Posted by jason on Sunday, April 22, 2012 at 5:29pm.
Moseley's law
1/λ = R (Z σ)^2(1/n^2 -1/m^2)
R=1.110^7 m^-1 is the Rydberg constant,
Z = 74,
σ = 1 for K-series,
for K-β: n = 1, m = 3.
ε = hc/ λ,
where
h = 6.6310^-34 Js is the Plancks constant,
c = 310^8 m/s is the speed of light.
So
ε = hc /λ = 6.6310^-34 310^8 1.110^7 (74 1)^2(1 - 1/9) = 1.03610^-14 J =6.4710^4 eV.
distance = speed * time
1.94*10^-11 = 3 * 10^8 * T
f = 1/T = 3*10^8/1.94*10^-11
E=h f =6.6*10^-34 * 3 *10^8/(1.94*10^-11)
in Joules
Electron volt - 1.6 * 10^-19 Joules
Distance in energy is
ΔE which is equal to the energy of the quantum ε
the wavelength was given.
You had me puzzled there for a while.
I've tried to give the method of determination, especially taking into account another Jason's question
I see, cool :)
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