If a fluorescent dye absorbs a photon of light at 485 nm and emits a photon at 540 nm, how much energy is lost as heat?

To determine how much energy is lost as heat when a fluorescent dye absorbs and then emits a photon, you first need to calculate the energy of both the absorbed and emitted photons using the formula for photon energy:

[ E = \frac{hc}{\lambda} ]

where ( E ) is the energy of the photon, ( h ) is Planck's constant (approximately ( 6.626 \times 10^{-34} , \text{J s} )), ( c ) is the speed of light (approximately ( 3.00 \times 10^8 , \text{m/s} )), and ( \lambda ) is the wavelength in meters.

1. Energy of the absorbed photon:

For the absorbed photon at a wavelength of 485 nm (which is ( 485 \times 10^{-9} , \text{m} )):

[

E_{absorbed} = \frac{(6.626 \times 10^{-34} , \text{J s})(3.00 \times 10^8 , \text{m/s})}{485 \times 10^{-9} , \text