The `K_(alpha)` X-ray emission line of lungsten accurs at `lambda = 0.021 nm`. What is the energy difference between `K`and `L` levels in the atom?

Options

A. `0.51 meV`
B. `1.2 meV`
C. `59 meV`
D. `13.6 meV`

Answer

A

Explanation

`E = (hc)/(lambda) = [(6.63 xx 10^(-34) xx 3 xx 10^(8))/(0.021 xx 10^(-9))] J`
`= (6.63 xx 10^(-34) xx 3 xx 10^(8))/(0.021 xx 10^(-9) xx 1.6 xx 10^(-13)) MeV`
`= 591.96 xx 10^(-4) MeV = 59.196 keV`