 nm * (1 x 10^-9 m / 1 nm) = 590 x 10^-9 meters
= 5.90 x 10^-7 meters
Now we can rearrange the formula to solve for the frequency:
f = c / λ
f = (3.00 x 10^8 m/s) / (5.90 x 10^-7 m)
= 5.08 x 10^14 Hz
Therefore, the natural resonant frequency of the sodium atom that produces the emission is approximately 5.08 x 10^14 Hz (hertz).
(b) To calculate the energy difference between the two levels responsible for this emission, we can use the equation:
ΔE = h * f
where ΔE is the energy difference in joules, h is Planck's constant (6.626 x 10^-34 joule-seconds), and f is the frequency in hertz.
ΔE = (6.626 x 10^-34 J·s) * (5.08 x 10^14 Hz)
= 3.361 x 10^-19 Joules
So the energy of a photon emitted by the sodium atom is approximately 3.361 x 10^-19 Joules.