Question

Hey! Assuming that the emission line used for zinc determinations is 214 nm, and there are 4.50*10^10 atoms of zinc emitting light in an instrument flame at any instant, what energy (joules) must the flame continuously provide to achieve this level of emission? I know you know you want to answer it :p

Answer 1

when you heat an atom sufficiently some of its electrons will jump to a higher energy
called an excited state
then the electrons will drop back to ground state releasing the same energy that caused the jump
the energy is released as a photon
atomic emission spectroscope is the instrument used to measure the wavelength and intensity of these emissions
the energy of a single photon is found by E=h*f where h is Planck's constant and f is frequency
but since c = f L, then f = c/L where c is speed of light, f is frequency and L is wavelength
this makes the energy equation E = h*c/L
so
the energy of a single photon at 214 nm or 214E-9 m (we must use meters) is
E = 3.00E8(6.626E-34)/214E-9 = 9.29E-19 J
that's for just one photon
so
9.29E-19 J/photon * 4.50E10 photons = 4.18E-8 J (assuming one electron jump per atom)

Answer 2

i answered your question so could you answer mine? the internet isnt very helpful

Answer 3

haha, I gave it a shot...I'm not too pro at Vectors so I tried a simple approach