What is the energy of a photon with the wavelength 300 NM?

Answer 1

To answer you first question: First we need the energy of one photon. Assuming you are familiar with the two formulae E=hf and c=fl where: c=speed of light /m.s^-1 | h= Planck constant / Js l=wavelength /m | f=frequency / Hz E= Energy / J we can substitute to get E=hc/l l = 320nm = 320 x10^-9 m | E = 6.626 x10^-34 Js | c=3 x10^8 m.s^-1 so E=(6.626 x10^-34 x 3 x10^8)/(320 x10^-9) = 6.211875 x10^-19 [seems believable so far] One mole = 6.02 x10^23 So total energy of a mole of the photons is: 6.02 x10^23 x 6.211875 x10^-9 = 373954.875 J = 374 kJ (3 s.f.) --------- As for the second question: http://en.wikipedia.org/wiki/Ultraviolet#Subtypes there you can see the relative wavelenghts of different types of UV light. UVA = 400-320nm | UVB = 320-200nm Now, looking again at the equation E=hc/l It is clear that to maximise E, we want the smallest l (to divide hc by the smallest number) Therefore shorter wavelength photons have the highest energy, so the answer is UV-B

Answer 2

We can use the Plank's Law:

E = hf = hc / λ

Where

h is Plank's constant (7 x 10-34 J/s)[approx.]

c is the speed of light in a vacuum (3 x 108 m/s)

λ is the wavelength, in meters, of the photon you're measuring

E = [ (7 x 10-34)(3 x 108) ] / 3 x 10-9 m

E = [ 21 x 10-26 ] / 3 x 10-9 m

E = 7 x 10-19 J