The Kâ characteristic X-ray line for tungsten has a wavelength of 1.94 10-11 m. What is the difference in energy between the two energy levels that give rise to this line? Express this in each of the following units.

Question

(a) joules

(b) electron volts

Answer 1

Moseley's law
1/λ = R• (Z –σ)^2•(1/n^2 -1/m^2)
R=1.1•10^7 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 = 6.63•10^-34 J•s is the Planck’s constant,
c = 3•10^8 m/s is the speed of light.
So
ε = hc /λ = 6.63•10^-34• 3•10^8 • 1.1•10^7 • (74 –1)^2•(1 - 1/9) = 1.036•10^-14 J =6.47•10^4 eV.

Answer 2

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

Answer 3

Distance in energy is
ΔE which is equal to the energy of the quantum ε

Answer 4

the wavelength was given.
You had me puzzled there for a while.

Answer 5

I've tried to give the method of determination, especially taking into account another Jason's question

Answer 6

I see, cool :)