Question

An atomic transition produces a photon with a wavelength of 410 nm. What is the energy of this photon in eV? a) 3 eV. b) 4 eV. c) 5 eV. d) 6 eV. e) 7 eV

Answer 1

E = h ν energy of photon

E = h c / λ ν = c / λ

E = 6.63E-34 * 3.00E8 / 410E-9

E = 6.63 * 3 / 4.10 E-19 = 4.85E-19 V

1 eV = 1.60E-19 * 1 = 1.60E-19

4.85 / 1.60 = 3 eV (a)