For a cartain metal, the `K` obsorption edge is at `0.72 Å`. The wavelength of `K_(alpha), K_(beta). And K_(gamma)` lines of of `K` series are `0.210 Å , 0.192 Å, and 0.180 Å, respectively. The eneggies of `K, L and M` orbit are `E_(K), E_(L) and E_(M)`, respectively. Then

Options

A. `E_(K) = - 13.04 keV`
B. `E_(L) = - 7.52 keV`
C. `E_(M) = - 3.21 keV`
D. `E_(K) = - 13.04 keV`

Answer

A, B, C

Explanation

Energy of `K` absorption edge
`E_(K) = (1242 eV nm)/(0.0172 nm) = 72.21 xx 10^(3) eV = 72.21 KeV`
Energy of `K_(alpha)` line is
`E_(K_alpha) = (hc)/(e lambda_(alpha)) = (1242 eV nm)/(0.021 nm) = 59.14 KeV`
Similarly, `E_(K_beta) = (1242)/(0.0192) = 64.69 KeV`
`E_(K_gamma) = (1242)/(0.0180) = 69 KeV`
Energy of `K` shell ` = (E_(K_alpha) - E_(K))`
`= (59.14 - 72.21) KeV = - 13.04 KeV`
Energy of `L` shell ` = E_(K_beta) - 72.21 KeV`
`= 64.69 KeV - 72.21 KeV = - 7.52 KeV`
Energy of `M` shell ` = (E_(K_gamma) - E_(K)) = (1242 eV nm)/(0.018 nm) - 72.21 KeV`
`= 69 KeV - 72.21KeV = - 3.21 KeV`