Hello Jhanard,
The energy E of a photon can be obtained from
(Eq.1) E = hf,
where h is Planck's constant, and f is the frequency. But also recall that
(Eq.2) fλ = c,
where λ is the wavelength and c is the speed of light. Solving the Equation 2 for f and substituting into the equation 1 gives us the result we can use here:
(Eq.3) E = hc/λ.
You wrote that the wavelength is 470no, but I will assume you mean nanometers, which are abbreviated nm. (Please let me know if this assumption is incorrect.) One nanometer equals 1 billionth of a meter. That is, 1nm = 1 X 10-9m, so the frequency is f = 470 X 10-9m = 4.70 X 10-7m. Also, Planck's constant is h=6.626 X 10-34J⋅s. Thus, we have
E = (6.626 X 10-34J⋅s)(3.00X108m/s)/(4.70 X 10-7m)
E = 4.23 X 10-19J, to 3 significant digits. (J = joules.)
We can also give the energy in electron volts (eV), which is a common unit used in atomic and subatomic physics. Using the fact that 1eV = 1.602 X 10-19J, we can convert the above answer to
E = 2.64 eV.
Let me know if you need any more help!
William
