In the process of fluorescence, a molecule in its ground state will absorb a photon with a certain energy Eex, called the excitation energy, and then emit a photon with energy Eem, the emission energy. Obviously, the molecule cannot emit more energy than was absorbed. a) You wrote in your notebook that you excited a fluorescent bead at λ = 640 nm and found that it fluoresced (emitted) at λ = 520 nm. Or was it the other way around? Do a calculation and see if those figures should be switched.
b) You shine a laser on your bead at the excitation wavelength. The laser has a power of 1 mW. How many photons are emitted by the laser in one second?
c) You wish to build a microscope that allows you to excite the bead at its excitation wavelength and detect the fluorescence at the emission wavelength. This requires a pair of filters, one that allows only the excitation wavelength to pass through, and the other that allows only the emission wavelength to pass through. What color will these filters appear (i. e. what color light will they transmit?) You may have to look at an electromagnetic spectrum to figure this out…

a) the excitation energy is E₂ λ = 520 nm

the emission energy is E₁,  λ= 640 nm

b)   #_photons = 2.6 10¹⁸ photons,  

c) he excitation wavelength   λ = 520 nm is green, therefore the filter is also green

the emission wavelength is lam = 640 nm is orange

Explanation:

a) the energy of a photo is given by the planck relation

           E = h f

the speed of light is

          c = λ f

          f = c /λ

we substitute

          E = hc /λ

let's calculate the energy for the two photons

λ = 640 nm = 640 10⁻⁰ m

            E₁ = 6.63 10⁻³⁴ 3 10⁸/640 10⁻⁹

            E₁ = 3.1 10⁻¹⁹ J

λ = 520 nm = 520 10⁻⁹ m

             E₂ = 6.63 10⁻³⁴ 3 10⁸/520 10⁻⁹

            E₂ = 3.825 10⁻¹⁹  J

therefore the excitation energy is E₂ λ = 520 nm

the emission energy is E₁,  λ= 640 nm

b) For this part let's use a direct proportion rule (rule of three). If a photon (lam = 520 nm) has an energy of 3.825 10⁻¹⁹ J, how many photons have an energy of E = 1 10-3 J. Remember that the power is the energy per unit of time

        #_photons = 1 10⁻³ J (1 photon / 3.825 10⁻¹⁹ J)

        #_photons = 2.6 10¹⁸ photons

c) the excitation wavelength   λ = 520 nm is green, therefore the filter is also green

the emission wavelength is lam = 640 nm is orange

trig 15.3 cos