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.

(a) joules

(b) electron volts

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.

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 :)

To determine the energy difference between the two energy levels that give rise to the Kâ characteristic X-ray line for tungsten, we can use the equation:

ΔE = hc/λ

where ΔE is the energy difference, h is Planck's constant (6.626 × 10^-34 J·s), c is the speed of light (3.00 × 10^8 m/s), and λ is the wavelength (1.94 × 10^-11 m).

(a) To express the energy difference in joules, we can use the given equation:

ΔE = (6.626 × 10^-34 J·s)(3.00 × 10^8 m/s) / (1.94 × 10^-11 m)

Calculating this expression will give you the energy difference in joules.

(b) To express the energy difference in electron volts (eV), we can use the conversion factor between joules and electron volts. 1 eV is equal to 1.602 × 10^-19 J.

So, once you have the energy difference in joules from part (a), you can convert it to electron volts by dividing the value by 1.602 × 10^-19 J.

By following these steps, you can find the energy difference between the two energy levels in both joules and electron volts.